The next year

Geometry Level 3

Given that P P is a point inside an equilateral triangle of side 2016 units. Find the sum of the lengths of the perpendiculars drawn from P P to the sides.

If the answer will be in the form of a b a\sqrt{b} for positive integers a a and b b with b b square free, type your answer as a + b a+ b .


The answer is 1011.

Akshat Sharda by Akshat Sharda , 21, India

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2 solutions

Deepak Kumar
Jul 14, 2015

Quick approach: Take P as centroid.Proper approach : let the perpendicular lengths be p,q,r .Join P with vertices and equate sum areas of the three triangles formed to the area of original triangle.

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Isaac Lu
Mar 6, 2017

1008 3 = 1008 3 ⋅ 3 3 = 336 3 \displaystyle\frac{1008}{\sqrt{3}}=\frac{1008}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}}=336\sqrt{3}

Since there are three perpendicular lines, we have: 3 × 336 3 = 1008 3 3\times336\sqrt{3}=1008\sqrt{3}

Hence, a = 1008 a=1008 and b = 3 b=3 ...giving a + b = 1011 a+b=1011

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