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

ISI BStat 2017 Subjective 2

I.S.I. 2017 geometry problem

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

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