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18.10: Schwarz-Pick theorem - Mathematics

18.10: Schwarz-Pick theorem - Mathematics


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The following theorem shows that the metric in the conformal disc model naturally appears in other branches of mathematics. We do not give its proof, but it can be found in any textbook on geometric complex analysis.

Suppose that (mathbb{D}) denotes the unit disc in the complex plane centered at (0); that is, a complex number (z) belongs to (mathbb{D}) if and only if (|z|<1).

Let us use the disc (mathbb{D}) as a h-plane in the conformal disc model; the h-distance between (z, winmathbb{D}) will be denoted by (d_h(z,w)); that is,

(d_h(z,w) := ZW_h,)

where (Z) and (W) are h-points with complex coordinates (z) and (w) respectively.

A function (f:mathbb{D} o mathbb{C}) is called holomorphic if for every (zin mathbb{D}) there is a complex number (s) such that

(f(z+w)=f(z)+scdot w+o(|w|).)

In other words, (f) is complex-differentiable at any (zinmathbb{D}). The complex number (s) is called the derivative of (f) at (z), or briefly (s=f'(z)).

Theorem (PageIndex{1}) Schwarz-Pick theorem

Schwarz–Pick theorem Assume (f: mathbb{D} o mathbb{D}) is a holomorphic function. Then

(d_h(f(z),f(w))le d_h(z,w))

for any (z,win mathbb{D}).

If the equality holds for one pair of distinct numbers (z,win mathbb{D}), then it holds for any pair. In this case (f) is a fractional linear transformation as well as a motion of the h-plane.

Exercise (PageIndex{1})

Show that if a fractional linear transformation (f) appears in the equality case of Schwarz–Pick theorem, then it can be written as

(f(z)=dfrac{vcdot z+ar w}{wcdot z+ar v}.)

where (v) and (w) are complex constants such that (|v|>|w|).

Hint

Note that (f = dfrac{a cdot z + b}{c cdot z + d}) preserves the unit circle (|z| = 1). Use Corollary 10.6.1 and Proposition 18.12 to show that (f) commutes with the inversion (z mapsto 1/ar{z}). In other words, (1/f(z) = f(1/ar{z})) or

(dfrac{ar{c} cdot ar{z} + ar{d}}{ar{a} cdot ar{z} + ar{b}} = dfrac{a/ar{z} + b}{c/ar{z} + d})

for any (z in ar{mathbb{C}}). The latter identity leads to the required statements. The condition (|w| < |v|) follows since (f(0) in mathbb{D}).

Exercise (PageIndex{2})

Recall that hyperbolic tangent ( anh) is defined on page . Show that

( anh [ frac12cdot d_h(z,w)]=left|frac{z-w}{1-zcdotar w} ight|.)

Conclude that the inequality in Schwarz–Pick theorem can be rewritten as

(left|frac{z'-w'}{1-z'cdotar w'} ight|leleft|frac{z-w}{1-zcdotar w} ight|,)

where (z'=f(z)) and (w'=f(w)).

Hint

Note that the inverses of the points (z) and (w) have complex coordinates (1/ar{z}) and (1/ar{w}). Apply Exercise 12.9.2 and simplify.

The second part follows since the function (x mapsto anh (dfrac{1}{2} cdot x)) is increasing.

Exercise (PageIndex{3})

Show that the Schwarz lemma stated below follows from Schwarz–Pick theorem.

Hint

Apply Schwarz-Pick theorem for a function (f) such that (f(0) = 0) and then apply Lemma 12.3.2.

Lemma (PageIndex{1}) Schwarz lemma

Let (f: mathbb{D} o mathbb{D}) be a holomorphic function and (f(0)=0). Then (|f(z)|le |z|) for any (zin mathbb{D}).

Moreover, if equality holds for some (z e 0), then there is a unit complex number (u) such that (f(z)=ucdot z) for any (zinmathbb{D}).


MATH 580 / CS 571

Lec 1, 8/24 Mon, Sec 1.1: Course details, general principles of enumeration, counting of words & subsets, binomial theorem, multisets/compositions.
Lec 2, 8/26 Wed, Sec 1.2: Lattice paths, basic identities, extended binomial coefficient, summing polynomials, Delannoy numbers, Hamming ball/Delannoy correspondence.
Lec 3, 8/28 Fri, Sec 1.3: Counting graphs and trees, multinomial coefficients (trees by degrees, Fermat's Little Theorem), Ballot problem, central binomial convolution.
Lec 4, 8/31 Mon, Sec 1.3-2.1: Catalan numbers (generalization, bijections, recurrence), Fibonacci numbers and 1,2-lists, derangements.

Lec 5, 9/2 Wed, Sec 2.1-2: Recurrences in two indices (distribution problems, Delannoy numbers), characteristic equation method (through repeated roots).
Lec 6, 9/4 Fri, Sec 2.2: Characteristic equation method (inhomogeneous terms), generating function method (linear w. constant coefficients, relation to char.eqn. method)
Lec 7, 9/9 Wed, Sec 2.3-3.1: Catalan solution, substitution method (factorials, derangements, Stirling's approximation), generating functions (sum/product operations, multisets)

Lec 8, 9/11 Fri, Sec 3.1: Generating functions: multisets with restricted multiplicity, functions in two variables (skipped), permutation statistics (by #inversions, #cycles), Eulerian numbers (#runs, Worpitzky's Identity by barred permutations)
Lec 9, 9/14 Mon, Sec 3.2: Generating function manipulations: sum & product (A(n,k) formula by inversion from Worpitzky), shifted index, differentiation & evaluation at special values, summing initial coefficients, summation by convolutions. First Snake Oil example.
Lec 10, 9/16 Wed, Sec 3.3: More Snake Oil, exponential generating functions: products of EGFs (words), examples and applications of EGFs (flags on poles, restricted words, Stirling numbers)
Lec 11, 9/19 Fri, Sec 3.3: EGF applications (binomial inversion, derangements), the Exponential Formula (graphs, partitions, permutations, recurrence), Lagrange Inversion Formula (statement and application to trees).
Lec 12, 9/21 Mon, Sec 3.4: Partitions of integers (basic generating functions, bounds on coefficients). combinatorics of partitions (Ferrers diagrams, conjugate, Fallon's Identity (mentioned), congruence classes of triangles, Euler's Identity).

Lec 13, 9/23 Wed, Sec 4.1: Basic inclusion-exclusion formula, applications (totients, Stirling numbers, alternating sums, skipped Eulerian numbers)
Lec 14, 9/25 Fri, Sec 4.1: Permutations with restricted positions (rook polynomials), OGF by number of properties (skipped probleme des menages), signed involutions (inclusion-exclusion as special case, partitions into distinct odd parts).
Lec 15, 9/28 Mon, Sec 4.1-2: Disjoint-path systems in digraphs, application to lattice paths and rhombus tilings. Examples for counting under symmetry, Lagrange's Theorem, Burnside's Lemma.
Lec 16, 9/30 Wed, Sec 4.2-3: Examples for Burnside's Lemma, Cycle indices, symmetries of cube, pattern inventory (Polya's Theorem), counting isomorphism classes of graphs. Young tableaux (brief presentation of Hook-length formula, RSK correspondence, and consequences of RSK correspondence).

