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 4)
(Made a solution with diagram. Coauthoring with s214425 to make the solution with diagram.)
Line 34: Line 34:
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
−
== Solution 1==
+
==Solution 1==
 +
 
 +
 
 +
~s214425
 +
 
 +
~MRENTHUSIASM
 +
 
 +
== Solution 2==
 
Since the triangle is a right isosceles triangle, <math>\angle B = \angle C = 45^\circ</math>.
 
Since the triangle is a right isosceles triangle, <math>\angle B = \angle C = 45^\circ</math>.
  
Line 56: Line 63:
 
~SAHANWIJETUNGA
 
~SAHANWIJETUNGA
  
−
== Solution 2==
+
== Solution 3==
 
Since the triangle is a right isosceles triangle, <math>\angle B = \angle C = 45^\circ</math>.
 
Since the triangle is a right isosceles triangle, <math>\angle B = \angle C = 45^\circ</math>.
  
Line 73: Line 80:
 
~SAHANWIJETUNGA
 
~SAHANWIJETUNGA
  
−
==Solution 3==
 
−
Let <math>\theta=m\angle PAB = m\angle PBC = m\angle PCA.</math> From that, we can get <math>m\angle PAC=90^{\circ}-\theta</math> and <math>m\angle PBA = m\angle PCB = 45^{\circ}-\theta.</math> From that, we can also know that <math>\triangle APC</math> is a right triangle.
 
−
 
−
With this information, we can deduce that <math>\triangle PAB\sim\triangle PBC.</math> Due to properties of <math>45-45-90</math> triangles, we know that <math>BC = \sqrt{2}\cdot AB.</math> This means that <math>\triangle PBC</math> is <math>\triangle PAB</math> scaled by a factor of <math>\sqrt{2}.</math> Using similar triangles ratios that <math>\frac{PB}{PA}=\frac{PC}{PB}=\frac{BC}{AB}=\sqrt{2},</math> we first find that <math>PB = 10\sqrt{2},</math> then that <math>PC=20.</math>
 
−
 
−
Now using the Pythagorean Theorem on right <math>\triangle APC,</math> we find that the length of the legs of this triangle are <math>10\sqrt{5}.</math>
 
−
Using that to solve for the area, we get our answer: <cmath>[ABC]=\frac{(10\sqrt{5})^2}{2}=\frac{500}{2}=\boxed{250}.</cmath>
 
−
 
−
~s214425
 
 
==Solution 4==
 
==Solution 4==
 
Since the triangle is a right isosceles triangle, <math>\angle B = \angle C = 45^\circ</math>.
 
Since the triangle is a right isosceles triangle, <math>\angle B = \angle C = 45^\circ</math>.

Revision as of 03:04, 17 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

[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, P; A = origin; B = (0,10*sqrt(5)); C = (10*sqrt(5),0); P = intersectionpoints(Circle(A,10),Circle(C,20))[0]; dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*SE,linewidth(4)); dot("$P$",P,1.5*NE,linewidth(4)); markscalefactor=0.125; draw(rightanglemark(B,A,C,10),red); draw(anglemark(P,A,B,25),red); draw(anglemark(P,B,C,25),red); draw(anglemark(P,C,A,25),red); add(pathticks(anglemark(P,A,B,25), n = 1, r = 0.1, s = 10, red)); add(pathticks(anglemark(P,B,C,25), n = 1, r = 0.1, s = 10, red)); add(pathticks(anglemark(P,C,A,25), n = 1, r = 0.1, s = 10, red)); draw(A--B--C--cycle^^P--A^^P--B^^P--C); label("$10$",midpoint(A--P),dir(-30),red); [/asy] ~MRENTHUSIASM

Solution 1

~s214425

~MRENTHUSIASM

Solution 2

Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$.

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

Note that $\angle ABP = 45^\circ-\theta$, 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 sine subtraction formula, and solve for $\sin\theta$: \begin{align*} \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}}. \end{align*} Substitute this to find that $AC=10\sqrt{5}$, thus the area is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$.

~SAHANWIJETUNGA

Solution 3

Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$.

Do some angle chasing yielding:

  • $\angle APB = \angle BPC = 135^\circ$
  • $\angle APC=90^\circ$

We have $AC=\frac{10}{\sin\theta}$ since $\triangle APC$ is a right triangle. Since $\triangle ABC$ is a $45^\circ$-$45^\circ$-$90^\circ$ triangle, $AB=\frac{10}{\sin\theta}$, and $BC=\frac{10\sqrt{2}}{\sin\theta}$.

Note that $\triangle APB \sim \triangle BPC$ by a factor of $\sqrt{2}$. Thus, $BP = 10\sqrt{2}$, and $PC = 20$.

From Pythagorean theorem, $AC=10\sqrt{5}$ so the area of $\triangle ABC$ is $\frac{(10\sqrt{5})^2}{2}=\boxed{250}$.

~SAHANWIJETUNGA

Solution 4

Since the triangle is a right isosceles triangle, $\angle B = \angle C = 45^\circ$.

Notice that in triangle $PBC$, $\angle PBC + 45-\angle PCA = 45^\circ$, so $\angle BPC = 135^\circ$. Similar logic shows $\angle APC = 135^\circ$.

Now, we see that $\triangle APB \sim \triangle BPC$ with ratio $1:\sqrt{2}$ (as $\triangle ABC$ is a $45^\circ$-$45^\circ$-$90^\circ$ triangle). Hence, $\overline{PB}=10\sqrt{2}$. We use the Law of Cosines to find $AB$. \begin{align*} AB^2&=BP^2+AP^2-2ab\cos(APB) \\ &=100+200-2(10)(10\sqrt{2}\cos(135^\circ)) \\ &=300+200\cdot\sqrt{2}\cdot\frac{1}{\sqrt{2}} \\ &=500. \end{align*} Since $\triangle ABC$ is a right triangle, the area is $\frac{AB^2}{2}=\frac{500}{2}=\boxed{250}$.

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_II_Problems/Problem_3&oldid=189830"