2021 Fall AMC 10A Problems/Problem 15

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 $C.$ Note that $\angle ABO_1 = \angle ACO_1 = 90^\circ.$ 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: $[asy] /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, O1, O2; A = (0,2sqrt(26)); O1 = (0,0); B = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[0]; C = intersectionpoints(Circle(A,3sqrt(6)),Circle(O1,5sqrt(2)))[1]; O2 = midpoint(A--O1); fill(A--B--C--cycle, yellow); dot("$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),scale(0.5)*rotate(90)*dir(midpoint(A--B)--A),red+fontsize(10)); label("$3\sqrt6$",midpoint(A--C),scale(0.5)*rotate(90)*dir(midpoint(A--C)--C),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 $AO_1 = \sqrt{AB^2 + BC^2} = 2\sqrt{26}.$ Therefore, the area of $\odot O_2$ is $\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{\textbf{(C) }26\pi}.$

~MRENTHUSIASM ~kante314 ~akpi2

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)

Solution 4 (Geometric Mean)

Let $D$ be the center of the circle with radius $5\sqrt{2}$ . Let the intersection of $AD$ and $BC$ be $E$ . And let $BC = 2a, AE = b$ and $ED = c$ . Since $ABDC$ is a kite $AD$ bisects $BC$ so $BE = EC = a$ . $a = \sqrt{bc}$ by the Geometric Mean Theorem. $a^{2} + b^{2} = 54, a^{2} + c^{2} = 50$ substituting $a = \sqrt{bc}$ gives $bc + b^{2} = 54, bc + c^{2} = 50$ factoring and dividing both equations gives $\frac{b}{c} = \frac{54}{50} \Rightarrow c =\frac{25}{27}$ . Substituting $c =\frac{25}{27}$ back into the 2nd equation and factoring gives $\frac{25}{27}b \cdot \frac{52}{27}=50$ so $b = \frac{27\sqrt{26}}{26}$ . The radius of the circumcircle of a triangle is $\frac{abc}{4A}$ where $a, b, c$ are the sides of the triangle and $A$ is the area. So we just need to find $\frac{a^{2}b^{2}c^{2}}{16A^{2}}\pi$ and plugging our variables and simplifying gives $\frac{54 \cdot 54 \cdot 4}{b^{2}}$ . Plugging all the variable into the formula gives us $\frac{54^{2} \cdot 4}{(\frac{27\sqrt{26}}{26})^{2}} \pi \Rightarrow \frac{54^{2} \cdot 4}{\frac{27^{2}}{26}}\pi \Rightarrow \boxed{\textbf{(C) }26\pi}$ .

~LuisFonseca123

Video Solution by The Power of Logic

https://youtu.be/2lDDbOAmW18

~math2718281828459

Video Solution by TheBeautyofMath

https://youtu.be/zq3UPu4nwsE?t=1674

~IceMatrix

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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