Euler is Everywhere! #3

Geometry Level 4

Triangle A B C ABC with area 25 25 is inscribed in a circle with radius 5. 5. A point P P is selected such that it is 3 3 units away from the center of this circle. Let A , B , C A', B', C' be the feet of the perpendiculars dropped from P P to the sides of A B C . ABC.

Find the area of A B C . A'B'C'.


The answer is 4.

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1 solution

Steven Yuan
Jan 24, 2015

We call A B C A'B'C' the pedal triangle with respect to P . P. Let R R be the length of the radius of the circle and O O be its center. A theorem from Euler states that the area of A B C A'B'C' equals

R 2 O P 2 4 R 2 \left | \frac{R^2 - OP^2}{4R^2} \right |

times the area of A B C . ABC. Since R = 5 , O P = 3 , R = 5, OP = 3, and [ A B C ] = 25 , [ABC] = 25,

[ A B C ] = 5 2 3 2 4 ( 5 2 ) ( 25 ) = 4 25 ( 25 ) = 4 . \begin{aligned} [A'B'C'] &= \left | \frac{5^2 - 3^2}{4(5^2)} \right |(25) \\ &= \frac{4}{25}(25) \\ &= \boxed{4}. \end{aligned}

Can you please explain a proof of this theorem.

Prayas Rautray - 3 years, 8 months ago

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