Difference between revisions of "2023 AIME II Problems/Problem 3"
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<cmath>[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot [APC] = \boxed{250}.</cmath> | <cmath>[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot [APC] = \boxed{250}.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Denote <math>\angle PCA = \theta</math>. Then, by trig Ceva's: | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\ | ||
+ | \sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\ | ||
+ | 2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\theta)\right)^2 \\ | ||
+ | 2\sin^2(\theta) &= \cot(\theta) \cdot \left(1 - 2\sin(\theta)\cos(\theta)\right) \\ | ||
+ | 2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\ | ||
+ | \cot(\theta) &= 2 \\ | ||
+ | \sin(\theta) &= \frac{\sqrt{5}}{5} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Note that <math>\angle APC</math> is a right angle. Therefore: | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | \sin(\theta) &= \frac{AP}{AC} \\ | ||
+ | AC &= \frac{10}{\frac{\sqrt{5}}{5}} \\ | ||
+ | &= 10\sqrt{5} \\ | ||
+ | |ABC| &= \frac{AC^2}{2} \\ | ||
+ | &= \boxed{250} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | ~ ConcaveTriangle | ||
==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== |
Revision as of 05:48, 13 July 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
This solution refers to the Diagram section.
Let and
from which
and
By the Pythagorean Theorem on right
we have
Moreover, we have as shown below:
Note that
by the AA Similarity. The ratio of similitude is
or
From
we get
It follows that from
we get
Finally, the area of is
~s214425
~MRENTHUSIASM
Solution 2
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 3
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 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
Solution 5
Denote the area of by
As in previous solutions, we see that
with ratio
vladimir.shelomovskii@gmail.com, vvsss
Solution 6
Denote . Then, by trig Ceva's:
Note that is a right angle. Therefore:
~ ConcaveTriangle
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=APSUN-9Z_AU
Video Solution 2 by Piboy
https://www.youtube.com/watch?v=-WUhMmdXCxU&t=26s&ab_channel=Piboy
Video Solution by The Power of Logic(#3 and #4)
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.