Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
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MRENTHUSIASM (talk | contribs) (Removed unnecessary edits, as the original solution is clear enough that AO1 = 2r) |
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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 | + | 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. |
− | < | + | |
− | + | 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> | ||
− | + | ~MRENTHUSIASM ~kante314 | |
==Solution 2 (Similar Triangles)== | ==Solution 2 (Similar Triangles)== | ||
− | |||
<asy> | <asy> | ||
− | |||
import olympiad; | import olympiad; | ||
unitsize(50); | unitsize(50); | ||
− | pair A,B,C,D,E,I,O; | + | pair A,B,C,D,E,I,F,G,O; |
− | A=origin; B=(2, | + | 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 | O=circumcenter(A,B,C); // olympiad - circumcenter | ||
I=incenter(A,D,E); | I=incenter(A,D,E); | ||
Line 23: | Line 48: | ||
dot(O); | dot(O); | ||
dot(I); | dot(I); | ||
+ | dot(F); | ||
+ | dot(G); | ||
draw(circumcircle(A,B,C)); // olympiad - circumcircle | draw(circumcircle(A,B,C)); // olympiad - circumcircle | ||
draw(incircle(A,D,E)); | draw(incircle(A,D,E)); | ||
− | label("$O$",O, | + | 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("$A$",A,S); | ||
− | label("$B$",B, | + | label("$B$",B,N); |
label("$C$",C,W); | label("$C$",C,W); | ||
− | label("$3\sqrt{6}$",(1. | + | label("$D$",F,S); |
− | label("$3\sqrt{6}$",(-1.25,1),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>. | ||
+ | |||
+ | ~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) | ||
+ | |||
+ | ==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 | ||
− | Solution | + | ==Video Solution == |
+ | https://youtu.be/zq3UPu4nwsE?t=1674 | ||
− | + | == 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 13:43, 5 May 2024
Contents
Problem
Isosceles triangle has
, and a circle with radius
is tangent to line
at
and to line
at
. What is the area of the circle that passes through vertices
,
, and
Solution 1 (Cyclic Quadrilateral)
Let be the circle with radius
that is tangent to
at
and to
at
Note that
Since the opposite angles of quadrilateral
are supplementary, quadrilateral
is cyclic.
Let be the circumcircle of quadrilateral
It follows that
is also the circumcircle of
as shown below:
By the Inscribed Angle Theorem, we conclude that
is the diameter of
By the Pythagorean Theorem on right
we have
Therefore, the area of
is
~MRENTHUSIASM ~kante314
Solution 2 (Similar Triangles)
Because circle
is tangent to
at
. Because
is the circumcenter of
is the perpendicular bisector of
, and
, so therefore
by AA similarity. Then we have
. We also know that
because of the perpendicular bisector, so the hypotenuse of
is
This is the radius of the circumcircle of
, so the area of this circle is
.
~KingRavi
Solution 3 (Trigonometry)
Denote by the center of the circle that is tangent to line
at
and to line
at
.
Because this circle is tangent to line at
, we have
and
.
Because this circle is tangent to line at
, we have
and
.
Because ,
,
, we get
. Hence,
.
Let and
meet at point
.
Because
,
,
, we get
. Hence,
and
.
Denote . Hence,
.
Denote by the circumradius of
.
In
, following from the law of sines,
.
Therefore, the area of the circumcircle of is
~Steven Chen (www.professorchenedu.com)
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution by The Power of Logic
~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.