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

(→‎Solution 1: Fixed notations in LaTeX.)
(→‎Solution 2)
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== Solution 2==
 
== Solution 2==
−
Since the triangle is a right isosceles triangle, angles B and C are <math>45^\circ</math>
+
Since the triangle is a right isosceles triangle, <math>\angle B = \angle C = 45^\circ</math>.
  
 
Do some angle chasing yielding:
 
Do some angle chasing yielding:
  
 +
* <math>\angle APB = \angle BPC = 135^\circ</math>
  
−
- APB=BPC=<math>135^\circ</math>
+
* <math>\angle APC=90^\circ</math>
  
 +
We have <math>AC=\frac{10}{\sin\theta}</math> since <math>\triangle APC</math> is a right triangle. Since <math>\triangle ABC</math> is a <math>45^\circ</math>-<math>45^\circ</math>-<math>90^\circ</math> triangle, <math>AB=\frac{10}{\sin\theta}</math>, and <math>BC=\frac{10\sqrt{2}}{\sin\theta}</math>.
  
−
- APC=<math>90^\circ</math>
+
Note that <math>\triangle APB \sim \triangle BPC</math> by a factor of <math>\sqrt{2}</math>. Thus, <math>BP = 10\sqrt{2}</math>, and <math>PC = 20</math>.
  
 +
From Pythagorean theorem, <math>AC=10\sqrt{5}</math> so the area of <math>\triangle ABC</math> is <math>\frac{(10\sqrt{5})^2}{2}=\boxed{250}</math>.
  
−
AC=<math>\frac{10}{\sin\theta}</math> due to APC being a right triangle. Since ABC is a 45-45-90 triangle, AB=<math>\frac{10}{\sin\theta}</math>, and BC=<math>\frac{10\sqrt{2}}{\sin\theta}</math>.
 
−
 
−
 
−
Note that triangle APB is similar to BPC, by a factor of <math>\sqrt{2}</math>. Thus, PC=<math>10\sqrt{2}</math>
 
−
 
−
 
−
From Pythagorean theorem, AC=<math>10\sqrt{5}</math> so the area of ABC is <math>\frac{(10\sqrt{5})^2}{2}=\boxed{250}</math>
 
 
~SAHANWIJETUNGA
 
~SAHANWIJETUNGA
  

Revision as of 18:23, 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

[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

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 2

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

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

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