Difference between revisions of "2023 AIME II Problems/Problem 3"
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Notice that in triangle <math>PBC</math>, <math>\angle PBC + 45-\angle PCA = 45^\circ</math>, so <math>\angle BPC = 135^\circ</math>. Similar logic shows <math>\angle APC = 135^\circ</math>. | Notice that in triangle <math>PBC</math>, <math>\angle PBC + 45-\angle PCA = 45^\circ</math>, so <math>\angle BPC = 135^\circ</math>. Similar logic shows <math>\angle APC = 135^\circ</math>. | ||
− | Now, we see that <math>\triangle APB \sim \triangle BPC</math> with ratio <math>1:\sqrt{2}</math> (as ABC is a 45 | + | Now, we see that <math>\triangle APB \sim \triangle BPC</math> with ratio <math>1:\sqrt{2}</math> (as <math>\triangle ABC</math> is a <math>45^\circ</math>-<math>45^\circ</math>-<math>90^\circ</math> triangle). Hence, <math>\overline{PB}=10\sqrt{2}</math>. We use the Law of Cosines to find <math>AB</math>. |
− | <cmath>AB^2=BP^2+AP^2-2ab\cos(APB) | + | <cmath>\begin{align*} |
− | + | AB^2&=BP^2+AP^2-2ab\cos(APB) \\ | |
− | + | &=100+200-2(10)(10\sqrt{2}\cos(135^\circ)) \\ | |
− | + | &=300+200\sqrt{2}\frac{1}{\sqrt{2}} \\ | |
+ | AB^2&=500. | ||
+ | \end{align*}</cmath> | ||
Since <math>\triangle ABC</math> is a right triangle, the area is <math>\frac{AB^2}{2}=\frac{500}{2}=\boxed{250}</math>. | Since <math>\triangle ABC</math> is a right triangle, the area is <math>\frac{AB^2}{2}=\frac{500}{2}=\boxed{250}</math>. | ||
Revision as of 03:57, 17 February 2023
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
Solution 3
Let From that, we can get
and
From that, we can also know that
is a right triangle.
With this information, we can deduce that Due to properties of
triangles, we know that
This means that
is
scaled by a factor of
Using similar triangles ratios that
we first find that
then that
Now using the Pythagorean Theorem on right we find that the length of the legs of this triangle are
Using that to solve for the area, we get our answer:
~s214425
Solution 4
Since the triangle is a right isosceles triangle, .
Notice that in triangle ,
, so
. Similar logic shows
.
Now, we see that with ratio
(as
is a
-
-
triangle). Hence,
. We use the Law of Cosines to find
.
Since
is a right triangle, the area is
.
~Kiran
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.