Difference between revisions of "2021 AMC 12A Problems/Problem 24"
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+ | ==Solution 4 (Similar Triangles)== | ||
+ | Let <math>O_{1}</math> be the center of <math>\odot\Gamma, O_2</math> be the center of <math>\odot\Omega,</math> and <math>M</math> be the midpoint of <math>\overline{QR}.</math> We have <math>O_{1}M=\sqrt{7^2-\left(\frac{3\sqrt3}{2}\right)^2}=\frac{13}{2}</math> and by Extended Law of Sines, the radius of <math>\odot\Omega</math> is <math>\frac{3\sqrt3}{2\sin 60^\circ}=3</math> so <math>O_{2}M=3\cos 60^\circ=\frac{3}{2}.</math> Therefore <math>O_{1}O_{2}=O_{1}M-O_{2}M=5</math> and <math>O_{1}P=\sqrt{5^2 - 3^2}=4.</math> | ||
+ | |||
+ | Let <math>X=\overline{AB}\cap\overline{QR}.</math> Obviously <cmath>\angle O_{1}PO_{2}=\angle O_{1}MX=90^\circ~\text{and}~\angle PO_{1}O_{2}=\angle MO_{1}X=\arcsin\left(\frac{3}{5}\right)</cmath> so <math>\triangle PO_{1}O_{2}\sim\triangle MO_{1}X</math> with ratio <math>\frac{PO_{1}}{MO_{1}}=\frac{4}{\tfrac{13}{2}}=\frac{8}{13}.</math> Therefore <math>O_1X=\frac{13}{8}\cdot O_{1}O_{2}=\frac{13}{8}\cdot5=\frac{65}{8}</math> and <math>MX=\frac{13}{8}\cdot PO_{2}=\frac{13}{8}\cdot3=\frac{39}{8}.</math> | ||
+ | |||
+ | Let <math>H</math> denote the foot of the altitude from <math>P</math> to <math>\overline{QR}.</math> Because <math>\overline{PH}\parallel\overline{O_{1}M},</math> it follows that <math>\triangle PHX\sim\triangle O_{1}MX.</math> This similarity has ratio <cmath>\frac{PX}{O_{1}X}=1-\frac{O_{1}P}{O_{1}X}=1-\frac{4}{\tfrac{65}{8}}=1-\frac{32}{65}=\frac{33}{65}.</cmath> We therefore have <math>PH=\frac{33}{65}\cdot O_{1}M=\frac{33}{65}\cdot\frac{13}{2}=\frac{33}{10}.</math> | ||
+ | |||
+ | Finally, the area of <math>\triangle PQR</math> is <cmath>\frac{1}{2}\cdot QR\cdot PH=\frac{1}{2}\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},</cmath> so the answer is <math>99+3+20=\boxed{\textbf{(D)}~122}.</math> | ||
+ | |||
+ | ~inventivedant | ||
+ | |||
+ | ==Solution 5 (Law of Sine and Power of a Point)== | ||
+ | |||
+ | [[File:2021AMC12AP24.png|center|400px]] | ||
+ | |||
+ | By the Law of Sine in <math>\triangle PQR</math> and its circumcircle <math>\odot \Omega</math>, <math>2r_{\Omega} = \frac{QR}{ \sin 60^{\circ} } = \frac{ 3\sqrt{3} }{ \frac{ \sqrt{3} }{2} } = 6</math>, <math>r_{\Omega} = 3</math> | ||
+ | |||
+ | <cmath>\Gamma \Omega = \sqrt{r_{\Gamma}^2 - \left( \frac{ PQ }{2}\right)^2} - \sqrt{r_{\Omega}^2 - \left( \frac{PQ}{2}\right)^2} = \sqrt{7^2 - \left( \frac{ 3 \sqrt{3} }{2}\right)^2} - \sqrt{3^2 - \left( \frac{ 3 \sqrt{3} }{2}\right)^2} = \frac{13}{2} - \frac32 = 5, \quad \Gamma P = \sqrt{5^2 - 3^2} = 4</cmath> | ||
+ | |||
+ | By Power of a Point in <math>\odot \Gamma</math> , <math>PQ \cdot PS = PA \cdot PB = (7+4)(7-4) = 33</math>. | ||
+ | |||
+ | By the Law of Sine in <math>\triangle PRS</math>, <math>\frac{PR}{PS} = \frac{ \sin \angle PSR }{ \sin \angle PRS }</math> | ||
+ | |||
+ | By the Law of Sine in <math>\triangle QRS</math> and its circumcircle <math>\odot \Gamma</math>, <math>\frac{QR}{ \sin \angle PSR } = 14</math>, <math>\frac{ 3\sqrt{3} }{ \sin \angle PSR } = 14</math>, <math>\sin \angle PSR = \frac{ 3\sqrt{3} }{14}</math>, <math>\cos \angle PSR = \frac{ 13 }{14}</math> | ||
+ | |||
+ | <cmath>\sin \angle PRS = \sin ( 60^{\circ} - \angle PSR ) = \sin 60^{\circ} \cos \angle PSR - \sin \angle PSR \cos 60^{\circ} = \frac{ \sqrt{3} }{2} \cdot \frac{ 13 }{14} - \frac{ 3\sqrt{3} }{14} \cdot \frac12 = \frac{ 5\sqrt{3} }{14}</cmath> | ||
+ | |||
+ | <cmath>\frac{PR}{PS} = \frac{ \frac{ 3\sqrt{3} }{14} }{ \frac{ 5\sqrt{3} }{14} } = \frac35, \quad PQ \cdot PR = PQ \cdot PS \cdot \frac{PR}{PS} = 33 \cdot \frac35 = \frac{99}{5}</cmath> | ||
+ | |||
+ | <cmath>[PQR] = \frac12 \cdot \sin 60^{\circ} \cdot PQ \cdot PR = \frac12 \cdot \frac12 \cdot \frac{99}{5} = \frac{ 99\sqrt{3} }{20}, \quad 99 + 3 + 20 = \boxed{\textbf{(D)}~122}</cmath> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 6 (Analytic Geometry)== | ||
+ | Following Solution 4, We have <math>O_{1}</math> (0,0) , <math>O_{2}</math> (4,3). | ||
+ | We can write the equation of the two circles as: <cmath>\odot\Gamma : x^{2}+y^{2}=7^{2}...(1)</cmath> | ||
+ | <cmath>\odot\Omega : (x-4)^{2}+(y-3)^{2}=3^{2}...(2)</cmath> | ||
