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Difference between revisions of "2023 AIME II Problems/Problem 3"
(→‎Diagram)
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== Diagram ==
 
== Diagram ==
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<asy>
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/* Made by MRENTHUSIASM */
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 +
size(200);
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pair A, B, C, P;
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A = origin;
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B = (0,10*sqrt(5));
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C = (10*sqrt(5),0);
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P = intersectionpoints(Circle(A,10),Circle(C,20))[0];
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dot("$A$",A,1.5*SW,linewidth(4));
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dot("$B$",B,1.5*NW,linewidth(4));
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dot("$C$",C,1.5*SE,linewidth(4));
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dot("$P$",P,1.5*NE,linewidth(4));
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markscalefactor=0.125;
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draw(rightanglemark(B,A,C,10),red);
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draw(anglemark(P,A,B,25),red);
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draw(anglemark(P,B,C,25),red);
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draw(anglemark(P,C,A,25),red);
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add(pathticks(anglemark(P,A,B,25), n = 1, r = 0.1, s = 10, red));
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add(pathticks(anglemark(P,B,C,25), n = 1, r = 0.1, s = 10, red));
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add(pathticks(anglemark(P,C,A,25), n = 1, r = 0.1, s = 10, red));
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draw(A--B--C--cycle^^P--A^^P--B^^P--C);
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label("$10$",midpoint(A--P),dir(-30),red);
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</asy>
 +
~MRENTHUSIASM
  
 
== Solution 1==
 
== Solution 1==

Revision as of 16:34, 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

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

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