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

Revision as of 16:49, 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, 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

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

@@ Line 57: / Line 57: @@
 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
+== Solution 2==
+Since the triangle is a right isosceles triangle, angles B and C are <math>45^\circ</math>
+Do some angle chasing yielding:
+- APB=BPC=<math>135^\circ</math>
+- APC=<math>90^\circ</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

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