Lec 17, 10/2 Fri, Sec 5.1-3 highlights, Sec 6.1: Properties of Petersen graph, degree-sum formula and rectangle partition, characterization of bipartite graphs, Eulerian circuits (Chapter 5, First Concepts for Graphs, for background reading). Bipartite Matching (Hall's Theorem).
Lec 18, 10/5 Mon, Sec 6.1: Marriage Theorem, orientations with specified outdegrees. Min/max relations (Ore's defect formula, Konig-Egervary Theorem, Gallai's Theorem, Konig's Other Theorem).
Lec 19, 10/7 Wed, Sec 6.2: General Matching: Tutte's 1-Factor Theorem from Berge-Tutte Formula, 1-factors in regular graphs, Petersen's 2-Factor Theorem (via Eulerian circuit and Hall's Theorem), augmenting paths, reduction of f-factor to 1-factor in blowup (mentioned only briefly)

Lec 20, 10/9 Fri, Sec 7.1: Connectivity (definitions, Harary graphs, cartesian products). Edge-connectivity (definitions, Whitney's Theorem, edge cuts, diameter 2). Bonds and blocks skipped.
Lec 21, 10/12 Mon, Sec 7.2: k-Connected Graphs (Independent x,y-paths, linkage and blocking sets, Pym's Theorem, Menger's Theorems (8 versions), Expansion and Fan Lemmas, cycles through specified vertices),
Lec 22, 10/14 Wed, Sec 7.2-3: Ford-Fulkerson CSDR, ear decomposition and Robbins' Theorem. Spanning cycles: necessary condition, Ore & Dirac conditions, closure, statement of Chvatal condition & sketch.
Lec 23, 10/16 Fri, Sec 7.3-8.1: Chvatal-Erdos Theorem, comments on Lu's Theorem, Fan's Theorem, regularity, pancyclicity, statements for circumference. Vertex coloring: examples, easy bounds, greedy coloring, interval graphs, degree bounds, start Minty's Theorem.

Lec 24, 10/19 Mon, Sec 8.1-2: Triangle-free graphs (Mycielski's construction, &radicn bound), color-critical graphs (minimum degree, edge-connectivity). List coloring for complete bipartite graphs.
Lec 25, 10/21 Wed, Sec 8.2-3: List coloring (degree choosability and extension of Brooks' Theorem), edge-coloring (complete graphs, Petersen graph, bipartite graphs).
Lec 26, 10/23 Fri, Sec 8.3: Edge-coloring (color fans and Vizing's Theorem for graphs, Anderson-Goldberg generalization of Vizing's Theorem for multigraphs), brief mention of perfect graphs (chordal graphs, PGT).

Lec 27, 10/26 Mon, Sec 9.1: Planar graphs and their duals, cycles vs bonds, bipartite plane graphs, Euler's Formula and edge bound, brief application to regular polyhedra, skipped outerplanar graphs
Lec 28, 10/28 Wed, Sec 9.2: Kuratowski's Theorem and convex embeddings, 6-coloring of planar graphs
Lec 29, 10/30 Fri, Sec 9.3: Coloring of planar graphs (5-choosability, Kempe), Discharging (approach to 4CT, planar graphs with forbidden face lengths), Tait's Theorem (skipped Grinberg's Theorem)

Lec 30, 11/2 Mon, Sec 10.1: Applications of pigeonhole principle (covering by bipartite graphs, divisible pairs, domino tilings, paths in cubes, monotone sublists, increasing trails, girth 6 with high chromatic number).
Lec 31, 11/4 Wed, Sec 10.2: Ramsey's Theorem and applications (convex m-gons, table storage).
Lec 32, 11/6 Fri, Sec 10.3: Ramsey numbers, graph Ramsey theory (tree vs complete graph), Schur's Theorem, Van der Waerden Theorem (statement and example)

Lec 33, 11/9 Mon, Sec 12.1: Partially ordered sets (definitions and examples, comparability graphs and cover graphs), Dilworth's Theorem, equivalence of Dilworth and Konig-Egervary, relation to PGT.
Lec 34, 11/11 Wed, Sec 12.2: graded posets & Sperner property, symmetric chain decompositions for subsets and products, bracketing decomposition, application to monotone Boolean functions.
Lec 35, 11/13 Fri, Sec 12.2: LYM posets (Sperner's Theorem via LYM, equivalence with regular covering and normalized matching, LYM and symmetric unimodal rank-sizes => symmetric chain decomposition, statement of log-concavity & product result).

Lec 36, 11/16 Mon, Sec 14.1: existence arguments (Ramsey number, 2-colorability of k-uniform hypergraphs), pigeonhole property of expectation (linearity and indicator variables, Caro-Wei bound on independence number, application of Caro-Wei to Turan's Theorem, pebbling in hypercubes)
Lec 37, 11/18 Wed, Sec 14.2: Crossing number (expectation application), Deletion method (Ramsey numbers, dominating sets, large girth and chromatic number)
Lec 38, 11/20 Fri, Sec 14.2-3: Symmetric Local Lemma & applications (Ramsey number, list coloring, Mutual Independence Principle). Random graph models, almost-always properties, connectedness for constant p, Markov's Inequality.
Lec 39, 11/30 Mon, Sec 14.3: Second moment method, threshold functions for disappearance of isolated vertices and appearance of balanced graphs, comments on evolution of graphs, comments on connectivity/cliques/coloring of random graphs.

Lec 40, 12/2 Wed, Sec 13.1: Latin squares (4-by-4 example, MOLS(n,k), upper bound, complete families, Moore-MacNeish construction). Block designs (examples, elementary constraints on parameters, Fisher's Inequality).
Lec 41, 12/4 Fri, Sec 13.1: Symmetric designs (Bose), necessary conditions (example of Bruck-Chowla-Ryser), Hadamard matrices (restriction on order, relation to designs, relation to coding theory).
Lec 42, 12/7 Mon, Sec 13.2: Projective planes (equivalence with (q^2+q+1,q+1,1)-designs, relation to Latin squares, polarity graph with application to extremal problems).
Lec 43, 12/9 Wed, Sec 13.2-3: difference sets and multipliers, Steiner triple systems.


Proof

The proof is a straightforward application of the maximum modulus principle on the function

which is holomorphic on the whole of D, including at the origin (because f is differentiable at the origin and fixes zero). Now if Dr = <z : |z| ≤ r> denotes the closed disk of radius r centered at the origin, then the maximum modulus principle implies that, for r < 1, given any z in Dr, there exists zr on the boundary of Dr such that

As r → 1 we get |g(z)| ≤ 1.

Moreover, suppose that |f(z)| = |z| for some non-zero z in D, or |f′(0)| = 1. Then, |g(z)| = 1 at some point of D. So by the maximum modulus principle, g(z) is equal to a constant a such that |a| = 1. Therefore, f(z) = az, as desired.


18.10: Schwarz-Pick theorem - Mathematics


Math 218a / Stat 205A. Probability Theory.

Fall 2017, TTh 9:30a-11:00a in Etcheverry 3108.

GENERAL INFORMATION. (## = berkeley dot edu)
Please include "MATH218A/STAT205A" and your SID in the subject line of all emails.
Instructor: Nike Sun (nikesun at ##). Office hours Tue 2:15p-4:15p in Evans 389.
GSI: Zsolt Bartha (bartha at ##). Office hours Mon 11:30a-1:00p and Wed 1:30p-3:00p in Evans 428.
Discussion section Fri 11:00a-12:00n in Evans 330.

Some materials, including homework assignments, will be posted to bcourses. If you cannot access the site, email the instructor with your @berkeley.edu address.

REFERENCES. References marked * are available electronically through lib.berkeley.edu.
Main reference:
*Probability: Theory and Examples (4th ed.) by R. Durrett (PTE).
Additional references:
Lecture notes by A. Dembo, for a similar course taught at Stanford.
Probability and Measure (3rd ed.) by P. Billingsley.
*Convergence of Probability Measures by P. Billingsley.
*Real Analysis and Probability by R. M. Dudley.
*Measure Theory by P. R. Halmos.
Foundations of Modern Probability by O. Kallenberg.
Analysis by E. H. Lieb and M. Loss.
Probability on Trees and Networks by R. Lyons and Y. Peres (link).
Complex Analysis by E. Stein and R. Shakarchi.

HOMEWORK. Please be considerate of the grader: write solutions neatly, and staple.
At the top of the first page, write clearly the following information: your name, student ID number, Berkeley email address, the course name (MATH218A/STAT205A), and the names of any other students with whom you discussed the assignment. Please also indicate if you are currently waitlisted.
Collaboration policy. You should first attempt to solve homework problems on your own. If you are having trouble, you may discuss with others. You are expected to write down your solutions alone.

GRADING. Your final grade will consist of 50% homework + 50% final.
No late homework will be accepted. Lowest two homework scores will be dropped.
Missing the final will automatically result in an F grade there will be no rescheduling (early/late/repeat). All students will be graded under the same scheme, regardless of late enrollment.


