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

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

(→‎Solution 1 (Cyclic Quadrilateral))
(Removed unnecessary edits, as the original solution is clear enough that AO1 = 2r)
 
(11 intermediate revisions by 6 users not shown)
Line 32: Line 32:
 
draw(A--B--O1--C--cycle^^B--C^^circumcircle(A,B,C));
 
draw(A--B--O1--C--cycle^^B--C^^circumcircle(A,B,C));
 
</asy>
 
</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 + BC^2} = 2\sqrt{26}.</cmath>
+
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>
 
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 ~akpi2
+
~MRENTHUSIASM ~kante314
  
 
==Solution 2 (Similar Triangles)==
 
==Solution 2 (Similar Triangles)==
Line 84: Line 84:
 
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>.
 
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
Line 121: Line 120:
 
~Steven Chen (www.professorchenedu.com)
 
~Steven Chen (www.professorchenedu.com)
  
−
== Solution 4 (Geometric Mean) ==
+
==Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/T2VFw2lEthY
  
−
Let <math>D</math> be the center of the circle with radius <math>5\sqrt{2}</math>. Let the intersection of <math>AD</math> and <math>BC</math> be <math>E</math>. And let <math>BC = 2a, AE = b</math> and <math>ED = c</math>. Since <math>ABDC</math> is a kite <math>AD</math> bisects <math>BC</math> so <math>BE = EC = a</math>. <math>a = \sqrt{bc}</math> by the Geometric Mean Theorem. <math>a^{2} + b^{2} = 54, a^{2} + c^{2} = 50</math> substituting <math>a = \sqrt{bc}</math> gives <math>bc + b^{2} = 54, bc + c^{2} = 50</math> factoring and dividing both equations gives <math>\frac{b}{c} = \frac{54}{50} \Rightarrow c =\frac{25}{27}</math>. Substituting <math>c =\frac{25}{27}</math> back into the 2nd equation and factoring gives <math>\frac{25}{27}b \cdot \frac{52}{27}=50</math> so <math>b = \frac{27\sqrt{26}}{26}</math>. The radius of the circumcircle of a triangle is <math>\frac{abc}{4A}</math> where <math>a, b, c</math> are the sides of the triangle and <math>A</math> is the area. So we just need to find <math>\frac{a^{2}b^{2}c^{2}}{16A^{2}}\pi</math> and plugging our variables and simplifying gives <math>\frac{54 \cdot 54 \cdot 4}{b^{2}}</math>. Plugging all the variable into the formula gives us <math>\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}</math>.
+
~Education, the Study of Everything
−
 
 
−
~LuisFonseca123
 
  
 
==Video Solution by The Power of Logic==
 
==Video Solution by The Power of Logic==
Line 131: Line 129:
  
 
~math2718281828459
 
~math2718281828459
−
==Video Solution by TheBeautyofMath==
+
 
 +
==Video Solution ==
 
https://youtu.be/zq3UPu4nwsE?t=1674
 
https://youtu.be/zq3UPu4nwsE?t=1674
  
−
~IceMatrix
+
== Video Solution ==
 +
https://youtu.be/DVuf-uXjfzY?t=211
 +
 
 
==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}}

Latest revision as of 12:43, 5 May 2024

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)

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/T2VFw2lEthY

~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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_10A_Problems/Problem_15&oldid=219023"