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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"

(→‎Solution 3)
(Ordered the solution based on efficiency. Removed nonrigorous stuffs.)
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<math>\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi</math>
 
<math>\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi</math>
  
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==Solution 1 (Similar Triangles)==
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==Solution 1 (Cyclic Quadrilateral)==
  
 +
~MRENTHUSIASM ~kante314
 +
 +
==Solution 2 (Similar Triangles)==
 
<asy>
 
<asy>
−
 
 
import olympiad;
 
import olympiad;
 
unitsize(50);
 
unitsize(50);
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draw(rightanglemark(A,F,O));  
 
draw(rightanglemark(A,F,O));  
 
draw(rightanglemark(A,G,O));
 
draw(rightanglemark(A,G,O));
−
 
  
 
label("$O$",O,W);
 
label("$O$",O,W);
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add(pathticks(A--G,1,0.5,0,2));
 
add(pathticks(A--G,1,0.5,0,2));
 
add(pathticks(G--C,1,0.5,0,2));
 
add(pathticks(G--C,1,0.5,0,2));
−
 
−
 
−
 
 
</asy>
 
</asy>
−
 
−
 
 
Because circle <math>I</math> is tangent to <math>\overline{AB}</math> at <math>B, \angle{ABI} \cong 90^{\circ}</math>. Because O is the circumcenter of <math>\bigtriangleup ABC, \overline{OD}</math> is the perpendicular bisector of <math>\overline{AB}</math>, and <math>\angle{BAI} \cong \angle{DAO}</math>, so therefore <math>\bigtriangleup ADO \sim \bigtriangleup ABI</math> by AA similarity. Then we have <math>\frac{AD}{AB} = \frac{DO}{BI} \implies \frac{1}{2} = \frac{r}{5\sqrt{2}} \implies r = \frac{5\sqrt{2}}{2}</math>. We also know that <math>\overline{AD} = \frac{3\sqrt{6}}{2}</math> because of the perpendicular bisector, so the hypotenuse of <math>\bigtriangleup ADO = \sqrt{(\frac{5\sqrt{2}}{2})^2+(\frac{3\sqrt{6}}{2})^2} = \sqrt{\frac{25}{2}+\frac{27}{2}} = \sqrt{26}</math>.
 
Because circle <math>I</math> is tangent to <math>\overline{AB}</math> at <math>B, \angle{ABI} \cong 90^{\circ}</math>. Because O is the circumcenter of <math>\bigtriangleup ABC, \overline{OD}</math> is the perpendicular bisector of <math>\overline{AB}</math>, and <math>\angle{BAI} \cong \angle{DAO}</math>, so therefore <math>\bigtriangleup ADO \sim \bigtriangleup ABI</math> by AA similarity. Then we have <math>\frac{AD}{AB} = \frac{DO}{BI} \implies \frac{1}{2} = \frac{r}{5\sqrt{2}} \implies r = \frac{5\sqrt{2}}{2}</math>. We also know that <math>\overline{AD} = \frac{3\sqrt{6}}{2}</math> because of the perpendicular bisector, so the hypotenuse of <math>\bigtriangleup ADO = \sqrt{(\frac{5\sqrt{2}}{2})^2+(\frac{3\sqrt{6}}{2})^2} = \sqrt{\frac{25}{2}+\frac{27}{2}} = \sqrt{26}</math>.
 
This is the radius of the circumcircle of <math>\bigtriangleup ABC</math>, so the area of this circle is <math>26\pi = \boxed{C}</math>
 
This is the radius of the circumcircle of <math>\bigtriangleup ABC</math>, so the area of this circle is <math>26\pi = \boxed{C}</math>
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~KingRavi
 
~KingRavi
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==Solution 2 (Geometric Assumption)==
 
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Let the center of the first circle be <math>O.</math> By Pythagorean Theorem,
 
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<cmath>AO=\sqrt{(3\sqrt{6})^2+(5\sqrt{2})^2}=2 \sqrt{26}</cmath>
 
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Now, notice that since <math>\angle ABO</math> is <math>90</math> degrees, so arc <math>AO</math> is <math>180</math> degrees and <math>AO</math> is the diameter. Thus, the radius is <math>\sqrt{26},</math> so the area is <math>\boxed{26\pi}.</math>
 
−
 
−
- kante314
 
−
 
−
==Video Solution by The Power of Logic==
 
−
https://youtu.be/2lDDbOAmW18
 
−
 
−
~math2718281828459
 
  
 
== Solution 3 ==
 
== Solution 3 ==
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\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
 +
Therefore, the answer is <math>\boxed{\textbf{(C) }26\pi}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
  
−
Therefore, the answer is <math>\boxed{\textbf{(C) }}26 \pi</math>.
+
==Video Solution by The Power of Logic==
 +
https://youtu.be/2lDDbOAmW18
  
−
~Steven Chen (www.professorchenedu.com)
+
~math2718281828459
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=14|num-a=16}}
 
{{AMC10 box|year=2021 Fall|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:09, 28 November 2021

Problem

Isosceles triangle $ABC$ has $AB = AC = 3\sqrt6$, and a circle with radius $5\sqrt2$ is tangent to line $AB$ at $B$ and to line $AC$ at $C$. What is the area of the circle that passes through vertices $A$, $B$, and $C?$

