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

(→‎Solution 1)
(→‎Solution 1)
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Moreover, we have <math>\angle PBA = \angle PCB = 45^\circ-\theta,</math> as shown below:
 
Moreover, we have <math>\angle PBA = \angle PCB = 45^\circ-\theta,</math> 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(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);
 +
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("$x$",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("$\sqrt{x^2-100}$",midpoint(P--C),dir(225),blue);
 +
label("$\theta$",A,9*dir(76),red);
 +
label("$\theta$",C,9*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);
 +
</asy>
 
Note that <math>\triangle PAB \sim \triangle PBC</math> by the AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> or <cmath>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}} = \frac{x}{x\sqrt2} = \frac{1}{\sqrt2}.</cmath>
 
Note that <math>\triangle PAB \sim \triangle PBC</math> by the AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> or <cmath>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}} = \frac{x}{x\sqrt2} = \frac{1}{\sqrt2}.</cmath>
 
From <math>\frac{10}{PB} = \frac{1}{\sqrt2},</math> we get <math>PB=10\sqrt2.</math> It follows that from <math>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}},</math> we get <math>x^2=500.</math>
 
From <math>\frac{10}{PB} = \frac{1}{\sqrt2},</math> we get <math>PB=10\sqrt2.</math> It follows that from <math>\frac{10}{PB} = \frac{PB}{\sqrt{x^2-100}},</math> we get <math>x^2=500.</math>

Revision as of 14:27, 17 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),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(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); 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("$x$",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("$\sqrt{x^2-100}$",midpoint(P--C),dir(225),blue); label("$\theta$",A,9*dir(76),red); label("$\theta$",C,9*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); [/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

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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All AIME Problems and Solutions

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