An Acute Possibility

Probability Level 4

A unit equilateral triangle △ A B C \triangle ABC is centered in a circle with a radius of 3 3 . A point P P is chosen randomly inside the circle, and the three lengths A P AP , B P BP , and C P CP are used as sides to form a new triangle △ D E F \triangle DEF . If the probability that △ D E F \triangle DEF is acute is a b \frac{a}{b} , where a a and b b are co-prime positive integers, find a + b a + b .


The answer is 5.

David Vreken by David Vreken , 42, USA

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2 solutions

David Vreken
Aug 2, 2019

Center the equilateral triangle at the origin so its coordinates are A ( 0 , 3 3 ) A(0, \frac{\sqrt{3}}{3}) , B ( 1 2 , − 3 6 ) B(\frac{1}{2}, -\frac{\sqrt{3}}{6}) , and C ( − 1 2 , − 3 6 ) C(-\frac{1}{2}, -\frac{\sqrt{3}}{6}) . Let P P have coordinates ( x , y ) (x, y) .

By the distance formula, A P 2 = ( x − 3 3 ) 2 + y 2 AP^2 = (x - \frac{\sqrt{3}}{3})^2 + y^2 , B P 2 = ( x + 1 2 ) 2 + ( y + 3 6 ) 2 BP^2 = (x + \frac{1}{2})^2 + (y + \frac{\sqrt{3}}{6})^2 , and C P 2 = ( x − 1 2 ) 2 + ( y + 3 6 ) 2 CP^2 = (x - \frac{1}{2})^2 + (y + \frac{\sqrt{3}}{6})^2 . To form an obtuse triangle, either A P 2 + B P 2 < C P 2 AP^2 + BP^2 < CP^2 , A P 2 + C P 2 < B P 2 AP^2 + CP^2 < BP^2 , or B P 2 + C P 2 < A P 2 BP^2 + CP^2 < AP^2 , which leads to equations x 2 + ( y + 2 3 3 ) 2 < 1 x^2 + (y + \frac{2\sqrt{3}}{3})^2 < 1 , ( x − 1 ) 2 + ( y − 3 3 ) 2 < 1 (x - 1)^2 + (y - \frac{\sqrt{3}}{3})^2 < 1 , and ( x + 1 ) 2 + ( y − 3 3 ) 2 < 1 (x + 1)^2 + (y - \frac{\sqrt{3}}{3})^2 < 1 , which are three tangential unit circles.

The probability that the triangle formed is acute is therefore 9 π − π − π − π 9 π = 2 3 \frac{9\pi - \pi - \pi - \pi}{9\pi} = \frac{2}{3} , so a = 2 a = 2 , b = 3 b = 3 , and a + b = 5 a + b = 5 .

2 Helpful 2 Interesting 2 Brilliant 1 Confused

3 small circles there are inside big circle here. But there is triangle with two circles inside big circle. Find it.

Yuriy Kazakov - 1 year, 10 months ago

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Nice follow-up question! If we change the triangle's coordinates to ( 0 , 1 ) (0, 1) , ( k , 0 ) (k, 0) , and ( − k , 0 ) (-k, 0) , so that we can adjust k k to examine all isosceles triangles, we will find that the three circles can be expressed as x 2 + ( y + 1 ) 2 < 2 − 2 k 2 x^2 + (y + 1)^2 < 2 - 2k^2 , ( x − 2 k ) 2 + ( y − 1 ) 2 < 4 k 2 (x - 2k)^2 + (y - 1)^2 < 4k^2 , and ( x + 2 k ) 2 + ( y − 1 ) 2 < 4 k 2 (x + 2k)^2 + (y - 1)^2 < 4k^2 . The first circle will disappear when 2 − 2 k 2 ≤ 0 2 - 2k^2 \leq 0 , or when k ≥ 1 k \geq 1 . In other words, there will only be two circles for a right isoceles triangle or for an obtuse isosceles triangle.

David Vreken - 1 year, 10 months ago

Simple example is triangle with lengths 3,4,5.

Yuriy Kazakov - 1 year, 10 months ago

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