Real Analysis 18.100B

Lecturer: Kiril Datchev, room 2-173, [email protected]

Class meetings: Tuesdays and Thursdays 9:30-11:00 in room 4-163.

Textbook: Walter Rudin, Principles of Mathematical Analysis.

Recommended reading: G. H. Hardy, A Course of Pure Mathematics. Edmund Landau, Foundations of Analysis. Tom M. Apostol, Mathematical Analysis. See also these notes on the text by George Bergman. For inspirational reading, consult The Study of Mathematics by Bertrand Russell.

  1. ten problem sets, worth 20 points each, due on 9/13, 9/20, 9/27, 10/11, 10/18, 10/25, 11/1, 11/15, 11/29, 12/6,
  2. two midterms, worth 100 points each, one on October 2nd and one on November 6th,
  3. one final exam, worth 200 points, on Thursday, December 20th, 1:30pm-4:30pm in Johnson Track (upstairs).

Problem sets are due Thursdays at 4:00 in room 2-285. Late problem sets are not accepted.


Contents

Via Euler's formula Edit

One proof of this theorem uses Euler's polyhedral formula to reduce the problem to the case of a triangle with three integer vertices and no other integer points. Such a triangle can tile the plane by copies of itself, rotated by 180° around the midpoint of each edge. The triangles of this tiling are twice as dense as the integer points (each integer point belongs to six triangles, while each triangle touches only three integer points) from which it follows that their area is exactly 1 2 <2>>> , as Pick's theorem (once it has been proved) would also imply. [4] It is also possible to use Minkowski's theorem on lattice points in symmetric convex sets to prove that these triangles have area 1 2 <2>>> . [8]

It is also possible to go the other direction, using Pick's theorem (proved in a different way) as the basis for a proof of Euler's formula. [5] [10]

Other proofs Edit

Alternative proofs of Pick's theorem that do not use Euler's formula include the following.

  • One can recursively decompose the given polygon into triangles, allowing some triangles of the subdivision to have area larger than 1/2. Both the area and the counts of points used in Pick's formula add together in the same way, so the truth of Pick's formula for general polygons follows from its truth for triangles. Any triangle subdivides its bounding box into the triangle itself and additional right triangles, and the areas of both the bounding box and the right triangles are easy to compute. Combining these area computations gives Pick's formula for triangles, and combining triangles gives Pick's formula for arbitrary polygons. [6][7][11]
  • The Voronoi diagram of the integer grid subdivides the plane into squares, centered around each grid point. One can compute the area of any polygon as the sum of its areas within each cell of this diagram. For each interior grid point of the polygon, the entire Voronoi cell is covered by the polygon. Grid points on an edge of the polygon have half of their Voronoi cell covered. The Voronoi cells of corner points are covered by amounts whose differences from half a square (using an argument based on turning number) total to the − 1 correction term in Pick's formula. [7]
  • Alternatively, instead of using grid squares centered on the grid points, it is possible to use grid squares having their vertices at the grid points. These grid squares cut the given polygon into pieces, which can be rearranged (by matching up pairs of squares along each edge of the polygon) into a polyomino with the same area. [12]
  • Pick's theorem may also be proved based on complex integration of a doubly periodic function related to Weierstrass's elliptic functions. [13]
  • Applying the Poisson summation formula to the characteristic function of the polygon leads to another proof. [14]

Generalizations to Pick's theorem to non-simple polygons are possible, but are more complicated and require more information than just the number of interior and boundary vertices. [2] [15] For instance, a polygon with h holes bounded by simple integer polygons, disjoint from each other and from the boundary, has area [16]

The Reeve tetrahedra in three dimensions have four integer points as vertices and contain no other integer points. However, they do not all have the same volume as each other. Therefore, there can be no analogue of Pick's theorem in three dimensions that expresses the volume of a polytope as a function only of its numbers of interior and boundary points. [17] However, these volumes can instead be expressed using Ehrhart polynomials. [18] [19]


Math/Stat 431 Introduction to the Theory of Probability

Math 431 is an introduction to the theory of probability, the part of mathematics that studies random phenomena. We model simple random experiments mathematically and learn techniques for studying these models. Topics covered include axioms of probability, random variables, the most important discrete and continuous probability distributions, expectations, moment generating functions, conditional probability and conditional expectations, multivariate distributions, Markov's and Chebyshev's inequalities, laws of large numbers, and the central limit theorem.

Math 431 is not a course in statistics. Statistics is a discipline mainly concerned with analyzing and representing data. Probability theory forms the mathematical foundation of statistics, but the two disciplines are separate.

From a broad intellectual perspective, probability is one of the core areas of mathematics with its own distinct style of reasoning. Among the other core areas are analysis, algebra, geometry/topology, logic and computation.

To go beyond 431 in probability you should take next Math 521 - Analysis, and after that one or both of these: Math 632 - Introduction to Stochastic Processes and Math 635 - Introduction to Brownian Motion and Stochastic Calculus. Those who would like a proof based introduction to probability could consider taking Math 531 - Probability Theory (531 requires a proof based course as a prerequisite).

Where is probability used?

Probability theory is ubiquitous in natural science, social science and engineering, so this course can be valuable in conjunction with many different majors. Aside from being a beautiful subject in and of itself, is used throughout the sciences and industry. For example, in biology many models of cellular phenomena are now modeled probabilistically as opposed to deterministically. As for industry, many models used by insurance and financial companies are probabilistic in nature. Thus, those wishing to go into actuarial science or finance need to have a solid understanding of probability. Probabilistic models show up in the study of networks, making probability theory useful for those interested in computer science and information technology.

Prerequisites

To be technically prepared for Math 431 one needs to be comfortable with the language of sets and calculus, including multivariable calculus, and be ready for abstract reasoning. Basic techniques of counting is also useful, but we will review these along the way. Probability theory can seem very hard in the beginning, even after success in past math courses.

[email protected]

We will use the [email protected] website of the course to post homework assignments and solutions. The lecture notes will also be posted there.

Textbook

  • A first course in Probability, by Sheldon Ross
  • Probability, by Jim Pitman.
  • Probability, statistics, and stochastic processes, by Peter Olofsson.
  • Probability and random processes, by Geoffrey Grimmett and David Stirzaker.
  • Midterm exam 1: Thursday, October 13 , 7:15-8:45 PM, 1310 Sterling
  • Midterm exam 2: Wednesday, November 30, 7:15-8:45 PM, 1310 Sterling
  • Final exam: Tuesday, December 20, 12:25PM - 2:25PM, Room TBA

Quizzes

  • 1st quiz: September 13, Tuesday, end of class.
  • 2nd quiz: September 20, Tuesday, end of class.
  • 3rd quiz: September 27, Tuesday, end of the class
  • 4th quiz: October 4, Tuesday, end of the class

There will be no more quizzes in the semester.

Homework

Homework assignments will be posted on the [email protected] site of the course. Weekly homework assignments are usually due on Thursday at the beginning of the class. You can also submit your solution in an electronic form via [email protected], but it has to arrive by 9:30AM on the due date. Note that there is a (short) homework assignment due on the first Thursday (September 8).
Some homework assignments will contain bonus problems for those who would like extra challenge. The points from the bonus problems will be converted into extra credit at the end of the semester.

Instructions for homework

  • Observe rules of academic integrity. Handing in plagiarized work, whether copied from a fellow student or off the web, is not acceptable. Plagiarism cases will lead to sanctions.
  • Homework is collected at the beginning of the class period on the due date. No late assignments will be accepted. You can bring the homework earlier to the instructor's office.
  • Working in groups on homework assignments is strongly encouraged however, every student must write their own assignments.
  • Organize your work neatly. Use proper English. Write in complete English or mathematical sentences. Answers should be simplified as much as possible. If the answer is a simple fraction or expression, a decimal answers from a calculator is not necessary. For some exercises you will need a calculator to get the final answer.
  • Answers to some exercises are in the back of the book, so answers alone carry no credit. It's all in the reasoning you write down.
  • Put problems in the correct order and staple your pages together.
  • Do not use paper torn out of a binder.
  • Be neat. There should not be text crossed out.
  • Recopy your problems. Do not hand in your rough draft or first attempt.
  • Papers that are messy, disorganized or unreadable cannot be graded.
  • I strongly encourage you to type up your solutions (perhaps using Latex).

