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

m (→‎Solution 1 (Cyclic Quadrilateral))
m (→‎Solution 3 (Trigonometry): Shortened sol a bit.)
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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>.
 
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>AB</math> at <math>B</math>, we have <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>.
+
Because this circle is tangent to line <math>AC</math> at <math>C</math>, we have <math>OC \perp AC</math> and <math>OC = 5 \sqrt{2}</math>.
−
Hence, <math>\angle BAO = \angle CAO</math>.
+
 
 +
Because <math>AB = AC</math>, <math>OB = OC</math>, <math>AO = AO</math>, we get <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>.
 
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>.
+
Because <math>AB = AC</math>, <math>\angle BAO = \angle CAO</math>, <math>AD = AD</math>, we get <math>\triangle ABD \cong \triangle ACD</math>. Hence, <math>BD = CD</math> and <math>\angle ADB = \angle ADC = 90^\circ</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 <math>\theta = \angle BAO</math>. Hence, <math>\angle BAC = 2 \theta</math>.
Line 117: Line 115:
 
& = \pi \left( \frac{AO}{2} \right)^2 \\
 
& = \pi \left( \frac{AO}{2} \right)^2 \\
 
& = \frac{\pi}{4} \left( AB^2 + OB^2 \right) \\
 
& = \frac{\pi}{4} \left( AB^2 + OB^2 \right) \\
−
& = 26 \pi .
+
& = \boxed{\textbf{(C) }26\pi}.
 
\end{align*}
 
\end{align*}
 
</cmath>
 
</cmath>
−
Therefore, the answer is <math>\boxed{\textbf{(C) }26\pi}</math>.
 
−
 
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
  

Revision as of 12:55, 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)

Let $\odot O_1$ be the circle with radius $5\sqrt2$ that is tangent to $\overleftrightarrow{AB}$ at $B$ and to $\overleftrightarrow{AC}$ at $C.$ Since the opposite angles of quadrilateral $ABO_1C$ are supplementary, quadrilateral $ABO_1C$ is cyclic.

Let $\odot O_2$ be the circumcircle of quadrilateral $ABO_1C.$ It follows that $\odot O_2$ is also the circumcircle of $\triangle ABC,$ as shown below: [asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O1, O2; A = (0,2sqrt(26)); O1 = (0,0); B = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[0]; C = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[1]; O2 = midpoint(A--O1); dot("$A$",A,1.5*N,linewidth(4)); dot("$B$",B,1.5*W,linewidth(4)); dot("$C$",C,1.5*E,linewidth(4)); dot("$O_1$",O1,1.5*S,linewidth(4)); dot("$O_2$",O2,1.5*N,linewidth(4)); label("$3\sqrt6$",midpoint(A--B),0.5*NW,red+fontsize(10)); label("$3\sqrt6$",midpoint(A--C),0.5*NE,red+fontsize(10)); label("$5\sqrt2$",midpoint(O1--B),0.5*SW,red+fontsize(10)); label("$5\sqrt2$",midpoint(O1--C),0.5*SE,red+fontsize(10)); markscalefactor=0.05; draw(rightanglemark(A,B,O1)^^rightanglemark(A,C,O1),red); draw(A--B--O1--C--cycle^^B--C^^circumcircle(A,B,C)); [/asy] By the Inscribed Angle Theorem, we conclude that $\overline{AO_1}$ is the diameter of $\odot O_2.$ By the Pythagorean Theorem on right $\triangle ABO_1,$ we have \[AO_1 = \sqrt{AB^2 + BO_1^2} = 2\sqrt{26}.\] Therefore, the area of $\odot O_2$ is $\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{\textbf{(C) }26\pi}.$

~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 $\bigtriangleup ADO$ is \[\sqrt{\left(\frac{5\sqrt{2}}{2}\right)^2+\left(\frac{3\sqrt{6}}{2}\right)^2} = \sqrt{\frac{25}{2}+\frac{27}{2}} = \sqrt{26}.\] This is the radius of the circumcircle of $\bigtriangleup ABC$, so the area of this circle is $\boxed{\textbf{(C) }26\pi}$.

Solution in Progress

~KingRavi

Solution 3 (Trigonometry)

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 $B$, we have $OB \perp AB$ and $OB = 5 \sqrt{2}$.

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

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

Let $AO$ and $BC$ meet at point $D$. Because $AB = AC$, $\angle BAO = \angle CAO$, $AD = AD$, we get $\triangle ABD \cong \triangle ACD$. 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) \\ & = \boxed{\textbf{(C) }26\pi}. \end{align*} ~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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