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 6)
Line 164: Line 164:
 
2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\
 
2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\
 
\cot(\theta) &= 2 \\
 
\cot(\theta) &= 2 \\
−
\sin(\theta) &= \frac{\sqrt{5}}{5}.
+
\sin(\theta) &= \frac{\sqrt{5}}{5}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
 +
 
Note that <math>\angle APC</math> is a right angle. Therefore:  
 
Note that <math>\angle APC</math> is a right angle. Therefore:  
 +
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\sin(\theta) &= \frac{AP}{AC} \\
 
\sin(\theta) &= \frac{AP}{AC} \\
Line 172: Line 174:
 
&= 10\sqrt{5} \\
 
&= 10\sqrt{5} \\
 
|ABC| &= \frac{AC^2}{2} \\
 
|ABC| &= \frac{AC^2}{2} \\
−
&= \boxed{250}.
+
&= \boxed{250}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
~ConcaveTriangle
+
 
 +
~ ConcaveTriangle
  
 
==Video Solution 1 by SpreadTheMathLove==
 
==Video Solution 1 by SpreadTheMathLove==

Revision as of 21:00, 13 July 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),blue); [/asy] ~MRENTHUSIASM

Solution 1

This solution refers to the Diagram section.

Let $AB=AC=x$ and $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $BC=x\sqrt2, \angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$ By the Pythagorean Theorem on right $\triangle APC,$ we have $PC=\sqrt{x^2-100}.$

Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(250); 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(rightanglemark(A,P,C,10),red); draw(anglemark(P,A,B,25),red); draw(anglemark(P,B,C,25),red); draw(anglemark(P,C,A,25),red); draw(anglemark(A,B,P,25),green); draw(anglemark(B,C,P,25),green); draw(anglemark(C,A,P,25),green); 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),blue); label("$\sqrt{x^2-100}$",midpoint(P--C),dir(225),blue); label("$x$",midpoint(A--B),1.5*W,blue); label("$x$",midpoint(A--C),1.5*S,blue); label("$x\sqrt2$",midpoint(B--C),1.5*NE,blue); label("$\theta$",A,9.5*dir(76),red); label("$\theta$",C,9.5*dir(168),red); label("$\theta$",B,9*dir(305),red); label("$45^\circ-\theta$",B,6*dir(235),green); label("$45^\circ-\theta$",C,6*dir(85),green); label("$90^\circ-\theta$",A,2*dir(-40),green); [/asy] Note that $\triangle PAB \sim \triangle PBC$ by the AA Similarity. The ratio of similitude is $\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},$ or \[\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}} = \frac{x}{x\sqrt2} = \frac{1}{\sqrt2}.\] From $\frac{10}{PB} = \frac{1}{\sqrt2},$ we get $PB=10\sqrt2.$ It follows that from $\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}},$ we get $x^2=500.$

Finally, the area of $\triangle ABC$ is \[\frac12\cdot AB\cdot AC = \frac12\cdot x^2 = \boxed{250}.\]

~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

Solution 5

Denote the area of $X$ by $[X].$ As in previous solutions, we see that $\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB$ with ratio $k = \sqrt{2}\implies$ \[\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.\] \[[BPC] = k^2 [APB] = 2 [APB].\] \[AB = BC, \angle PCA = \angle PAB \implies \frac {[APC]}{[APB]} = \frac {PC}{PA} = 2 \implies\] \[[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot [APC] = \boxed{250}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution 6

Denote $\angle PCA = \theta$. Then, by trig Ceva's: \begin{align*} \frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\ \sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\ 2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\theta)\right)^2 \\ 2\sin^2(\theta) &= \cot(\theta) \cdot \left(1 - 2\sin(\theta)\cos(\theta)\right) \\ 2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\ \cot(\theta) &= 2 \\ \sin(\theta) &= \frac{\sqrt{5}}{5} \end{align*}

Note that $\angle APC$ is a right angle. Therefore:

\begin{align*} \sin(\theta) &= \frac{AP}{AC} \\ AC &= \frac{10}{\frac{\sqrt{5}}{5}} \\ &= 10\sqrt{5} \\ |ABC| &= \frac{AC^2}{2} \\ &= \boxed{250} \end{align*}

~ ConcaveTriangle

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)

https://youtu.be/dS9K1o4gCA0

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=195681"