2023 AIME II Problems/Problem 3
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 from which and
Moreover, we have as shown below: Note that by AA Similarity. The ratio of similitude is so and thus Similarly, we can figure out that .
Finally, , so 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
Solution 7
Notice that point is one of the two Brocard Points of . (The angle equalities given in the problem are equivalent to the definition of a Brocard point.) By the Brocard point formula, , where is equal to .(This is also called the Brocard angle of triangle ABC). Because the triangle is an isosceles right triangle, the cotangents are easy to compute: By definition, . By the Pythagorean identity, and . Consider triangle . By the problem condition, , so Now, we can use the Law of Sines. Therefore, the answer is ~ewei12
Solution 8
Notice that , , and (from Solution 4). Now let and . Then by the angle restrictions and for some . Since , , or . Therefore , , and .
~aliz
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.