Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
MRENTHUSIASM (talk | contribs) (→Solution 1 (Cyclic Quadrilateral)) |
MRENTHUSIASM (talk | contribs) (→Solution 1 (Cyclic Quadrilateral)) |
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==Solution 1 (Cyclic Quadrilateral)== | ==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: | 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: | ||
− | + | <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); | ||
+ | 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),0.5*NW,red+fontsize(10)); | ||
+ | label("$3\sqrt6$",midpoint(A--C),0.5*NE,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> | 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> |
Revision as of 04:50, 28 November 2021
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
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 O 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
.
Solution in Progress
~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
,
and
.
Because this circle is tangent to line
at
,
and
.
Because ,
,
,
.
Hence,
.
Let and
meet at point
.
Because
,
,
,
.
Hence, and
.
Denote . Hence,
.
Denote by the circumradius of
.
In
, following from the law of sines,
.
Therefore, the area of the circumcircle of is
Therefore, the answer is
.
~Steven Chen (www.professorchenedu.com)
Video Solution by The Power of Logic
~math2718281828459
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.