1967 AHSME Problems/Problem 40
Problem
Located inside equilateral triangle is a point such that , , and . To the nearest integer the area of triangle is:
Solution 1
Notice that That makes us want to construct a right triangle.
Rotate about A. Note that , so
Therefore, is equilateral, so , which means
Let Notice that and
Applying the Law of Cosines to (remembering ):
We want to find the area of , which is
~pfalcon
Solution 2 (Magic Formula)
Fun formula: Given a point whose distances from the vertices of an equilateral triangle are , , and , the side length of the triangle is:
Given that the area of an equilateral triangle is , the answer is:
is not a choice, therefore the answer is .
(Note that the answer is actually the solution for when point is exterior to .)
~proloto
Solution 3
Rotate and CCW around , becoming and . Rotate and CCW around , becoming and . Rotate and CCW around , becoming and :
Notice that since , , and , then
Now the area of the big hexagon is easy to compute since it's comprised of 3 equilateral triangle and 3 right triangles:
~proloto
See also
1967 AHSC (Problems • Answer Key • Resources) | ||
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