+ | By substituting (1) into (2), we get <cmath>8x+6y-65=0...(3)</cmath> | ||
+ | Notice (3) is the relationship between <math>x</math> value and <math>y</math> value, in other words, (3) is the linear equation that go through <math>R</math> and <math>Q</math>. | ||
+ | Let the height drops from <math>P</math> to <math>QR</math> at <math>H</math>. Therefore, we have <cmath>Area \triangle PQR =\frac{1}{2}\cdot{QR}\cdot{PH}</cmath> | ||
+ | So <cmath>QR=3\sqrt{3}</cmath> | ||
+ | And by distance formula, <math>PH</math> is the distance from <math>P</math> (4,0) to <math>\overline{QR}</math>. <cmath>PH={\frac{|8\cdot4+6\cdot0-65|}{\sqrt{8^{2}+6^{2}}}=\frac{33}{10}}</cmath> | ||
+ | Thus, We get <cmath>Area \triangle PQR =\frac{1}{2}\cdot3\sqrt{3}\cdot\frac{33}{10}=\frac{99\sqrt{3}}{20}</cmath> | ||
+ | So the answer is <math>99+3+20=\boxed{\textbf{(D)}~122}.</math> | ||
+ | |||
+ | ~ERiccc | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtube.com/watch?v=UJ_M_cjul1Q | ||
+ | |||
+ | ~r00tsOfUnity | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://youtube.com/watch?v=cEHF5iWMe9c | https://youtube.com/watch?v=cEHF5iWMe9c | ||
+ | |||
+ | This is invalid. ~r00tsOfUnity | ||
== Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines ) == | == Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines ) == | ||
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~pi_is_3.14 | ~pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/VUvSH-8AxzM | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
==See also== | ==See also== |
Latest revision as of 16:24, 15 June 2024
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Possible Without Trigonometry)
- 4 Solution 2 (Trigonometry)
- 5 Solution 3 (Weighted Averages and Similar Triangles)
- 6 Solution 4 (Similar Triangles)
- 7 Solution 5 (Law of Sine and Power of a Point)
- 8 Solution 6 (Analytic Geometry)
- 9 Video Solution by MOP 2024
- 10 Video Solution by Punxsutawney Phil
- 11 Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines )
- 12 Video Solution
- 13 See also
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
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, we obtain
and
By the Pythagorean Theorem on right
we get
and
By the Pythagorean Theorem on right
we get
Let be the foot of the perpendicular from
to
and
be the foot of the perpendicular from
to
as shown below:
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
Solution 3 (Weighted Averages and Similar Triangles)
Define points as shown above, where
. The area of
is simply
it remains to compute the value of
. Note that
is simply a weighted average of
and
it is
times closer to
than it is to
. Observe that
since the radius of
is
as its diameter is
. Note also by the Extended Law of Sines the radius of
is
so
. Since
and
are collinear by symmetry we have
so
and
. Therefore,
is a
right triangle;
since
and
. Therefore
so
. Since
we have
. Therefore
so
is
times as close to
as to
we can compute
. The area of
is
and
.
~sugar_rush
Solution 4 (Similar Triangles)
Let be the center of
be the center of
and
be the midpoint of
We have
and by Extended Law of Sines, the radius of
is
so
Therefore
and
Let Obviously
so
with ratio
Therefore
and
Let denote the foot of the altitude from
to
Because
it follows that
This similarity has ratio
We therefore have
Finally, the area of is
so the answer is
~inventivedant
Solution 5 (Law of Sine and Power of a Point)
By the Law of Sine in and its circumcircle
,
,
By Power of a Point in ,
.
By the Law of Sine in ,
By the Law of Sine in and its circumcircle
,
,
,
,
Solution 6 (Analytic Geometry)
Following Solution 4, We have (0,0) ,
(4,3).
We can write the equation of the two circles as:
By substituting (1) into (2), we get
Notice (3) is the relationship between
value and
value, in other words, (3) is the linear equation that go through
and
.
Let the height drops from
to
at
. Therefore, we have
So
And by distance formula,
is the distance from
(4,0) to
.
Thus, We get
So the answer is
~ERiccc
Video Solution by MOP 2024
https://youtube.com/watch?v=UJ_M_cjul1Q
~r00tsOfUnity
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=cEHF5iWMe9c
This is invalid. ~r00tsOfUnity
Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines )
~pi_is_3.14
Video Solution
~MathProblemSolvingSkills.com
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.