Difference between revisions of "2023 AIME II Problems/Problem 3"
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~Kiran | ~Kiran | ||
+ | ==Solution 5== | ||
+ | Denote the area of <math>X</math> by <math>[X].</math> As in previous solutions, we see that <math>\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB</math> with ratio <math>k = \sqrt{2}\implies</math> <cmath>\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.</cmath> | ||
+ | <cmath>[BPC] = k^2 [APB] = 2 [APB].</cmath> | ||
+ | <cmath>AB = BC, \angle PCA = \angle PAB \implies \frac {[APC]}{[APB]} = \frac {PC}{PA} = 2 \implies</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''' | ||
==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== |
Revision as of 04:45, 5 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
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.