1950 AMC 12 Problem 48

Attempt Problem 48 of the 1950 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1950 AMC 12 solutions, or check the answer key.

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48.

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:

Least when the point is the center of gravity of the triangle

Greater than the altitude of the triangle

Equal to the altitude of the triangle

One-half the sum of the sides of the triangle

Greatest when the point is the center of gravity

Answer: C
Concepts:Viviani’s Theoremequilateral trianglearea decomposition
Difficulty rating: 1920
Small Hint:

Join the interior point to all three vertices

Big Hint:

Add the areas of the three smaller triangles using the common side length as their bases

Solution:

Let the equilateral triangle have side length ss and altitude h,h, and let the three perpendicular distances be d1,d2,d3.d_1,d_2,d_3. Splitting the triangle at the interior point gives 12s(d1+d2+d3)=12sh. \frac12s(d_1+d_2+d_3)=\frac12sh. Therefore d1+d2+d3=h,d_1+d_2+d_3=h, independent of the selected point.

Thus, the correct answer is C.

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Problem 48 in Other Years

1951 AMC 12 · 1952 AMC 12 · 1953 AMC 12 · 1954 AMC 12 · 1955 AMC 12 · 1956 AMC 12 · 1957 AMC 12 · 1958 AMC 12 · 1959 AMC 12