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

m (→‎Diagram)
(→‎Solution 8)
 
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draw(A--B--C--cycle^^P--A^^P--B^^P--C);
 
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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label("$10$",midpoint(A--P),dir(-30),blue);
 
</asy>
 
</asy>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
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==Solution 1==
 
==Solution 1==
  
−
Let <math>AB=AC=x</math> and <math>\angle PAB = \angle PBC = \angle PCA = \theta,</math> from which <math>BC=x\sqrt2, \angle PAC = 90^\circ-\theta,</math> and <math>\angle APC = 90^\circ.</math> By the Pythagorean Theorem on right <math>\triangle APC,</math> we have <math>PC=\sqrt{x^2-100}.</math>
+
This solution refers to the <b>Diagram</b> section.
 +
 
 +
Let <math>\angle PAB = \angle PBC = \angle PCA = \theta,</math> from which <math>\angle PAC = 90^\circ-\theta,</math> and <math>\angle APC = 90^\circ.</math>
  
 
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));
  
−
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>
+
markscalefactor=0.125;
−
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>
+
draw(rightanglemark(B,A,C,10),red);
 +
draw(rightanglemark(A,P,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);
 +
draw(anglemark(C,A,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);
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label("$10$",midpoint(A--P),dir(-30),blue);
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label("$\theta$",A,9.5*dir(76),red);
 +
label("$\theta$",C,9.5*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);
 +
label("$90^\circ-\theta$",A,2*dir(-40),green);
 +
</asy>
 +
Note that <math>\triangle PAB \sim \triangle PBC</math> by AA Similarity. The ratio of similitude is <math>\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},</math> so <math>\frac{10}{PB} = \frac{1}{\sqrt2}</math> and thus <math>PB=10\sqrt2.</math> Similarly, we can figure out that <math>PC=20</math>.
  
−
Finally, the area of <math>\triangle ABC</math> is <cmath>\frac12\cdot AB\cdot AC = \frac12\cdot x^2 = \boxed{250}.</cmath>
+
Finally, <math>AC=\sqrt{10^2+20^2}=10\sqrt{5}</math>, so the area of <math>\triangle ABC</math> is <cmath>\frac12\cdot AB\cdot AC = \frac12\cdot (10\sqrt{5})^2 = \boxed{250}.</cmath>
  
 
~s214425
 
~s214425
Line 103: Line 141:
  
