Understand the problem

Consider a circle of radius 6 as given in the diagram below. Let \(B,C,D\) and \(E\) be points on the circle such that \(BD\) and \(CE\), when extended, intersect at \(A\). If \(AD\) and \(AE\) have length 5 and 4 respectively, and \(DBC\) is a right angle, then show that the length of \(BC\) is \(\frac{12+\sqrt{15}}{5}\).

Diagram : click here.

Source of the problem

I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2017

Topic

Plane Geometry

Difficulty Level

6 out of 10

Suggested Book

‘An excursion in Mathematics’ published by Bhaskaracharya Pratishthana

Start with hints

Do you really need a hint? Try it first!

Given, \(AD=5, AE=4\) and \(\angle DBC=90^\circ \).

As \(D,B,C\) are points on the circle having radius 6.

Therefore \(DC\) is the diameter of the circle

\(\Rightarrow DC=6×2=12\).

Now DC is diameter of the circle \(\Rightarrow \angle DEC=90^\circ \).

Therefore \(\angle DEA\) is also right angle.

The length of \(DE=\sqrt{5^2-4^2}=3\).

And the length of \(EC=\sqrt{12^2-3^2}=3\sqrt{15}\).

Therefore \(AC=AE+EC=4+3\sqrt{15}\).

From \(∆AED\) and \(∆ABC\) we have,

\(\frac{AD}{DE}=\frac{AC}{BC} \Rightarrow BC=\frac{AC\cdot DE}{AD}=\frac{(4+3\sqrt{15})\cdot 3}{5}\).

Therefore the length of \(BC\) is \(\frac{12+9\sqrt{15}}{5}\). (Ans.)

Watch the video ( Coming Soon … )

Connected Program at Cheenta

I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

Similar Problems

Polynomial Functional Equation – Random Olympiad Problem

This beautiful application of Functional Equation is related to the concepts of Polynomials. Sequential hints are given to work out the problem and to revisit the concepts accordingly.

Number Theory – Croatia MO 2005 Problem 11.1

This beautiful application from Croatia MO 2005, Problem 11.1 is based on the concepts of Number Theory. Sequential hints are given to work the problem accordingly.

SMO(senior)-2014 Problem 2 Number Theory

This beautiful application from SMO(senior)-2014 is based on the concepts of Number Theory . Sequential hints are provided to understand and solve the problem .

SMO (senior) -2014/problem-4 Number Theory

This beautiful application from SMO(senior)-2014/Problem 4 is based Number Theory . Sequential hints are provided to understand and solve the problem .

The best exponent for an inequality

Understand the problemLet be positive real numbers such that .Find with proof that is the minimal value for which the following inequality holds:Albania IMO TST 2013 Inequalities Medium Inequalities by BJ Venkatachala Start with hintsDo you really need a hint?...

A functional inequation

Understand the problemFind all functions such thatholds for all . Benelux MO 2013 Functional Equations Easy Functional Equations by BJ Venkatachala Start with hintsDo you really need a hint? Try it first!Note that the RHS does not contain $latex y$. Thus it should...

Mathematical Circles Inequality Problem

A beautiful inequality problem from Mathematical Circles Russian Experience . we provide sequential hints . key idea is to use arithmetic mean , geometric mean inequality.

RMO 2019

Regional Math Olympiad (RMO) 2019 is the second level Math Olympiad Program in India involving Number Theory, Geometry, Algebra and Combinatorics.

AMC 2019 12A Problem 15 Diophantine Equation

Beautiful application of Logarithm and Diophantine Equation in American Mathematics Competition (2019) 12A

Costa Rica NMO 2010, Final Round, Problem 4 – Number Theory

The problem is a beautiful application of the techniques in Diophantine Equation from Costa Rica Math Olympiad 2010.