Difference between revisions of "2023 AIME II Problems/Problem 3"
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== Solution 2== | == Solution 2== | ||
− | Since the triangle is a right isosceles triangle, | + | Since the triangle is a right isosceles triangle, <math>\angle B = \angle C = 45^\circ</math>. |
Do some angle chasing yielding: | Do some angle chasing yielding: | ||
+ | * <math>\angle APB = \angle BPC = 135^\circ</math> | ||
− | + | * <math>\angle APC=90^\circ</math> | |
+ | We have <math>AC=\frac{10}{\sin\theta}</math> since <math>\triangle APC</math> is a right triangle. Since <math>\triangle ABC</math> is a <math>45^\circ</math>-<math>45^\circ</math>-<math>90^\circ</math> triangle, <math>AB=\frac{10}{\sin\theta}</math>, and <math>BC=\frac{10\sqrt{2}}{\sin\theta}</math>. | ||
− | + | Note that <math>\triangle APB \sim \triangle BPC</math> by a factor of <math>\sqrt{2}</math>. Thus, <math>BP = 10\sqrt{2}</math>, and <math>PC = 20</math>. | |
+ | From Pythagorean theorem, <math>AC=10\sqrt{5}</math> so the area of <math>\triangle ABC</math> is <math>\frac{(10\sqrt{5})^2}{2}=\boxed{250}</math>. | ||
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~SAHANWIJETUNGA | ~SAHANWIJETUNGA | ||
Revision as of 19:23, 16 February 2023
Contents
Problem
Let be an isosceles triangle with
There exists a point
inside
such that
and
Find the area of
Diagram
~MRENTHUSIASM
Solution 1
Since the triangle is a right isosceles triangle, .
Let the common angle be . Note that
, thus
. From there, we know that
.
Note that , so from law of sines we have
Dividing by
and multiplying across yields
From here use the sine subtraction formula, and solve for
:
Substitute this to find that
, thus the area is
.
~SAHANWIJETUNGA
Solution 2
Since the triangle is a right isosceles triangle, .
Do some angle chasing yielding:
We have since
is a right triangle. Since
is a
-
-
triangle,
, and
.
Note that by a factor of
. Thus,
, and
.
From Pythagorean theorem, so the area of
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.