Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

← Older editNewer edit →
MRENTHUSIASM (talk | contribs)
editor
9,986 edits
mNo edit summary
No edit summary
Line 75: Line 75:


~math2718281828459
~math2718281828459
== Solution 3 ==
Denote by <math>O</math> the center of the circle that is tangent to line <math>AB</math> at <math>B</math> and to line <math>AC</math> at <math>C</math>.
Because this circle is tangent to line <math>AB</math> at <math>B</math>, <math>OB \perp AB</math> and <math>OB = 5 \sqrt{2}</math>.
Because this circle is tangent to line <math>AC</math> at <math>C</math>, <math>OC \perp AC</math> and <math>OC = 5 \sqrt{2}</math>.
Because <math>AB = AC</math>, <math>OB = OC</math>, <math>AO = AO</math>, <math>\triangle ABO \cong \triangle ACO</math>.
Hence, <math>\angle BAO = \angle CAO</math>.
Let <math>AO</math> and <math>BC</math> meet at point <math>D</math>.
Because <math>AB = AC</math>, <math>\angle BAO = \angle CAO</math>, <math>AD = AD</math>, <math>\triangle ABD \cong \triangle ACD</math>.
Hence, <math>BD = CD</math> and <math>\angle ADB = \angle ADC = 90^\circ</math>.
Denote <math>\theta = \angle BAO</math>. Hence, <math>\angle BAC = 2 \theta</math>.
Denote by <math>R</math> the circumradius of <math>\triangle ABC</math>.
In <math>\triangle ABC</math>, following from the law of sines, <math>2 R = \frac{BC}{\sin \angle BAC}</math>.
Therefore, the area of the circumcircle of <math>\triangle ABC</math> is
<cmath>
\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*}
</cmath>
Therefore, the answer is <math>\boxed{\textbf{(C) }}26 \pi</math>.
~Steven Chen (www.professorchenedu.com)


==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 20:12, 25 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 (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 2 (Geometric Assumption)

Let the center of the first circle be $O.$ By Pythagorean Theorem, \[AO=\sqrt{(3\sqrt{6})^2+(5\sqrt{2})^2}=2 \sqrt{26}\] Now, notice that since $\angle ABO$ is $90$ degrees, so arc $AO$ is $180$ degrees and $AO$ is the diameter. Thus, the radius is $\sqrt{26},$ so the area is $\boxed{26\pi}.$

- kante314

Video Solution by The Power of Logic

https://youtu.be/2lDDbOAmW18

~math2718281828459

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)

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

These problems are copyrighted © by the Mathematical Association of America.

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_15&oldid=166399"