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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
(→‎Solution 4 (Circumradius of a Right Triangle))
 
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 +
== Problem ==
 +
 
Isosceles triangle <math>ABC</math> has <math>AB = AC = 3\sqrt6</math>, and a circle with radius <math>5\sqrt2</math> is tangent to line <math>AB</math> at <math>B</math> and to line <math>AC</math> at <math>C</math>. What is the area of the circle that passes through vertices <math>A</math>, <math>B</math>, and <math>C?</math>
 
Isosceles triangle <math>ABC</math> has <math>AB = AC = 3\sqrt6</math>, and a circle with radius <math>5\sqrt2</math> is tangent to line <math>AB</math> at <math>B</math> and to line <math>AC</math> at <math>C</math>. What is the area of the circle that passes through vertices <math>A</math>, <math>B</math>, and <math>C?</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>
 
<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>
  
−
==Solution 1==
+
==Solution 1 (Cyclic Quadrilateral)==
−
Let the center of the first circle be <math>O.</math> By Pythagorean Theorem,
+
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> Note that <math>\angle ABO_1 = \angle ACO_1 = 90^\circ.</math> Since the opposite angles of quadrilateral <math>ABO_1C</math> are supplementary, quadrilateral <math>ABO_1C</math> is cyclic.
−
<cmath>AO=\sqrt{(3\sqrt{6})^2+(5\sqrt{2})^2}=2 \sqrt{26}</cmath>
+
 
−
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>
+
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:
 +
<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 <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>
  
−
- kante314
+
~MRENTHUSIASM ~kante314
−
==See Also==
 
−
{{AMC10 box|year=2021 Fall|ab=A|num-b=14|num-a=16}}
 
−
{{MAA Notice}}
 
  
 
==Solution 2 (Similar Triangles)==
 
==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 <math>I</math> is tangent to <math>\overline{AB}</math> at <math>B, \angle{ABI} \cong 90^{\circ}</math>. Because <math>O</math> 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>\boxed{\textbf{(C) }26\pi}</math>.
  
−
Solution in Progress
 
  
 
~KingRavi
 
~KingRavi
 +
 +
== Solution 3 (Trigonometry) ==
 +
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>, 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>, we have <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>, 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>.
 +
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>.
 +
 +
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) \\
 +
& = \boxed{\textbf{(C) }26\pi}.
 +
\end{align*}
 +
</cmath>
 +
~Steven Chen (www.professorchenedu.com)
 +
 +
== Solution 4 (Circumradius of a Right Triangle) ==
 +
Label the center of the circle with radius <math> 5\sqrt{2} </math> as <math> O_2 </math>. Observe that both <math>\triangle ABO_2 </math> and <math>\triangle ABO_2 </math> are right triangles, as radii connecting the center of a circle to the point of tangency of a line are perpendicular to the line. Furthermore, these two triangles are congruent by SSS. Consider the midpoint of <math>\overline{AO_2} </math>, letting it be <math> O_1 </math>. We prove that <math>O_2</math> is the circumcenter of <math>\triangle ABC</math>.
 +
 +
The radii of the circumcenters of <math>\triangle ABO_2 </math> and <math>\triangle ABO_2 </math> must be equal, as the two triangles are congruent. We recall that the midpoint of the hypotenuse of a right triangle is it's circumcenter, so <math> \overline{BO_1}</math> and <math>\overline{CO_2}</math> - as the radii of the circumcircles of their respective triangles - must be congruent. Furthermore, we have <math> \overline{BO_1} \cong \overline{AO_1} </math> and <math>\overline{CO_1} \cong \overline{AO_1}</math>, from the same fact, so <math> \overline{BO_1} \cong \overline{AO_1} \cong \overline{CO_1} </math>. Thus, <math>O_1 </math> is the circumcenter of <math>\triangle ABC </math>.
 +
 +
By the Pythagorean Theorem, <math> \overline{AO_2} = \sqrt{ (3\sqrt{6})^2 + (5\sqrt{2})^2 } = 2\sqrt{26} </math>, so <math>\overline{AO_1}</math>, the radius of the circle through points <math>A</math>, <math>B</math>, and <math>C</math> is <math> \sqrt{26}</math>, and thus the area of the circle is <math> \boxed{ (C) \ 26\pi }</math>.
 +
 +
~LeonQS
 +
 +
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/T2VFw2lEthYqrt
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by The Power of Logic==
 +
https://youtu.be/2lDDbOAmW18
 +
 +
~math2718281828459
 +
 +
==Video Solution ==
 +
https://youtu.be/zq3UPu4nwsE?t=1674
 +
 +
== Video Solution ==
 +
https://youtu.be/DVuf-uXjfzY?t=211
 +
 +
==See Also==
 +
{{AMC10 box|year=2021 Fall|ab=A|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 01:54, 18 August 2025

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 \[AO_1 = \sqrt{AB^2 + BO_1^2} = 2\sqrt{26}.\] 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 $\boxed{\textbf{(C) }26\pi}$.


~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 (Circumradius of a Right Triangle)

Label the center of the circle with radius $5\sqrt{2}$ as $O_2$. Observe that both $\triangle ABO_2$ and $\triangle ABO_2$ are right triangles, as radii connecting the center of a circle to the point of tangency of a line are perpendicular to the line. Furthermore, these two triangles are congruent by SSS. Consider the midpoint of $\overline{AO_2}$, letting it be $O_1$. We prove that $O_2$ is the circumcenter of $\triangle ABC$.

The radii of the circumcenters of $\triangle ABO_2$ and $\triangle ABO_2$ must be equal, as the two triangles are congruent. We recall that the midpoint of the hypotenuse of a right triangle is it's circumcenter, so $\overline{BO_1}$ and $\overline{CO_2}$ - as the radii of the circumcircles of their respective triangles - must be congruent. Furthermore, we have $\overline{BO_1} \cong \overline{AO_1}$ and $\overline{CO_1} \cong \overline{AO_1}$, from the same fact, so $\overline{BO_1} \cong \overline{AO_1} \cong \overline{CO_1}$. Thus, $O_1$ is the circumcenter of $\triangle ABC$.

By the Pythagorean Theorem, $\overline{AO_2} = \sqrt{ (3\sqrt{6})^2 + (5\sqrt{2})^2 } = 2\sqrt{26}$, so $\overline{AO_1}$, the radius of the circle through points $A$, $B$, and $C$ is $\sqrt{26}$, and thus the area of the circle is $\boxed{ (C) \ 26\pi }$.

~LeonQS

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/T2VFw2lEthYqrt

~Education, the Study of Everything

Video Solution by The Power of Logic

https://youtu.be/2lDDbOAmW18

~math2718281828459

Video Solution

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

Video Solution

https://youtu.be/DVuf-uXjfzY?t=211

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, as part of the American Mathematics Competitions. AMC Logo.png

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