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

(Created page with "== Problem == Let <math>\triangle ABC</math> be an isosceles triangle with <math>\angle A = 90^\circ.</math> There exists a point <math>P</math> inside <math>\triangle ABC</m...")
 
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Let <math>\triangle ABC</math> be an isosceles triangle with <math>\angle A = 90^\circ.</math> There exists a point <math>P</math> inside <math>\triangle ABC</math> such that <math>\angle PAB = \angle PBC = \angle PCA</math> and <math>AP = 10.</math> Find the area of <math>\triangle ABC.</math>
 
Let <math>\triangle ABC</math> be an isosceles triangle with <math>\angle A = 90^\circ.</math> There exists a point <math>P</math> inside <math>\triangle ABC</math> such that <math>\angle PAB = \angle PBC = \angle PCA</math> and <math>AP = 10.</math> Find the area of <math>\triangle ABC.</math>
  
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== Solution ==
+
== Solution 1==
 +
Since the triangle is a right isosceles triangle, angles B and C are <math>45</math>
 +
 
 +
Let the common angle be <math>\theta</math>
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 +
Note that angle PAC is <math>90-\theta</math>, thus angle APC is <math>90</math>. From there, we know that AC is <math>\frac{10}{\sin\theta}</math>
 +
 
 +
Note that ABP is <math>45-\theta</math>, so from law of sines we have:
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<cmath>\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45-\theta)}</cmath>
 +
 
 +
Dividing by 10 and multiplying across yields:
 +
<cmath>\sqrt{2}\sin(45-\theta)=\sin\theta</cmath>
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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>
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<cmath>\sin\theta=\frac{1}{\sqrt{5}}</cmath>
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 +
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>
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~SAHANWIJETUNGA
  
 
== 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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:52, 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 $\triangle ABC.$

Solution 1

Since the triangle is a right isosceles triangle, angles B and C are $45$

Let the common angle be $\theta$

Note that angle PAC is $90-\theta$, thus angle APC is $90$. From there, we know that AC is $\frac{10}{\sin\theta}$

Note that ABP is $45-\theta$, so from law of sines we have: \[\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45-\theta)}\]

Dividing by 10 and multiplying across yields: \[\sqrt{2}\sin(45-\theta)=\sin\theta\]

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

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