 
~Kiran
 
~Kiran
 +
 +
==Solution 5==
 +
Denote the area of <math>X</math> by <math>[X].</math> As in previous solutions, we see that <math>\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB</math> with ratio <math>k = \sqrt{2}\implies</math> <cmath>\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.</cmath>
 +
<cmath>[BPC] = k^2 [APB] = 2 [APB].</cmath>
 +
<cmath>AB = BC, \angle PCA = \angle PAB \implies \frac {[APC]}{[APB]} = \frac {PC}{PA} = 2 \implies</cmath>
 +
<cmath>[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot  [APC] = \boxed{250}.</cmath>
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 +
==Solution 6==
 +
 +
Denote <math>\angle PCA = \theta</math>. Then, by trig Ceva's:
 +
<cmath>\begin{align*}
 +
\frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\
 +
\sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\
 +
2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\theta)\right)^2 \\
 +
2\sin^2(\theta) &= \cot(\theta) \cdot \left(1 - 2\sin(\theta)\cos(\theta)\right) \\
 +
2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\
 +
\cot(\theta) &= 2 \\
 +
\sin(\theta) &= \frac{\sqrt{5}}{5}.
 +
\end{align*}</cmath>
 +
Note that <math>\angle APC</math> is a right angle. Therefore:
 +
<cmath>\begin{align*}
 +
\sin(\theta) &= \frac{AP}{AC} \\
 +
AC &= \frac{10}{\frac{\sqrt{5}}{5}} \\
 +
&= 10\sqrt{5} \\
 +
|ABC| &= \frac{AC^2}{2} \\
 +
&= \boxed{250}.
 +
\end{align*}</cmath>
 +
~ConcaveTriangle
 +
 +
==Solution 7==
 +
 +
Notice that point <math>P</math> is one of the two Brocard Points of <math>\triangle ABC</math>. (The angle equalities given in the problem are equivalent to the definition of a Brocard point.) By the Brocard point formula,
 +
<cmath>\begin{align*} \cot(\phi) = \cot(A)+\cot(B)+\cot(C)  \end{align*}</cmath>
 +
, where <math>\phi</math> is equal to <math>\angle PAB</math>.(This is also called the Brocard angle of triangle ABC). Because the triangle is an isosceles right triangle, the cotangents are easy to compute:
 +
<cmath>\begin{align*}\cot(\phi) = 0 + 1 + 1 \\ \cot(\phi) = 2\end{align*}</cmath>
 +
By definition, <math>\cot(\phi) = \frac{\cos(\phi)}{\sin(\phi)}</math>. By the Pythagorean identity, <math>\cos(\phi)=\frac{2\sqrt{5}}{5}</math> and <math>\sin(\phi) = \cos(\phi)=\frac{\sqrt{5}}{5}</math>. Consider triangle <math>APB</math>. By the problem condition, <math>\angle PBA = 45-\phi</math>, so <math>\angle BPA = 135^{\circ}</math>
 +
<cmath>\begin{align*}\sin{45-\theta} = \sin{45}\cos{\phi}-\cos{45}\sin{\phi} = \frac{\sqrt{10}}{10}\end{align*}</cmath>
 +
Now, we can use the Law of Sines.
 +
<cmath>\begin{align*}
 +
\frac{AP}{\sin{45-\theta}}&=\frac{AB}{\sin{135}} \\
 +
10 \sqrt{10} &= \sqrt{2} AB \\
 +
AB &= 10 \sqrt{5}
 +
\end{align*}</cmath>
 +
Therefore, the answer is <cmath>[ABC] = \frac{1}{2} (AB)^2 = \boxed{250}.</cmath>
 +
~ewei12
 +
 +
==Solution 8==
 +
 +
Notice that <math>\angle APC = 90^{\circ}</math>, <math>\angle BPA = 135^{\circ}</math>, and <math>\angle CPB = 135^{\circ}</math> (from Solution 4). Now let <math>a = 0</math> and <math>p = 10i</math>. Then by the angle restrictions <math>c = m + 10i</math> and <math>b = -n + (n+10)i</math> for some <math>m, n</math>. Since <math>\angle BAC = 90^{\circ}</math>, <math>ci = b</math>, or <math>(m + 10i)i = -n + (n+10)i</math>. Therefore <math>n = 10</math>, <math>AB = \sqrt{10^2 + 20^2} = \sqrt{500}</math>, and <math>[ABC] = \frac{1}{2} AB^2 = \boxed{250}</math>.
 +
 +
~aliz
 +
 +
==Video Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=APSUN-9Z_AU
 +
 +
==Video Solution 2 by Piboy==
 +
https://www.youtube.com/watch?v=-WUhMmdXCxU&t=26s&ab_channel=Piboy
 +
 +
==Video Solution by The Power of Logic(#3 and #4)==
 +
https://youtu.be/dS9K1o4gCA0
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2023|num-b=2|num-a=4|n=II}}
 
{{AIME box|year=2023|num-b=2|num-a=4|n=II}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:42, 27 June 2024

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 $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$

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(rightanglemark(A,P,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); draw(anglemark(C,A,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("$\theta$",A,9.5*dir(76),red); label("$\theta$",C,9.5*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); label("$90^\circ-\theta$",A,2*dir(-40),green); [/asy] Note that $\triangle PAB \sim \triangle PBC$ by AA Similarity. The ratio of similitude is $\frac{PA}{PB} = \frac{PB}{PC} = \frac{AB}{BC},$ so $\frac{10}{PB} = \frac{1}{\sqrt2}$ and thus $PB=10\sqrt2.$ Similarly, we can figure out that $PC=20$.

Finally, $AC=\sqrt{10^2+20^2}=10\sqrt{5}$, so the area of $\triangle ABC$ is \[\frac12\cdot AB\cdot AC = \frac12\cdot (10\sqrt{5})^2 = \boxed{250}.\]

