1950 AHSME Problems/Problem 48

Problem

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:

$\textbf{(A)}\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ \textbf{(B)}\ \text{Greater than the altitude of the triangle} \qquad\\ \textbf{(C)}\ \text{Equal to the altitude of the triangle}\qquad\\ \textbf{(D)}\ \text{One-half the sum of the sides of the triangle} \qquad\\ \textbf{(E)}\ \text{Greatest when the point is the center of gravity}$

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 $ABC$ with $AB=BC=AC=s$ . We will call the aforementioned point $P$ . Call altitude from $P$ to $BC$ $PA'$ . Similarly, we will name the other two altitudes $PB'$ and $PC'$ . We can see that $\frac{1}{2}sPA'+\frac{1}{2}sPB'+\frac{1}{2}sPC'=\frac{1}{2}sh$ Where h is the altitude. Multiplying both sides by $2$ and dividing both sides by $s$ gives us $PA'+PB'+PC'=h$ The answer is $\textbf{C}$ .

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 47	Followed by Problem 49
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1950 AHSME Problems/Problem 48

Problem

Solution

See Also