Difference between revisions of "1967 AHSME Problems/Problem 40"
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== See also == | == See also == | ||
− | {{AHSME box|year=1967|num-b=39| | + | {{AHSME 40p box|year=1967|num-b=39|after=Last Problem}} |
[[Category: Intermediate Geometry Problems]] | [[Category: Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:31, 16 August 2023
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) | ||
Preceded by Problem 39 |
Followed by Last Problem | |
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 | ||
All AHSME Problems and Solutions |
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