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2023 AIME II Problems/Problem 3: Difference between revisions

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== Solution 1==
== Solution 1==
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>.


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


Note that angle PAC is <math>90^\circ-\theta</math>, thus angle APC is <math>90^\circ</math>. From there, we know that AC is <math>\frac{10}{\sin\theta}</math>
Note that <math>\angle ABP = 45^\circ-\theta</math>, so from law of sines we have
<cmath>\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45^\circ-\theta)}.</cmath>
Dividing by <math>10</math> and multiplying across yields
<cmath>\sqrt{2}\sin(45^\circ-\theta)=\sin\theta.</cmath>
From here use the sine subtraction formula, and solve for <math>\sin\theta</math>:
<cmath>\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*}</cmath>
Substitute this to find that <math>AC=10\sqrt{5}</math>, thus the area is <math>\frac{(10\sqrt{5})^2}{2}=\boxed{250}</math>.


Note that ABP is <math>45^\circ-\theta</math>, so from law of sines we have:
<cmath>\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45^\circ-\theta)}</cmath>
Dividing by 10 and multiplying across yields:
<cmath>\sqrt{2}\sin(45^\circ-\theta)=\sin\theta</cmath>
From here use the sin subtraction formula, and solve for <math>\sin\theta</math>
<cmath>\cos\theta-\sin\theta=\sin\theta</cmath>
<cmath>2\sin\theta=\cos\theta</cmath>
<cmath>4\sin^2\theta=cos^2\theta</cmath>
<cmath>4\sin^2\theta=1-\sin^2\theta</cmath>
<cmath>5\sin^2\theta=1</cmath>
<cmath>\sin\theta=\frac{1}{\sqrt{5}}</cmath>
Substitute this to find that AC=<math>10\sqrt{5}</math>, thus the area is <math>\frac{(10\sqrt{5})^2}{2}=\boxed{250}</math>
~SAHANWIJETUNGA
~SAHANWIJETUNGA


== Solution 2==
== Solution 2==

Revision as of 18:16, 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, angles B and C are $45^\circ$

Do some angle chasing yielding:


- APB=BPC=$135^\circ$


- APC=$90^\circ$


AC=$\frac{10}{\sin\theta}$ due to APC being a right triangle. Since ABC is a 45-45-90 triangle, AB=$\frac{10}{\sin\theta}$, and BC=$\frac{10\sqrt{2}}{\sin\theta}$.


Note that triangle APB is similar to BPC, by a factor of $\sqrt{2}$. Thus, PC=$10\sqrt{2}$


From Pythagorean theorem, AC=$10\sqrt{5}$ so the area of 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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