Weekly schedule

Here is a tentative weekly schedule, to be adjusted as we go. The numbers refer to sections in lecture notes that can be found at the [email protected] website.

The Math Club provides interesting lectures and other math-related events. Everybody is welcome. Summer Reading Suggestions


--> Notice: Prof. Kurtz will teach our class on Tuesday Oct. 29. We will not meet on Thursday Oct. 31. We can use this extra time later if we need it. -->


Schwarz-Pick Type Inequalities

This book discusses in detail the extension of the Schwarz-Pick inequality to higher order derivatives of analytic functions with given images. It is the first systematic account of the main results in this area. Recent results in geometric function theory presented here include the attractive steps on coefficient problems from Bieberbach to de Branges, applications of some hyperbolic characteristics of domains via Beardon-Pommerenke's theorem, a new interpretation of coefficient estimates as certain properties of the Poincaré metric, and a successful combination of the classical ideas of Littlewood, Löwner and Teichmüller with modern approaches. The material is complemented with historical remarks on the Schwarz Lemma and a chapter introducing some challenging open problems.

The book will be of interest for researchers and postgraduate students in function theory and hyperbolic geometry.

“The aim of this book is to give a unified presentation of some recent results in geometric function theory together with a consideration of their historical sources. The extensive historical references are … interesting, thorough and informative. … this book is filled with many challenging conjectures and suggested problems for exploring new research. In summary this is a delightful book that anyone interested in interrelating geometry and classical geometric function theory should read.”­­­ (Roger W. Barnard, Mathematical Reviews, Issue 2010 j)


Mathematics_part_ _II_(solutions) for Class 10 Math Chapter 2 - Pythagoras Theorem

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Page No 38:

Question 1:

Identify, with reason, which of the following are Pythagorean triplets.
(i) (3, 5, 4)
(ii) (4, 9, 12)
(iii) (5, 12, 13)
(iv) (24, 70, 74)
(v) (10, 24, 27)
(vi) (11, 60, 61)

Answer:

(i) In the triplet (3, 5, 4),
3 2 = 9, 5 2 = 25, 4 2 = 16 and 9 + 16 = 25

The square of the largest number is equal to the sum of the squares of the other two numbers.

&there4 (3, 5, 4) is a pythagorean triplet.

(ii) In the triplet (4, 9, 12),
4 2 = 16, 9 2 = 81, 12 2 = 144 and 16 + 81 = 97 &ne 144

The square of the largest number is not equal to the sum of the squares of the other two numbers.

&there4 (4, 9, 12) is not a pythagorean triplet.

(iii) In the triplet (5, 12, 13),
5 2 = 25, 12 2 = 144, 13 2 = 169 and 25 + 144 = 169

The square of the largest number is equal to the sum of the squares of the other two numbers.

&there4 (5, 12, 13) is a pythagorean triplet.

(iv) In the triplet (24, 70, 74),
24 2 = 576, 70 2 = 4900, 74 2 = 5476 and 576 + 4900 = 5476

The square of the largest number is equal to the sum of the squares of the other two numbers.

&there4 (24, 70, 74) is a pythagorean triplet.

(v) In the triplet (10, 24, 27),
10 2 = 100, 24 2 = 576, 27 2 = 729 and 100 + 576 = 676 &ne 729

The square of the largest number is not equal to the sum of the squares of the other two numbers.

&there4 (10, 24, 27) is not a pythagorean triplet.

(vi) In the triplet (11, 60, 61),
11 2 = 121, 60 2 = 3600, 61 2 = 3721 and 121 + 3600 = 3721

The square of the largest number is equal to the sum of the squares of the other two numbers.

&there4 (11, 60, 61) is a pythagorean triplet.

Page No 38:

Question 2:

In the given figure, &angMNP = 90°, seg NQ &perpseg MP, MQ = 9, QP = 4, find NQ.

Answer:

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

Page No 38:

Question 3:

In the given figure, &angQPR = 90°, seg PM &perp seg QR and Q&ndashM&ndashR, PM = 10, QM = 8, find QR.

Answer:

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

Page No 39:

Question 4:

In the given figure. Find RP and PS using the information given in ∆PSR.

Answer:

Page No 39:

Question 5:

For finding AB and BC with the help of information given in the figure, complete following activity.

Answer:

In ∆ABC,
&angB = 90 ∘ , AC = 8 , AB = BC, &there4 &angA = &angC = 45 ∘

Hence, the completed activity is

Page No 39:

Question 6:

Find the side and perimeter of a square whose diagonal is 10 cm.

Answer:


It is given that ABCD is a square.

According to Pythagoras theorem, in ∆ABD

AB 2 + AD 2 = BD 2 ⇒ a 2 + a 2 = 10 2 ⇒ 2 a 2 = 100 ⇒ a 2 = 50 ⇒ a = 50 ⇒ a = 5 2   cm

Hence, the side of the square is 5 2 cm.

Now,
Perimeter of a square = 4 × side
= 4 × a
= 4 × 5 2
= 20 2 cm

Hence, the perimeter of the square is 20 2 cm.

Page No 39:

Question 7:

In the given figure, &angDFE = 90°, FG &perp ED, If GD = 8, FG = 12, find (1) EG (2) FD and (3) EF

Answer:

We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

∴   GF 2 = EG × GD ⇒ 12 2 = EG × 8 ⇒ 144 = EG × 8 ⇒ EG = 144 8 ⇒ EG = 18

Now,
According to Pythagoras theorem, in ∆DGF

DG 2 + GF 2 = FD 2 ⇒ 8 2 + 12 2 = FD 2 ⇒ 64 + 144 = FD 2 ⇒ FD 2 = 208 ⇒ FD = 4 13

EG 2 + GF 2 = EF 2 ⇒ 18 2 + 12 2 = EF 2 ⇒ 324 + 144 = EF 2 ⇒ EF 2 = 468 ⇒ EF = 6 13

Hence, FD = 4 13 and EF = 6 13 .

Page No 39:

Question 8:

Find the diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

Answer:

According to Pythagoras theorem, in ∆ABC

AB 2 + BC 2 = AC 2 ⇒ 35 2 + 12 2 = AC 2 ⇒ 1225 + 144 = AC 2 ⇒ AC 2 = 1369 ⇒ AC = 37   cm

Hence, the length of the diagonal is 37 cm.

Page No 39:

Question 9:

In the given figure, M is the midpoint of QR. &angPRQ = 90°. Prove that, PQ 2 = 4PM 2 &ndash 3PR 2

Answer:

According to Pythagoras theorem,

PR 2 + RM 2 = PM 2 ⇒ RM 2 = PM 2 - PR 2                 . . . 1

PR 2 + RQ 2 = PQ 2 ⇒ PQ 2 = PR 2 + RM + MQ 2 ⇒ PQ 2 = PR 2 + RM + RM 2 ⇒ PQ 2 = PR 2 + 2 RM 2 ⇒ PQ 2 = PR 2 + 4 RM 2 ⇒ PQ 2 = PR 2 + 4 PM 2 - PR 2                   from   1 ⇒ PQ 2 = PR 2 + 4 PM 2 - 4 PR 2 ⇒ PQ 2 = 4 PM 2 - 3 PR 2

Page No 39:

Question 10:

Walls of two buildings on either side of a street are parellel to each other. A ladder 5.8 m long is placed on the street such that its top just reaches the window of a building at the height of 4 m. On turning the ladder over to the other side of the street , its top touches the window of the other building at a height 4.2 m. Find the width of the street.

Answer:

Let the length of the ladder be 5.8 m.

