Difference between revisions of "2023 AIME II Problems/Problem 3"
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Let <math>\triangle ABC</math> be an isosceles triangle with <math>\angle A = 90^\circ.</math> There exists a point <math>P</math> inside <math>\triangle ABC</math> such that <math>\angle PAB = \angle PBC = \angle PCA</math> and <math>AP = 10.</math> Find the area of <math>\triangle ABC.</math> | Let <math>\triangle ABC</math> be an isosceles triangle with <math>\angle A = 90^\circ.</math> There exists a point <math>P</math> inside <math>\triangle ABC</math> such that <math>\angle PAB = \angle PBC = \angle PCA</math> and <math>AP = 10.</math> Find the area of <math>\triangle ABC.</math> | ||
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== Solution 1== | == Solution 1== |
Revision as of 16:58, 16 February 2023
Contents
Problem
Let be an isosceles triangle with
There exists a point
inside
such that
and
Find the area of
Diagram
Solution 1
Since the triangle is a right isosceles triangle, angles B and C are
Let the common angle be
Note that angle PAC is , thus angle APC is
. From there, we know that AC is
Note that ABP is , so from law of sines we have:
Dividing by 10 and multiplying across yields:
From here use the sin subtraction formula, and solve for
Substitute this to find that AC=, thus the area is
~SAHANWIJETUNGA
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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