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

(→‎Solution)
(→‎Solution 1)
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== Solution 1==
 
== Solution 1==
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Since the triangle is a right isosceles triangle, angles B and C are <math>45</math>
+
Since the triangle is a right isosceles triangle, angles B and C are <math>45^\circ</math>
  
 
Let the common angle be <math>\theta</math>
 
Let the common angle be <math>\theta</math>
  
−
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 angle PAC is <math>90^\circ-\theta</math>, thus angle APC is <math>90^\circ</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:  
+
Note that ABP is <math>45^\circ-\theta</math>, so from law of sines we have:  
−
<cmath>\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45-\theta)}</cmath>
+
<cmath>\frac{10}{\sin\theta \cdot \frac{\sqrt{2}}{2}}=\frac{10}{\sin(45^\circ-\theta)}</cmath>
  
 
Dividing by 10 and multiplying across yields:
 
Dividing by 10 and multiplying across yields:
−
<cmath>\sqrt{2}\sin(45-\theta)=\sin\theta</cmath>
+
<cmath>\sqrt{2}\sin(45^\circ-\theta)=\sin\theta</cmath>
  
 
From here use the sin subtraction formula, and solve for <math>\sin\theta</math>
 
From here use the sin subtraction formula, and solve for <math>\sin\theta</math>

Revision as of 16:53, 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 $45^\circ$

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

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