Difference between revisions of "2021 AMC 12A Problems/Problem 24"
MRENTHUSIASM (talk | contribs) (Added back the original solution as Solution 2.) |
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==Solution 2== | ==Solution 2== | ||
+ | ===Diagram=== | ||
[[File:2021 AMC 12A Problem 24.png|center]] | [[File:2021 AMC 12A Problem 24.png|center]] | ||
+ | ~MRENTHUSIASM (by Geometry Expressions) | ||
+ | ===Solution=== | ||
Let <math>O=\Gamma</math> be the center of the semicircle, <math>X=\Omega</math> be the center of the circle, and <math>M</math> be the midpoint of <math>\overline{QR}.</math> By the Perpendicular Chord Theorem 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. | Let <math>O=\Gamma</math> be the center of the semicircle, <math>X=\Omega</math> be the center of the circle, and <math>M</math> be the midpoint of <math>\overline{QR}.</math> By the Perpendicular Chord Theorem 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. | ||
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By the SAS Congruence, we have <math>\triangle QXM\cong\triangle RXM,</math> both of which are <math>30^\circ</math>-<math>60^\circ</math>-<math>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 in <math>\triangle ORM,</math> we get <math>OM=\frac{13}{2}</math> and <math>OX=OM-XM=5.</math> By the Pythagorean Theorem on <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</math>-<math>60^\circ</math>-<math>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 in <math>\triangle ORM,</math> we get <math>OM=\frac{13}{2}</math> and <math>OX=OM-XM=5.</math> By the Pythagorean Theorem on <math>\triangle OXP,</math> we obtain <math>OP=4.</math> | ||
− | [[File:2021 AMC 12A Problem 24(2) (Revised).png]] | + | [[File:2021 AMC 12A Problem 24(2) (Revised).png|center]] |
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> |
Revision as of 20:09, 23 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
[asy]
draw(circle((7,0),7));
pair A = (0, 0);
pair B = (14, 0);
draw(A--B);
draw(circle((11,3),3));
label("", (7, 0), S);
label("
", (11, 3), E);
label("
", (11, 0), S);
pair C = (7, 0);
pair O = (11, 3);
pair P = (11, 0);
pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1];
pair R = intersectionpoints(circle(C, 7), circle(O, 3))[0];
draw(C--O);
draw(C--Q);
draw(C--R);
draw(Q--R);
draw(O--P);
draw(O--Q);
draw(O--R);
draw(P--Q);
draw(P--R);
label("
", Q, N);
label("
", R, E);
[/asy]
someone pls make this asymptote code work
Solution
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
Solution 2
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.