Difference between revisions of "1987 AIME Problems/Problem 9"
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<cmath>4x = 132 \Longrightarrow x = \boxed{033}.</cmath> | <cmath>4x = 132 \Longrightarrow x = \boxed{033}.</cmath> | ||
=== Note === | === Note === | ||
− | This is the Fermat point of the triangle. | + | This is the [[Fermat point]] of the triangle. |
== See also == | == See also == |
Revision as of 07:52, 4 March 2021
Contents
Problem
Triangle has right angle at
, and contains a point
for which
,
, and
. Find
.
Solution
Let . Since
, each of them is equal to
. By the Law of Cosines applied to triangles
,
and
at their respective angles
, remembering that
, we have
Then by the Pythagorean Theorem, , so
and
Note
This is the Fermat point of the triangle.
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.