ISI BStat 2017 Subjective 2 | Right angled triangle in a circle

Try this beautiful problem from ISI BStat 2017 Subjective 2 based on right-angled triangle in a circle.

Problem: ISI BStat 2017 Subjective 2

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 $$ \frac {12 + 9 \sqrt {15} }{5} $$

Discussion:

This is a simple application of the power of a point.  We have to find the power of point A.

Suppose B = (0,0), C = (x,0) and D = (0,y).

Since AD = 5, therefore, coordinate of A = (0, 5+y).

Using the property of the power of point A, we have: \(  AD \times AB = AE \times AC \) . Since AE is 4, this implies

\( 5 \times (5+y) = 4 \times \sqrt { x^2 + (5+y)^2 } \ \Rightarrow 25 (5+y)^2 = 16 (x^2 + (5+y)^2 ) \ \Rightarrow 9(5+y)^2 = 16x^2 \)

Since x and y are positive, we can take square roots on both sides. Hence we will have:

\( 3(5+y) = 4x  \ \Rightarrow y = \frac {4x}{3} - 5 \)

Now note that \( \angle DBC = 90^o \) implying DC is a diameter. Hence DC is 12 (as radius is given to be 6).

Hence \( x^2 + y^2 = 12^2 \).

Replacing y by \(  \frac {4x}{3} - 5 \) as found earlier, we have

\( \displaystyle {x^2 + \left (\frac {4x}{3} - 5 \right )^2 = 12^2  \ \Rightarrow 9x^2 + 16x^2 - 120x + 225 = 144 \times 9 \ \Rightarrow 25x^2 - 120x - 1071 = 0 \ \Rightarrow x = \frac{120 \pm \sqrt { 120^2 + 107100}}{50} \ \Rightarrow x = \frac {120 \pm \sqrt { 14400 + 107100}}{50} \ \Rightarrow  x = \frac {120 \pm 10 \sqrt {1215}}{50} \ \Rightarrow  x = \frac {12 \pm 9 \sqrt {15}}{5} } \)

Since x is positive we have \(\displaystyle { x = \frac {12 + 9 \sqrt {15}}{5} } \)

(proved)

Back to the question paper

Some Useful Links:

• ISI BStat Entrance - Solutions and Problems
• ISI CMI Entrance - BStat Program
• Solutions of Equations, ISI 2018 Problem 1 - video