According to Pythagoras theorem, in ∆EAB

EA 2 + AB 2 = EB 2 ⇒ 4 . 2 2 + AB 2 = 5 . 8 2 ⇒ 17 . 64 + AB 2 = 33 . 64 ⇒ AB 2 = 33 . 64 - 17 . 64 ⇒ AB 2 = 16 ⇒ AB = 4   m                     . . . 1

DC 2 + CB 2 = DB 2 ⇒ 4 2 + CB 2 = 5 . 8 2 ⇒ 16 + CB 2 = 33 . 64 ⇒ CB 2 = 33 . 64 - 16 ⇒ CB 2 = 17 . 64 ⇒ CB = 4 . 2   m                     . . . 2

Hence, the width of the street is 8.2 m.

Page No 43:

Question 1:

In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS =13, find QR.

Answer:

In ∆PQR, point S is the midpoint of side QR.

PQ 2 + PR 2 = 2 PS 2 + 2 QS 2           by   Apollonius   theorem ⇒ 11 2 + 17 2 = 2 13 2 + 2 QS 2 ⇒ 121 + 289 = 2 169 + 2 QS 2 ⇒ 410 = 338 + 2 QS 2 ⇒ 2 QS 2 = 410 - 338 ⇒ 2 QS 2 = 72 ⇒ QS 2 = 36 ⇒ QS = 6 ∴   QR = 2 × QS                       = 2 × 6                       = 12

Page No 43:

Question 2:

In ∆ABC, AB = 10, AC = 7, BC = 9 then find the length of the median drawn from point C to side AB

Answer:

In ∆ACB, point D is the midpoint of side AB.

CA 2 + CB 2 = 2 DC 2 + 2 AD 2           by   Apollonius   theorem ⇒ 7 2 + 9 2 = 2 DC 2 + 2 5 2 ⇒ 49 + 81 = 2 DC 2 + 2 25 ⇒ 130 = 2 DC 2 + 50 ⇒ 2 DC 2 = 130 - 50 ⇒ 2 DC 2 = 80 ⇒ DC 2 = 40 ⇒ DC = 2 10

Hence, the length of the median drawn from point C to side AB is 2 10 .

Page No 43:

Question 3:

In the given figure seg PS is the median of ∆PQR and PT &perp QR. Prove that,

(1) PR 2 = PS 2 + QR × ST + QR 2 2

(2) PQ 2 = PS 2 - QR × ST + QR 2 2

Answer:

According to Pythagoras theorem, in ∆PTQ

In ∆PQR, point S is the midpoint of side QR.

PQ 2 + PR 2 = 2 PS 2 + 2 QS 2           by   Apollonius   theorem ⇒ PR 2 = 2 PS 2 + 2 QS 2 - PQ 2 ⇒ PR 2 = PS 2 + PS 2 + QS 2 + QS 2 - PQ 2 ⇒ PR 2 = PS 2 + PT 2 + TS 2 + QS 2 + QR 2 2 - PT 2 + QT 2             From   1 ,   2   and   4 ⇒ PR 2 = PS 2 + QR 2 2 + PT 2 + TS 2 + QS 2 - PT 2 - QT 2 ⇒ PR 2 = PS 2 + QR 2 2 + QS 2 + TS 2 - QT 2 ⇒ PR 2 = PS 2 + QR 2 2 + QS 2 + TS + QT TS - QT ⇒ PR 2 = PS 2 + QR 2 2 + QS 2 + QS TS - QT ⇒ PR 2 = PS 2 + QR 2 2 + QS 2 + QS × TS - QS × QT ⇒ PR 2 = PS 2 + QR 2 2 + QS × TS + QS 2 - QS × QT ⇒ PR 2 = PS 2 + QR 2 2 + QS × TS + QS QS - QT ⇒ PR 2 = PS 2 + QR 2 2 + QS × TS + QS × TS ⇒ PR 2 = PS 2 + QR 2 2 + 2 QS × TS ⇒ PR 2 = PS 2 + QR 2 2 + QR × TS

Hence, PR 2 = PS 2 + QR × ST + QR 2 2 .

PQ 2 + PR 2 = 2 PS 2 + 2 QS 2           by   Apollonius   theorem ⇒ PQ 2 = 2 PS 2 + 2 QS 2 - PR 2 ⇒ PQ 2 = PS 2 + PS 2 + QS 2 + QS 2 - PR 2 ⇒ PQ 2 = PS 2 + PT 2 + TS 2 + QS 2 + QR 2 2 - PT 2 + RT 2             From   2 ,   3   and   4 ⇒ PQ 2 = PS 2 + QR 2 2 + PT 2 + TS 2 + QS 2 - PT 2 - RT 2 ⇒ PQ 2 = PS 2 + QR 2 2 + QS 2 - RT 2 - TS 2 ⇒ PQ 2 = PS 2 + QR 2 2 + QS 2 - TS + RT RT - TS ⇒ PQ 2 = PS 2 + QR 2 2 + QS 2 - TS + RT RS ⇒ PQ 2 = PS 2 + QR 2 2 + QS 2 - TS + RT QS ⇒ PQ 2 = PS 2 + QR 2 2 + QS 2 - QS × TS - QS × RT ⇒ PQ 2 = PS 2 + QR 2 2 - QS × TS + QS 2 - QS × RT ⇒ PQ 2 = PS 2 + QR 2 2 - QS × TS - QS RT - QS ⇒ PQ 2 = PS 2 + QR 2 2 - QS × TS - QS RT - SR ⇒ PQ 2 = PS 2 + QR 2 2 - QS × TS - QS × TS ⇒ PQ 2 = PS 2 + QR 2 2 - 2 QS × TS ⇒ PQ 2 = PS 2 + QR 2 2 - QR × TS

Hence, PQ 2 = PS 2 - QR × ST + QR 2 2 .

Page No 43:

Question 4:

In ∆ABC, point M is the midpoint of side BC.
If, AB 2 + AC 2 = 290 cm 2 , AM = 8 cm, find BC.

Answer:

In ∆ABC, point M is the midpoint of side BC.

AB 2 + AC 2 = 2 AM 2 + 2 BM 2           by   Apollonius   theorem ⇒ 290 = 2 8 2 + 2 BM 2 ⇒ 290 = 2 64 + 2 BM 2 ⇒ 290 = 128 + 2 BM 2 ⇒ 2 BM 2 = 290 - 128 ⇒ 2 BM 2 = 162 ⇒ BM 2 = 81 ⇒ BM = 9 ∴   BC = 2 × BM                       = 2 × 9                       = 18   cm

Page No 43:

Question 5:

In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS 2 + TQ 2 = TP 2 + TR 2 (As shown in the figure, draw seg AB || side SR and A-T-B)

Answer:

According to Pythagoras theorem, in ∆PAT

Now,
TS 2 + TQ 2 = AS 2 + AT 2 + QB 2 + BT 2                   From   2   and   3 ⇒ TS 2 + TQ 2 = BR 2 + AT 2 + PA 2 + BT 2                   ∵   AS = BR   and   PA = QB ⇒ TS 2 + TQ 2 = BR 2 + BT 2 + PA 2 + AT 2 ⇒ TS 2 + TQ 2 = TR 2 + PT 2                   From   1   and   4 ⇒ TS 2 + TQ 2 = TR 2 + PT 2

Hence, TS 2 + TQ 2 = TP 2 + TR 2 .

Page No 43:

Question 1:

Some questions and their alternative answers are given. Select the correct alternative.
(1) Out of the following which is the Pythagorean triplet?

Answer:

(1) (A) In the triplet (1, 5, 10),
1 2 = 1, 5 2 = 25, 10 2 = 100 and 1 + 25 = 26 &ne 100

The square of the largest number is not equal to the sum of the squares of the other two numbers.

&there4 (1, 5, 10) is not a pythagorean triplet.

(B) In the triplet (3, 4, 5),
3 2 = 9, 4 2 = 16, 5 2 = 25 and 9 + 16 = 25

The square of the largest number is equal to the sum of the squares of the other two numbers.

&there4 (3, 4, 5) is a pythagorean triplet.

(C) In the triplet (2, 2, 2),
2 2 = 4, 2 2 = 4, 2 2 = 4 and 4 + 4 = 8 &ne 4

The square of the largest number is not equal to the sum of the squares of the other two numbers.