~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

Solution 5

Denote the area of $X$ by $[X].$ As in previous solutions, we see that $\angle APC = 90 ^\circ, \triangle BPC \sim \triangle APB$ with ratio $k = \sqrt{2}\implies$ \[\frac {PC}{PB} = \frac {PB}{PA} = k \implies PC = k^2 \cdot AP = 20 \implies [APC] = \frac {AP \cdot PC}{2} = 100.\] \[[BPC] = k^2 [APB] = 2 [APB].\] \[AB = BC, \angle PCA = \angle PAB \implies \frac {[APC]}{[APB]} = \frac {PC}{PA} = 2 \implies\] \[[ABC] = [APB] + [APC] + [BPC] = [APC] \cdot (\frac {1}{2} + 1 + 2 \cdot \frac {1}{2}) = \frac {5}{2} \cdot [APC] = \boxed{250}.\] vladimir.shelomovskii@gmail.com, vvsss

Solution 6

Denote $\angle PCA = \theta$. Then, by trig Ceva's: \begin{align*} \frac{\sin^3(\theta)}{\sin(90-\theta) \cdot \left(\sin(45-\theta)\right)^2} &= 1 \\ \sin^3(\theta) &= \cos(\theta) \cdot \left(\sin(45) \cos(\theta) - \cos(45) \sin(\theta)\right)^2 \\ 2\sin^3(\theta) &= \cos(\theta) \cdot \left(\cos(\theta) - \sin(\theta)\right)^2 \\ 2\sin^2(\theta) &= \cot(\theta) \cdot \left(1 - 2\sin(\theta)\cos(\theta)\right) \\ 2\sin^2(\theta) &= \cot(\theta) - 2\cos^2(\theta) \\ \cot(\theta) &= 2 \\ \sin(\theta) &= \frac{\sqrt{5}}{5}. \end{align*} Note that $\angle APC$ is a right angle. Therefore: \begin{align*} \sin(\theta) &= \frac{AP}{AC} \\ AC &= \frac{10}{\frac{\sqrt{5}}{5}} \\ &= 10\sqrt{5} \\ |ABC| &= \frac{AC^2}{2} \\ &= \boxed{250}. \end{align*} ~ConcaveTriangle

Solution 7

Notice that point $P$ is one of the two Brocard Points of $\triangle ABC$. (The angle equalities given in the problem are equivalent to the definition of a Brocard point.) By the Brocard point formula, \begin{align*} \cot(\phi) = \cot(A)+\cot(B)+\cot(C) \end{align*} , where $\phi$ is equal to $\angle PAB$.(This is also called the Brocard angle of triangle ABC). Because the triangle is an isosceles right triangle, the cotangents are easy to compute: \begin{align*}\cot(\phi) = 0 + 1 + 1 \\ \cot(\phi) = 2\end{align*} By definition, $\cot(\phi) = \frac{\cos(\phi)}{\sin(\phi)}$. By the Pythagorean identity, $\cos(\phi)=\frac{2\sqrt{5}}{5}$ and $\sin(\phi) = \cos(\phi)=\frac{\sqrt{5}}{5}$. Consider triangle $APB$. By the problem condition, $\angle PBA = 45-\phi$, so $\angle BPA = 135^{\circ}$ \begin{align*}\sin{45-\theta} = \sin{45}\cos{\phi}-\cos{45}\sin{\phi} = \frac{\sqrt{10}}{10}\end{align*} Now, we can use the Law of Sines. \begin{align*} \frac{AP}{\sin{45-\theta}}&=\frac{AB}{\sin{135}} \\ 10 \sqrt{10} &= \sqrt{2} AB \\ AB &= 10 \sqrt{5} \end{align*} Therefore, the answer is \[[ABC] = \frac{1}{2} (AB)^2 = \boxed{250}.\] ~ewei12

Solution 8

Notice that $\angle APC = 90^{\circ}$, $\angle BPA = 135^{\circ}$, and $\angle CPB = 135^{\circ}$ (from Solution 4). Now let $a = 0$ and $p = 10i$. Then by the angle restrictions $c = m + 10i$ and $b = -n + (n+10)i$ for some $m, n$. Since $\angle BAC = 90^{\circ}$, $ci = b$, or $(m + 10i)i = -n + (n+10)i$. Therefore $n = 10$, $AB = \sqrt{10^2 + 20^2} = \sqrt{500}$, and $[ABC] = \frac{1}{2} AB^2 = \boxed{250}$.

~aliz

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=APSUN-9Z_AU

Video Solution 2 by Piboy

https://www.youtube.com/watch?v=-WUhMmdXCxU&t=26s&ab_channel=Piboy

Video Solution by The Power of Logic(#3 and #4)

https://youtu.be/dS9K1o4gCA0

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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