1967 AHSME Problems/Problem 40
Contents
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
Solution 4(Answer Choices, Approximation)
Let be the side length of Notice that by the triangle inequality. This means that This automatically rules out choices and Now, we will look at if the area is . By the equilateral triangle area formula, would equal This is very close to If and by the Pythagorean theorem and Pythagorean inequalities. Thus, needs to be and it probably cannot increase by more than by just adding to (more rigorous proof below) Thus, the only viable answer choice is
- In fact, for () still holds. As Thus, we know for sure that the nearest integer to the area cannot be
See also
1967 AHSC (Problems • Answer Key • Resources) | ||
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