Difference between revisions of "2023 AIME II Problems/Problem 3"
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Moreover, we have <math>\angle PBA = \angle PCB = 45^\circ-\theta,</math> as shown below: | Moreover, we have <math>\angle PBA = \angle PCB = 45^\circ-\theta,</math> as shown below: | ||
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Note that <math>\triangle PAB \sim \triangle PBC</math> by the AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> or <cmath>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}} = \frac{x}{x\sqrt2} = \frac{1}{\sqrt2}.</cmath> | Note that <math>\triangle PAB \sim \triangle PBC</math> by the AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> or <cmath>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}} = \frac{x}{x\sqrt2} = \frac{1}{\sqrt2}.</cmath> | ||
From <math>\frac{10}{PB} = \frac{1}{\sqrt2},</math> we get <math>PB=10\sqrt2.</math> It follows that from <math>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}},</math> we get <math>x^2=500.</math> | From <math>\frac{10}{PB} = \frac{1}{\sqrt2},</math> we get <math>PB=10\sqrt2.</math> It follows that from <math>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}},</math> we get <math>x^2=500.</math> |
Revision as of 14:37, 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
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
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.