Difference between revisions of "2021 AMC 12A Problems/Problem 24"
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Applying the Extended Law of Sines to <math>\triangle PQR,</math> we find the radius of <math>\odot X:</math> <cmath>XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.</cmath> | Applying the Extended Law of Sines to <math>\triangle PQR,</math> we find the radius of <math>\odot X:</math> <cmath>XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.</cmath> | ||
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Alternatively, by the Inscribed Angle Theorem, <math>\triangle QRX</math> is a <math>30^\circ\text{-}30^\circ\text{-}120^\circ</math> triangle with base <math>QR=3\sqrt3.</math> Dividing <math>\triangle QRX</math> into two congruent <math>30^\circ\text{-}60^\circ\text{-}90^\circ</math> triangles, we get that the radius of <math>\odot X</math> is <math>XQ=XR=3</math> by the side-length ratios. | Alternatively, by the Inscribed Angle Theorem, <math>\triangle QRX</math> is a <math>30^\circ\text{-}30^\circ\text{-}120^\circ</math> triangle with base <math>QR=3\sqrt3.</math> Dividing <math>\triangle QRX</math> into two congruent <math>30^\circ\text{-}60^\circ\text{-}90^\circ</math> triangles, we get that the radius of <math>\odot X</math> is <math>XQ=XR=3</math> by the side-length ratios. | ||
− | Let <math>M</math> be the midpoint of <math>\overline{QR}.</math> By the Perpendicular Chord | + | Let <math>M</math> be the midpoint of <math>\overline{QR}.</math> By the Perpendicular Chord Bisector Converse, we have <math>\overline{XM}\perp\overline{QR}</math> and <math>\overline{OM}\perp\overline{QR}.</math> Together, points <math>O, X,</math> and <math>M</math> must be collinear. |
By the SAS Congruence, we have <math>\triangle QXM\cong\triangle RXM,</math> both of which are <math>30^\circ\text{-}60^\circ\text{-}90^\circ</math> triangles. By the side-length ratios, <math>RM=\frac{3\sqrt3}{2}, RX=3,</math> and <math>MX=\frac{3}{2}.</math> By the Pythagorean Theorem on right <math>\triangle ORM,</math> we get <math>OM=\frac{13}{2}</math> and <math>OX=OM-XM=5.</math> By the Pythagorean Theorem on right <math>\triangle OXP,</math> we obtain <math>OP=4.</math> | By the SAS Congruence, we have <math>\triangle QXM\cong\triangle RXM,</math> both of which are <math>30^\circ\text{-}60^\circ\text{-}90^\circ</math> triangles. By the side-length ratios, <math>RM=\frac{3\sqrt3}{2}, RX=3,</math> and <math>MX=\frac{3}{2}.</math> By the Pythagorean Theorem on right <math>\triangle ORM,</math> we get <math>OM=\frac{13}{2}</math> and <math>OX=OM-XM=5.</math> By the Pythagorean Theorem on right <math>\triangle OXP,</math> we obtain <math>OP=4.</math> |
Revision as of 09:21, 30 April 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 1 (Possible Without Trigonometry)
Let be the center of the semicircle and
be the center of the circle.
Applying the Extended Law of Sines to we find the radius of
Alternatively, by the Inscribed Angle Theorem,
is a
triangle with base
Dividing
into two congruent
triangles, we get that the radius of
is
by the side-length ratios.
Let be the midpoint of
By the Perpendicular Chord Bisector Converse, we have
and
Together, points
and
must be collinear.
By the SAS Congruence, we have both of which are
triangles. By the side-length ratios,
and
By the Pythagorean Theorem on right
we get
and
By the Pythagorean Theorem on right
we obtain
As shown above, we construct altitude of
Since
and
we know that
We construct point
on
such that
Clearly, quadrilateral
is a rectangle. Since
by alternate interior angles, we have
by the AA Similarity, with the ratio of similitude
Therefore, we get
and
The area of is
from which the answer is
~MRENTHUSIASM
Solution 2 (Trigonometry)
Suppose we label the points as shown in the diagram above, where is the center of the semicircle and
is the center of the circle tangent to
. Since
, we have
and
is a
triangle, which can be split into two
triangles by the altitude from
. Since
we know
by
triangles. The area of this part of
is
. We would like to add this value to the sum of the areas of the other two parts of
.
To find the areas of the other two parts of using the
area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that
and triangles
and
are congruent as they share a side,
and
. Therefore
. Suppose
. Then
, and since
, this simplifies to
. This factors nicely as
, so
as
can't be
. Since
and
, we now know that
is a
right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of
.
Let . Then
and
. The sum of the areas of
and
is
which we will add to
to get the area of
. Observe that
and similarly
. Adding these two gives
and multiplying that by
gets us
which we add to
to get
. The answer is
~sugar_rush
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=cEHF5iWMe9c
Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines )
~pi_is_3.14
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.