Difference between revisions of "2021 Fall AMC 10A Problems/Problem 15"
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− | 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> | + | 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>. | 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>. | ||
Revision as of 18:58, 17 February 2022
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 .
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 , 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 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.