&there4 (2, 2, 2) is not a pythagorean triplet.

(D) In the triplet (5, 5, 2),
2 2 = 4, 5 2 = 25, 5 2 = 25 and 4 + 25 = 29 &ne 25

The square of the largest number is not equal to the sum of the squares of the other two numbers.

&there4 (5, 5, 2) is not a pythagorean triplet.

Hence, the correct option is (B).


(2) According to the pythagoras theorem,
Sum of the squares of the sides making right angle is equal to the square of the third side.

&there4 169 = square of the hypotenuse
&rArr Length of the hypotenuse = 169
= 13

Hence, the correct option is (B).


(3) (A) In the triplet 15/08/17,
15 2 = 225, 8 2 = 64, 17 2 = 289 and 225 + 64 = 289

The square of the largest number is equal to the sum of the squares of the other two numbers.

&there4 15/08/17 is a pythagorean triplet.

(B) In the triplet 16/08/16,
16 2 = 256, 8 2 = 64, 16 2 = 256 and 256 + 64 = 320 &ne 256

The square of the largest number is not equal to the sum of the squares of the other two numbers.

&there4 16/08/16 is not a pythagorean triplet.

(C) In the triplet 3/5/17,
3 2 = 9, 5 2 = 25, 17 2 = 289 and 9 + 25 = 34 &ne 289

The square of the largest number is not equal to the sum of the squares of the other two numbers.

&there4 3/5/17 is not a pythagorean triplet.

(D) In the triplet 4/9/15,
4 2 = 16, 9 2 = 81, 15 2 = 225 and 16 + 81 = 97 &ne 225

The square of the largest number is not equal to the sum of the squares of the other two numbers.

&there4 4/9/15 is not a pythagorean triplet.

Hence, the correct option is (A).


(4) In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle.

Hence, the correct option is (C).

It is given that ABCD is a square.

According to Pythagoras theorem, in ∆ABD

AB 2 + AD 2 = BD 2 ⇒ x 2 + x 2 = 10 2 2 ⇒ 2 x 2 = 200 ⇒ x 2 = 100 ⇒ x = 100 ⇒ x = 10   cm

Hence, the side of the square is 10 cm.

Now,
Perimeter of a square = 4 × side
= 4 × x
= 4 × 10
= 40 cm

Hence, the correct option is (D).


We know that,
In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

Hence, the correct option is (C).


(7) According to Pythagoras theorem,

Hence, the correct option is (B).


(8) In ∆ABC, AB = 6 3 cm, AC = 12 cm, BC = 6 cm.

AB 2 = ( 6 3 ) 2 = 108
AC 2 = (12) 2 = 144
BC 2 = (6) 2 = 36

In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle.

In a right angled triangle, if one side is half of the hypotenuse then the angle opposite to that side is 30°.
Here, BC is half of AC.

Hence, the correct option is (A).

Page No 44:

Question 2:

Solve the following examples.
(1) Find the height of an equilateral triangle having side 2a.
(2) Do sides 7 cm, 24 cm, 25 cm form a right angled triangle ? Give reason.
(3) Find the length a diagonal of a rectangle having sides 11 cm and 60 cm.
(4) Find the length of the hypotenuse of a right angled triangle if remaining sides are 9 cm and 12 cm.
(5) A side of an isosceles right angled triangle is x.Find its hypotenuse.
(6) In ∆PQR PQ = 8 , QR = 5 , PR = 3 . Is ∆PQR a right angled triangle ? If yes, which angle is of 90°?

Answer:


Since, ABC is an equilateral triangle, AD is the perpendicular bisector of BC.

Now, According to Pythagoras theorem,
In ∆ABD

AB 2 = AD 2 + BD 2 ⇒ 2 a 2 = AD 2 + a 2 ⇒ 4 a 2 - a 2 = AD 2 ⇒ AD 2 = 3 a 2 ⇒ AD = 3 a

Hence, the height of an equilateral triangle is 3 a .


(2) In the triplet (7, 24, 25),
7 2 = 49, 24 2 = 576, 25 2 = 625 and 49 + 576 = 625

The square of the largest number is equal to the sum of the squares of the other two numbers.

&there4 Sides 7 cm, 24 cm, 25 cm form a right angled triangle.


According to Pythagoras theorem,

AB 2 + BC 2 = AC 2 ⇒ 60 2 + 11 2 = AC 2 ⇒ 3600 + 121 = AC 2 ⇒ AC 2 = 3721 ⇒ AC = 61   cm

Hence, the length of a diagonal of the rectangle is 61 cm.


(4) In a right angled triangle,

According to Pythagoras theorem.

Hence, the length of the hypotenuse is 15 cm.


(5) It is given that, a side of an isosceles right angled triangle is x.

Then, the other side of the triangle is also x.

According to Pythagoras theorem.

Hence, its hypotenuse is 2 x .


(6) In ∆PQR, PQ = 8 , QR = 5 , PR = 3 .
( 8 ) 2 = 8, ( 5 ) 2 = 5, ( 3 ) 2 = 3 and 3 + 5 = 8

The square of the largest number is equal to the sum of the squares of the other two numbers.

&there4 ∆PQR form a right angled triangle, where angle R is of 90°.

Page No 44:

Question 3:

In ∆RST, &angS = 90°, &angT = 30°, RT = 12 cm then find RS and ST.

Answer:

Hence, RS = 6 cm and ST = 6 3 cm.

Page No 44:

Question 4:

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm.

Answer:

It is given that, area of rectangle is 192 sq.cm.

According to Pythagoras theorem,

AB 2 + BC 2 = AC 2 ⇒ 16 2 + 12 2 = AC 2 ⇒ 256 + 144 = AC 2 ⇒ AC 2 = 400 ⇒ AC = 20   cm

Hence, the length of a diagonal of the rectangle is 20 cm.

Page No 44:

Question 5:

Find the length of the side and perimeter of an equilateral triangle whose height is 3 cm.

Answer:

Since, ABC is an equilateral triangle, CD is the perpendicular bisector of AB.

Now, According to Pythagoras theorem,
In ∆ACD

AC 2 = AD 2 + CD 2 ⇒ 2 a 2 = a 2 + 3 2 ⇒ 4 a 2 - a 2 = 3 ⇒ 3 a 2 = 3 ⇒ a 2 = 1 ⇒ a = 1   cm

Hence, the length of the side of an equilateral triangle is 2 cm.

Now,
Perimeter of the triangle = (2 + 2 + 2) cm
= 6 cm

Hence, perimeter of an equilateral triangle is 6 cm.

Page No 44:

Question 6:

In ∆ABC seg AP is a median. If BC = 18, AB 2 + AC 2 = 260 Find AP.

Answer:

In ∆ABC, point P is the midpoint of side BC.

AB 2 + AC 2 = 2 AP 2 + 2 BP 2           by   Apollonius   theorem ⇒ 260 = 2 AP 2 + 2 9 2 ⇒ 260 = 2 AP 2 + 2 81 ⇒ 260 = 2 AP 2 + 162 ⇒ 2 AP 2 = 260 - 162 ⇒ 2 AP 2 = 98 ⇒ AP 2 = 49 ⇒ AP = 7

Page No 45:

Question 7:

∆ABC is an equilateral triangle. Point P is on base BC such that PC = 1 3 BC, if AB = 6 cm find AP.

Answer:

∆ABC is an equilateral triangle.

