Right angled triangle in a circle (B.Stat 2017, subjective 2)

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

I.S.I. 2017 geometry problem
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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