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

Revision as of 03:28, 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:

DIAGRAM WILL BE READY VERY SOON.

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 $B$ , $OB \perp AB$ 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

@@ Line 6: / Line 6: @@
 ==Solution 1 (Cyclic Quadrilateral)==
+Let <math>\odot O_1</math> be the circle with radius <math>5\sqrt2</math> that is tangent to <math>\overleftrightarrow{AB}</math> at <math>B</math> and to <math>\overleftrightarrow{AC}</math> at <math>C.</math> Since the opposite angles of quadrilateral <math>ABO_1C</math> are supplementary, quadrilateral <math>ABO_1C</math> is cyclic. Let <math>\odot O_2</math> be the circumcircle of quadrilateral <math>ABO_1C.</math> It follows that <math>\odot O_2</math> is also the circumcircle of <math>\triangle ABC,</math> as shown below:
+<b>DIAGRAM WILL BE READY VERY SOON.</b>
+By the Inscribed Angle Theorem, we conclude that <math>\overline{AO_1}</math> is the diameter of <math>\odot O_2.</math> By the Pythagorean Theorem on right <math>\triangle ABO_1,</math> we have <cmath>AO_1 = \sqrt{AB^2 + BO_1^2} = 2\sqrt{26}.</cmath>
+Therefore, the area of <math>\odot O_2</math> is <math>\pi\cdot\left(\frac{AO_1}{2}\right)^2=\boxed{\textbf{(C) }26\pi}.</math>
 ~MRENTHUSIASM ~kante314
@@ Line 53: / Line 58: @@
 add(pathticks(G--C,1,0.5,0,2));
 </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</math> is <cmath>\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}.</cmath>
-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>\boxed{\textbf{(C) }26\pi}</math>.
 Solution in Progress

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