Maths Olympiad Prep

Track / Stage 3 / 12 of 260 #12 of 1964

Problem 12

AMC 10/12, early questions
Geometry Difficulty 3.0 Find the answer

A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:
(A) Least when the point is the center of gravity of the triangle\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}(B) Greater than the altitude of the triangle\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle}(C) Equal to the altitude of the triangle\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}(D) One-half the sum of the sides of the triangle\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle}(E) Greatest when the point is the center of gravity\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity}

Multiple choice: answer with the letter of the option you want.

Official solution

Begin by drawing the triangle, the point, the altitudes from the point to the sides, and the segments connecting the point to the vertices. Let the triangle be ABCABC with AB=BC=AC=sAB=BC=AC=s. We will call the aforementioned point PP. Call altitude from PP to BCBC PA′PA'. Similarly, we will name the other two altitudes PB′PB' and PC′PC'. We can see that
12sPA′+12sPB′+12sPC′=12sh\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh
Where h is the altitude. Multiplying both sides by 22 and dividing both sides by ss gives us
PA′+PB′+PC′=hPA'+PB'+PC'=h
The answer is (C)\textbf{(C)}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.