It is given that,
PC = 1 3 BC ⇒ PC = 1 3 × 6 ⇒ PC = 2   cm ⇒ BP = 4   cm

Since, ABC is an equilateral triangle, OA is the perpendicular bisector of BC.
&there4 OC = 3 cm
&rArr OP = OC &minus PC
= 3 &minus 2
= 1 . (1)

Now, According to Pythagoras theorem,
In ∆AOB,

AB 2 = AO 2 + OB 2 ⇒ 6 2 = AO 2 + 3 2 ⇒ 36 - 9 = AO 2 ⇒ AO 2 = 27 ⇒ AO = 3 3   cm                     . . . 2

AP 2 = AO 2 + OP 2 ⇒ AP 2 = 3 3 2 + 1 2                         From   1   and   2 ⇒ AP 2 = 27 + 1 ⇒ AP 2 = 28 ⇒ AP = 2 7   cm

Page No 45:

Question 8:

From the information given in the figure, prove that PM = PN = 3 × a

Answer:

Since, ∆PQR is an equilateral triangle, PS is the perpendicular bisector of QR.
&there4 QS = SR = a 2 . (1)

Now, According to Pythagoras theorem,
In ∆PQS,

PQ 2 = QS 2 + PS 2 ⇒ a 2 = a 2 2 + PS 2 ⇒ PS 2 = a 2 - a 2 4 ⇒ PS 2 = 4 a 2 - a 2 4 ⇒ PS 2 = 3 a 2 4 ⇒ PS = 3 a 2                     . . . 2

PM 2 = MS 2 + PS 2 ⇒ PM 2 = a + a 2 2 + 3 2 a 2 ⇒ PM 2 = 3 a 2 2 + 3 2 a 2 ⇒ PM 2 = 9 a 2 4 + 3 a 2 4 ⇒ PM 2 = 12 a 2 4 ⇒ PM 2 = 3 a 2 ⇒ PM = 3 a                     . . . 3

PN 2 = NS 2 + PS 2 ⇒ PN 2 = a + a 2 2 + 3 2 a 2 ⇒ PN 2 = 3 a 2 2 + 3 2 a 2 ⇒ PN 2 = 9 a 2 4 + 3 a 2 4 ⇒ PN 2 = 12 a 2 4 ⇒ PN 2 = 3 a 2 ⇒ PN = 3 a                     . . . 4

From (3) and (4), we get
PM = PN = 3 × a

Hence, PM = PN = 3 × a.

Page No 45:

Question 9:

Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Answer:

Diagonals of a parallelogram bisect each other.
i.e. O is the mid point of AC and BD.

In ∆ABD, point O is the midpoint of side BD.

AB 2 + AD 2 = 2 AO 2 + 2 BO 2           by   Apollonius   theorem               . . . 1

In ∆CBD, point O is the midpoint of side BD.

AB 2 + AD 2 + CB 2 + CD 2 = 2 AO 2 + 2 BO 2 + 2 CO 2 + 2 BO 2 ⇒ AB 2 + AD 2 + CB 2 + CD 2 = 2 AO 2 + 4 BO 2 + 2 AO 2               ∵ OC = OA ⇒ AB 2 + AD 2 + CB 2 + CD 2 = 4 AO 2 + 4 BO 2 ⇒ AB 2 + AD 2 + CB 2 + CD 2 = 2 AO 2 + 2 BO 2 ⇒ AB 2 + AD 2 + CB 2 + CD 2 = AC 2 + BD 2 ⇒ AB 2 + AD 2 + CB 2 + CD 2 = AC 2 + BD 2

Hence, the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Page No 45:

Question 10:

Pranali and Prasad started walking to the East and to the North respectively, from the same point and at the same speed. After 2 hours distance between them was 15 2 km. Find their speed per hour.

Answer:

It is given that, Pranali and Prasad have same speed.
Thus, they cover same distance in 2 hours.
i.e. OA = OB

Let the speed be x km per hour.

According to Pythagoras theorem,
In ∆AOB

AB 2 = AO 2 + OB 2 ⇒ 15 2 2 = AO 2 + OA 2 ⇒ 450 = 2 AO 2 ⇒ AO 2 = 450 2 ⇒ AO 2 = 225 ⇒ AO = 15   km ⇒ BO = 15   km

Hence, their speed is 7.5 km per hour.

Page No 45:

Question 11:

In ∆ABC, &angBAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that:
4(BL 2 + CM 2 ) = 5 BC 2

Answer:


According to Pythagoras theorem,
In ∆ABC

In ∆ABC, point M is the midpoint of side BD.

AC 2 + BC 2 = 2 CM 2 + 2 AM 2           by   Apollonius   theorem               . . . 2

In ∆CBA, point L is the midpoint of side AC.

AC 2 + BC 2 + CB 2 + AB 2 = 2 CM 2 + 2 AM 2 + 2 BL 2 + 2 AL 2 ⇒ 2 CM 2 + 2 BL 2 = AC 2 + BC 2 + CB 2 + AB 2 - 2 AM 2 - 2 AL 2 ⇒ 2 CM 2 + BL 2 = AC 2 + 2 BC 2 + AB 2 - 2 AM 2 - 2 AL 2 ⇒ 2 CM 2 + BL 2 = AC 2 + AB 2 + 2 BC 2 - 2 1 2 AB 2 - 2 1 2 AC 2 ⇒ 2 CM 2 + BL 2 = BC 2 + 2 BC 2 - 1 2 AB 2 - 1 2 AC 2                         From   1 ⇒ 2 CM 2 + BL 2 = 3 BC 2 - 1 2 AB 2 + AC 2 ⇒ 2 CM 2 + BL 2 = 3 BC 2 - 1 2 BC 2 ⇒ 4 CM 2 + BL 2 = 2 3 BC 2 - 1 2 BC 2 ⇒ 4 CM 2 + BL 2 = 6 BC 2 - BC 2 ⇒ 4 CM 2 + BL 2 = 5 BC 2

Hence, 4(BL 2 + CM 2 ) = 5 BC 2 .

Page No 45:

Question 12:

Sum of the squares of adjacent sides of a parallelogram is 130 sq.cm and length of one of its diagonals is 14 cm. Find the length of the other diagonal.

Answer:

It is given that,
AB 2 + AD 2 = 130 sq. cm
BD = 14 cm

Diagonals of a parallelogram bisect each other.
i.e. O is the mid point of AC and BD.

In ∆ABD, point O is the midpoint of side BD.

AB 2 + AD 2 = 2 AO 2 + 2 BO 2           by   Apollonius   theorem ⇒ 130 = 2 AO 2 + 2 7 2 ⇒ 130 = 2 AO 2 + 2 × 49 ⇒ 130 = 2 AO 2 + 98 ⇒ 2 AO 2 = 130 - 98 ⇒ 2 AO 2 = 32 ⇒ AO 2 = 16 ⇒ AO = 4   cm

Sinec, point O is the midpoint of side AC.

Hence, the length of the other diagonal is 8 cm.

Page No 45:

Question 13:

In ∆ABC, seg AD &perp seg BC DB = 3CD. Prove that :
2AB 2 = 2AC 2 + BC 2

Answer:

It is given that,
DB = 3 CD


According to Pythagoras theorem,
In ∆ABD

AB 2 = AD 2 + DB 2 ⇒ AD 2 = AB 2 - DB 2                 . . . 2

AC 2 = AD 2 + CD 2 ⇒ AC 2 = AB 2 - DB 2 + CD 2               From   2   ⇒ AC 2 = AB 2 - 3 CD 2 + CD 2               Given ⇒ AC 2 = AB 2 - 9 CD 2 + CD 2 ⇒ AC 2 = AB 2 - 8 CD 2 ⇒ AB 2 = AC 2 + 8 CD 2 ⇒ AB 2 = AC 2 + 8 BC 4 2                     From   1 ⇒ AB 2 = AC 2 + BC 2 2 ⇒ 2 AB 2 = 2 AC 2 + BC 2

Page No 45:

Question 14:

In an isosceles triangle, length of the congruent sides is 13 cm and its base is 10 cm. Find the distance between the vertex opposite the base and the centroid.

Answer:

The centroid is located two third of the distance from any vertex of the triangle.

∴   Distance   between   the   vertex   and   the   centroid = 2 3 × 12 = 8   cm

Hence, the distance between the vertex opposite the base and the centroid is 8 cm.

Page No 46:

Question 15:

In a trapezium ABCD, seg AB || seg DC seg BD &perp seg AD, seg AC &perp seg BC, If AD = 15, BC = 15 and AB = 25. Find A(▢ABCD)

Answer:

According to Pythagoras theorem,
In ∆ABD

AB 2 = AD 2 + DB 2 ⇒ 25 2 = 15 2 + BD 2 ⇒ 625 = 225 + BD 2 ⇒ BD 2 = 625 - 225 ⇒ BD 2 = 400 ⇒ BD = 20

Therefore, height of the trapezium = 12.