$\textbf{(A) }24\pi\qquad\textbf{(B) }25\pi\qquad\textbf{(C) }26\pi\qquad\textbf{(D) }27\pi\qquad\textbf{(E) }28\pi$

Solution 1 (Cyclic Quadrilateral)

~MRENTHUSIASM ~kante314

Solution 2 (Similar Triangles)

[asy] import olympiad; unitsize(50); pair A,B,C,D,E,I,F,G,O; A=origin; B=(2,3); C=(-2,3); D=(4.3,6.3); E=(-4.3,6.3); F=(1,1.5); G=(-1,1.5); O=circumcenter(A,B,C); // olympiad - circumcenter I=incenter(A,D,E); draw(A--B--C--cycle); dot(O); dot(I); dot(F); dot(G); draw(circumcircle(A,B,C)); // olympiad - circumcircle draw(incircle(A,D,E)); draw(I--B); draw(I--C); draw(I--A); draw(rightanglemark(A,C,I)); draw(rightanglemark(A,B,I)); draw(O--F); draw(O--G); draw(rightanglemark(A,F,O)); draw(rightanglemark(A,G,O)); label("$O$",O,W); label("$A$",A,S); label("$B$",B,N); label("$C$",C,W); label("$D$",F,S); label("$E$",G,W); label("$3\sqrt{6}$",(1.5,1.5),S); label("$3\sqrt{6}$",(-1.5,1.5),S); label("$5\sqrt{2}$",(1,3.625),N); label("$5\sqrt{2}$",(-1,3.625),N); label("$I$",I,N); label("$r$",(-0.25,1.5),E); label("$r$",(0.5,2.125),S); add(pathticks(A--F,1,0.5,0,2)); add(pathticks(F--B,1,0.5,0,2)); add(pathticks(A--G,1,0.5,0,2)); add(pathticks(G--C,1,0.5,0,2)); [/asy] Because circle $I$ is tangent to $\overline{AB}$ at $B, \angle{ABI} \cong 90^{\circ}$. Because O is the circumcenter of $\bigtriangleup ABC, \overline{OD}$ is the perpendicular bisector of $\overline{AB}$, and $\angle{BAI} \cong \angle{DAO}$, so therefore $\bigtriangleup ADO \sim \bigtriangleup ABI$ by AA similarity. Then we have $\frac{AD}{AB} = \frac{DO}{BI} \implies \frac{1}{2} = \frac{r}{5\sqrt{2}} \implies r = \frac{5\sqrt{2}}{2}$. We also know that $\overline{AD} = \frac{3\sqrt{6}}{2}$ because of the perpendicular bisector, so the hypotenuse of $\bigtriangleup ADO = \sqrt{(\frac{5\sqrt{2}}{2})^2+(\frac{3\sqrt{6}}{2})^2} = \sqrt{\frac{25}{2}+\frac{27}{2}} = \sqrt{26}$. This is the radius of the circumcircle of $\bigtriangleup ABC$, so the area of this circle is $26\pi = \boxed{C}$

Solution in Progress

~KingRavi

Solution 3

Denote by $O$ the center of the circle that is tangent to line $AB$ at $B$ and to line $AC$ at $C$.

Because this circle is tangent to line $AB$ at $B$, $OB \perp AB$ and $OB = 5 \sqrt{2}$. Because this circle is tangent to line $AC$ at $C$, $OC \perp AC$ and $OC = 5 \sqrt{2}$.

Because $AB = AC$, $OB = OC$, $AO = AO$, $\triangle ABO \cong \triangle ACO$. Hence, $\angle BAO = \angle CAO$.

Let $AO$ and $BC$ meet at point $D$. Because $AB = AC$, $\angle BAO = \angle CAO$, $AD = AD$, $\triangle ABD \cong \triangle ACD$.

Hence, $BD = CD$ and $\angle ADB = \angle ADC = 90^\circ$.

Denote $\theta = \angle BAO$. Hence, $\angle BAC = 2 \theta$.

Denote by $R$ the circumradius of $\triangle ABC$. In $\triangle ABC$, following from the law of sines, $2 R = \frac{BC}{\sin \angle BAC}$.

Therefore, the area of the circumcircle of $\triangle ABC$ is \begin{align*} \pi R^2 & = \pi \left( \frac{BC}{2 \sin \angle BAC} \right)^2 \\ & = \pi \left( \frac{2 BD}{2 \sin \angle BAC} \right)^2 \\ & = \pi \left( \frac{BD}{\sin 2 \theta} \right)^2 \\ & = \pi \left( \frac{AB \sin \theta }{\sin 2 \theta} \right)^2 \\ & = \pi \left( \frac{AB \sin \theta }{2 \sin \theta \cos \theta} \right)^2 \\ & = \pi \left( \frac{AB }{2 \cos \theta} \right)^2 \\ & = \pi \left( \frac{AO}{2} \right)^2 \\ & = \frac{\pi}{4} \left( AB^2 + OB^2 \right) \\ & = 26 \pi . \end{align*} Therefore, the answer is $\boxed{\textbf{(C) }26\pi}$.

~Steven Chen (www.professorchenedu.com)

Video Solution by The Power of Logic

https://youtu.be/2lDDbOAmW18

~math2718281828459

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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