Difference between revisions of "2021 AMC 12A Problems/Problem 24"
MRENTHUSIASM (talk | contribs) (→Solution) |
MRENTHUSIASM (talk | contribs) (Answer should be (D) instead of (A), as all the supporting work indicate.) |
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As shown above, we construct an altitude <math>\overline{PC}</math> of <math>\triangle PQR.</math> Since <math>\overline{PC}\perp\overline{RQ}</math> and <math>\overline{OM}\perp\overline{RQ},</math> we know that <math>\overline{PC}\parallel\overline{OM}.</math> We construct <math>D</math> on <math>\overline{PC}</math> such that <math>\overline{XD}\perp\overline{PC}.</math> Clearly, <math>MXDC</math> is a rectangle. Since <math>\angle XPD=\angle OXP</math> by alternate interior angles, we have <math>\triangle XPD\sim\triangle OXP</math> by the AA Similarity, with ratio of similitude <math>\frac{XP}{OX}=\frac 35.</math> Therefore, we get that <math>PD=\frac 95</math> and <math>PC=PD+DC=PD+MX=\frac 95 + \frac 32 = \frac{33}{10}.</math> | As shown above, we construct an altitude <math>\overline{PC}</math> of <math>\triangle PQR.</math> Since <math>\overline{PC}\perp\overline{RQ}</math> and <math>\overline{OM}\perp\overline{RQ},</math> we know that <math>\overline{PC}\parallel\overline{OM}.</math> We construct <math>D</math> on <math>\overline{PC}</math> such that <math>\overline{XD}\perp\overline{PC}.</math> Clearly, <math>MXDC</math> is a rectangle. Since <math>\angle XPD=\angle OXP</math> by alternate interior angles, we have <math>\triangle XPD\sim\triangle OXP</math> by the AA Similarity, with ratio of similitude <math>\frac{XP}{OX}=\frac 35.</math> Therefore, we get that <math>PD=\frac 95</math> and <math>PC=PD+DC=PD+MX=\frac 95 + \frac 32 = \frac{33}{10}.</math> | ||
− | The area of <math>\triangle PQR</math> is <cmath>\frac12(RQ)(PC)=\frac12(3\sqrt3)(\frac{33}{10})=\frac{99\sqrt3}{20},</cmath> and the answer is <math>99+3+20=\boxed{\textbf{( | + | The area of <math>\triangle PQR</math> is <cmath>\frac12(RQ)(PC)=\frac12(3\sqrt3)(\frac{33}{10})=\frac{99\sqrt3}{20},</cmath> and the answer is <math>99+3+20=\boxed{\textbf{(D) } 122}.</math> |
~MRENTHUSIASM | ~MRENTHUSIASM |
Revision as of 04:25, 14 February 2021
Contents
Problem
Semicircle has diameter
of length
. Circle
lies tangent to
at a point
and intersects
at points
and
. If
and
, then the area of
equals
, where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. What is
?
Diagram
~MRENTHUSIASM (by Geometry Expressions)
Solution
Let be the center of the semicircle,
be the center of the circle, and
be the midpoint of
By the Perpendicular Chord Theorem Converse, we have
and
Together, points
and
must be collinear.
Applying the Extended Law of Sines on we have
in which the radius of
is
By the SAS Congruence, we have both of which are
-
-
triangles. By the side-length ratios,
and
By the Pythagorean Theorem in
we get
and
By the Pythagorean Theorem on
we obtain
As shown above, we construct an altitude of
Since
and
we know that
We construct
on
such that
Clearly,
is a rectangle. Since
by alternate interior angles, we have
by the AA Similarity, with ratio of similitude
Therefore, we get that
and
The area of is
and the answer is
~MRENTHUSIASM
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=cEHF5iWMe9c
Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines )
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.