Now,
According to Pythagoras theorem,
In ∆ADP

AD 2 = AP 2 + DP 2 ⇒ 15 2 = 12 2 + AP 2 ⇒ 225 = 144 + AP 2 ⇒ AP 2 = 225 - 144 ⇒ AP 2 = 81 ⇒ AP = 9

Hence, A(▢ABCD) = 192 sq. units.

Page No 46:

Question 16:

In the given figure, ∆PQR is an equilateral triangle. Point S is on seg QR such that
QS = 1 3 QR.
Prove that : 9 PS 2 = 7 PQ 2

Answer:

Let the side of equilateral triangle ∆PQR be x.
PT be the altitude of the ∆PQR.

We know that, in equilateral triangle, altitude divides the base in two equal parts.
&there4 QT = TR = 1 2 QR = x 2

∴   ST = QT - QS = x 2 - x 3 = x 6

According to Pythagoras theorem,
In ∆PQT

PQ 2 = QT 2 + PT 2 ⇒ x 2 = x 2 2 + PT 2 ⇒ x 2 = x 2 4 + PT 2 ⇒ PT 2 = x 2 - x 2 4 ⇒ PT 2 = 3 x 2 4 ⇒ PT = 3 x 2

PS 2 = ST 2 + PT 2 ⇒ PS 2 = x 6 2 + 3 x 2 2 ⇒ PS 2 = x 2 36 + 3 x 2 4 ⇒ PS 2 = x 2 + 27 x 2 36 ⇒ PS 2 = 28 x 2 36 ⇒ PS 2 = 7 x 2 9 ⇒ 9 PS 2 = 7 PQ 2

Page No 46:

Question 17:

Seg PM is a median of ∆PQR. If PQ = 40, PR = 42 and PM = 29, find QR.

Answer:

In ∆PQR, point M is the midpoint of side QR.

PQ 2 + PR 2 = 2 PM 2 + 2 QM 2           by   Apollonius   theorem ⇒ 40 2 + 42 2 = 2 29 2 + 2 QM 2 ⇒ 1600 + 1764 = 1682 + 2 QM 2 ⇒ 3364 - 1682 = 2 QM 2 ⇒ 1682 = 2 QM 2 ⇒ QM 2 = 841 ⇒ QM = 29 ⇒ QR = 2 × 29 ⇒ QR = 58

Page No 46:

Question 18:

Seg AM is a median of ∆ABC. If AB = 22, AC = 34, BC = 24, find AM

Answer:

In ∆ABC, point M is the midpoint of side BC.

AB 2 + AC 2 = 2 AM 2 + 2 BM 2           by   Apollonius   theorem ⇒ 22 2 + 34 2 = 2 AM 2 + 2 12 2 ⇒ 484 + 1156 = 2 AM 2 + 288 ⇒ 1640 - 288 = 2 AM 2 ⇒ 1352 = 2 AM 2 ⇒ AM 2 = 676 ⇒ AM = 26


Georg Alexander Pick

Georg Pick was born into a Jewish family. His mother was Josefa Schleisinger and his father was Adolf Josef Pick, the head of a private institute. Georg was educated at home by his father up to the age of eleven when he entered the fourth class of the Leopoldstaedter Communal Gymnasium. He sat his school leaving examinations in 1875 which qualified him for university entrance.

Pick entered the University of Vienna in 1875 . He published a mathematics paper in the following year when only seventeen years old. He studied mathematics and physics, graduating in 1879 with a qualification which would allow him to teach both of these subjects. In 1877 Leo Königsberger had moved from Technische Hochschule in Dresden to take up a chair at the University of Vienna. He became Pick's supervisor and, on 16 April 1880 , Pick was awarded his doctorate for his dissertation Über eine Klasse abelscher Integrale Ⓣ . Emil Weyr had been appointed as second examiner of the thesis.

After the award of his doctorate, Pick was appointed as an assistant to Ernest Mach at the Karl-Ferdinand University in Prague. Mach had moved from Graz, where he was professor of mathematics, to Prague in 1867 to take up the chair of physics there. He, like Pick, had studied at the University of Vienna and, by the time Pick became his assistant, he was regarded as one of the leading scientists in Europe. Pick now aimed at becoming a lecturer in Prague and in order to obtain the right to lecture he had to write an habilitation thesis. This he did quite quickly and received the right to lecture in Prague in 1881 with his habilitation thesis Über die Integration hyperelliptischer Differentiale durch Logarithmen Ⓣ .

Except for the academic year 1884 - 85 which Pick spent studying under Klein at the University of Leipzig, he remained in Prague for the rest of his career. He was promoted to extraordinary professor of mathematics in 1888 , then he was appointed as ordinary professor ( full professor ) in 1892 at the German University of Prague. His mathematical work was extremely broad and his 67 papers range across many topics such as linear algebra, invariant theory, integral calculus, potential theory, functional analysis, and geometry. However more than half of his papers were on functions of a complex variable, differential equations, and differential geometry. Terms such as 'Pick matrices', 'Pick-Nevanlinna interpolation', and the 'Schwarz-Pick lemma' are sometimes used today. He is best remembered, however, for Pick's theorem which appeared in his eight page paper of 1899 Geometrisches zur Zahlenlehre Ⓣ published in Prague in Sitzungber. Lotos, Naturwissen Zeitschrift.

Pick's theorem is on reticular geometry. The plane becomes a lattice on setting up two systems of parallel equally spaced straight lines in the plane. These Pick calls the 'main reticular lines' and their points of intersection are called 'reticular points'. A line joining any two reticular points is called a 'reticular line'. Notice that the main reticular lines are reticular lines but there are many other reticular lines. A polygon whose edges are reticular lines Pick calls a reticular polygon. Pick's theorem states that the area of a reticular polygon is L + 1 2 B − 1 L + largefrac<1><2> ormalsize B - 1 L + 2 1 ​ B − 1 where L L L is the number of reticular points inside the polygon and B B B is the number of reticular points on the edges of the polygon. The result did not receive much attention after Pick published it, but in 1969 Steinhaus included it in his famous book Mathematical Snapshots. From that time on Pick's theorem has attracted much attention and admiration for its simplicity and elegance.

At the German University of Prague Pick became dean of the philosophy faculty in 1900 - 01 . He supervised about 20 students for their doctorates, the most famous being Charles Loewner who worked under Pick's supervision and was awarded his doctorate for his thesis on geometric function theory in 1917 . There is another aspect of Pick's life which merits attention. In 1910 he was on a committee set up by the German University of Prague to consider appointing Einstein to the university. Pick was the driving force behind the appointment and Einstein was appointed to a chair of mathematical physics at the German University of Prague in 1911 . He held this post until 1913 and during these years the two were close friends. Not only did they share scientific interests, but they also shared a passionate interest in music. Pick, who played in a quartet, introduced Einstein into the scientific and musical societies of Prague. In fact Pick's quartet consisted of four professors from the university including Camillo Körner, the professor of mechanical engineering.

After Pick retired in 1927 he was named professor emeritus and returned to Vienna, the town of his birth. However, in 1938 he returned to Prague after the Anschluss on 12 March when German troops marched into Austria. At the end of September 1938 the Prague government was asked to give Germany all districts of Bohemia and Moravia with populations that were 50 percent or more German. The leaders of Czechoslovakia resigned rather than agree, but those who took over gave the regions to Germany. Hitler's armies invaded on 14 March 1939 and Hitler installed his representative in Prague to run the country. Pick had been elected as a member of the Czech Academy of Sciences and Arts, but after the Nazis took over Prague, Pick was excluded from the Academy. The Nazis set up a camp at Theresienstadt in Nordboehmen on 24 November 1941 to house elderly, privileged, and famous Jews. Of around 144 , 000 Jews sent to Theresienstadt about a quarter died there and around 60 % were sent on to Auschwitz or other death camps. Pick was sent to Theresienstadt on 13 July 1942 and he died there two weeks later aged 82 .