Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AIME II Problems/Problem 3"

(→‎Solution 1)
Line 2: Line 2:
  
 
Let <math>\triangle ABC</math> be an isosceles triangle with <math>\angle A = 90^\circ.</math> There exists a point <math>P</math> inside <math>\triangle ABC</math> such that <math>\angle PAB = \angle PBC = \angle PCA</math> and <math>AP = 10.</math> Find the area of <math>\triangle ABC.</math>
 
Let <math>\triangle ABC</math> be an isosceles triangle with <math>\angle A = 90^\circ.</math> There exists a point <math>P</math> inside <math>\triangle ABC</math> such that <math>\angle PAB = \angle PBC = \angle PCA</math> and <math>AP = 10.</math> Find the area of <math>\triangle ABC.</math>
 +
 +
== Diagram ==
  
 
== Solution 1==
 
== Solution 1==

Revision as of 15:58, 16 February 2023

Problem

Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$

Diagram

Solution 1

Since the triangle is a right isosceles triangle, angles B and C are $45^\circ$

Let the common angle be $\theta$

Note that angle PAC is $90^\circ-\theta$, thus angle APC is $90^\circ$. From there, we know that AC is $\frac{10}{\sin\theta}$

Note that ABP is $45^\circ-\theta$, so from law of sines we have: \[\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45^\circ-\theta)}\]

Dividing by 10 and multiplying across yields: \[\sqrt{2}\sin(45^\circ-\theta)=\sin\theta\]

From here use the sin subtraction formula, and solve for $\sin\theta$

\[\cos\theta-\sin\theta=\sin\theta\] \[2\sin\theta=\cos\theta\] \[4\sin^2\theta=cos^2\theta\] \[4\sin^2\theta=1-\sin^2\theta\] \[5\sin^2\theta=1\] \[\sin\theta=\frac{1}{\sqrt{5}}\]

Substitute this to find that AC=$10\sqrt{5}$, thus the area is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$ ~SAHANWIJETUNGA

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_II_Problems/Problem_3&oldid=189694"