Difference between revisions of "2016 AMC 12A Problems/Problem 15"
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− | Notice that we can find <math>[P'PQRR']</math> in two different ways: <math>[P'PQQ']+[Q'QRR']</math> and <math>[PQR]+[P'PRR']</math>, so <math>[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR']</math> | + | Notice that we can find <math>[P'PQRR']</math> in two different ways: <math>[P'PQQ']+[Q'QRR']</math> and <math>[PQR]+[P'PRR'].</math> Since we want <math>[PQR],</math> we use the latter method, so we have <math>[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].</math> |
<math>\break</math> | <math>\break</math> | ||
<math>P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}</math>. Additionally, <math>Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}</math>. Therefore, <math>[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}</math>. Similarly, <math>[Q'QRR']=5\sqrt6</math>. We can calculate <math>[P'PRR']</math> easily because <math>P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}</math>. <math>[P'PRR']=4\sqrt{2}+4\sqrt{6}</math>. <math>\newline</math> | <math>P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}</math>. Additionally, <math>Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}</math>. Therefore, <math>[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}</math>. Similarly, <math>[Q'QRR']=5\sqrt6</math>. We can calculate <math>[P'PRR']</math> easily because <math>P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}</math>. <math>[P'PRR']=4\sqrt{2}+4\sqrt{6}</math>. <math>\newline</math> | ||
− | Plugging into first equation, the two sums of areas, <math>3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR]</math> | + | Plugging into first equation, the two sums of areas, <math>3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR],</math> so therefore <math>[PQR]=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math> |
− | |||
− | <math>[PQR]=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math> | ||
==Solution 2== | ==Solution 2== | ||
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To make our calculations easier, let <math>\sqrt{9 + 4\sqrt{3}} = a</math>. The semi-perimeter of our triangle is <math>\frac{3 + 5 + 2a}{2} = 4 + a</math>. Symbolize the area of the triangle with <math>A</math>. Using Heron's formula, we have <cmath>A^2 = (4 + a)(4 + a - 2a)(4 + a - 3)(4 + a - 5) = (4 + a)(4 - a)(a + 1)(a - 1) = (16 - a^2)(a^2 - 1)</cmath> We can remove the outer root of a. <cmath>A^2 = (16 - 9 - 4\sqrt{3})(9 + 4\sqrt{3} - 1) = (7 - 4\sqrt{3})(8 + 4\sqrt{3}) = 8 - 4\sqrt{3} \rightarrow A = \sqrt{8 - 4\sqrt{3}}</cmath> | To make our calculations easier, let <math>\sqrt{9 + 4\sqrt{3}} = a</math>. The semi-perimeter of our triangle is <math>\frac{3 + 5 + 2a}{2} = 4 + a</math>. Symbolize the area of the triangle with <math>A</math>. Using Heron's formula, we have <cmath>A^2 = (4 + a)(4 + a - 2a)(4 + a - 3)(4 + a - 5) = (4 + a)(4 - a)(a + 1)(a - 1) = (16 - a^2)(a^2 - 1)</cmath> We can remove the outer root of a. <cmath>A^2 = (16 - 9 - 4\sqrt{3})(9 + 4\sqrt{3} - 1) = (7 - 4\sqrt{3})(8 + 4\sqrt{3}) = 8 - 4\sqrt{3} \rightarrow A = \sqrt{8 - 4\sqrt{3}}</cmath> | ||
− | We solve the nested root. We want to turn <math>8 - 4\sqrt{3}</math> into the square of something. If we have <math>(a - b) ^ 2 = 8 - 4\sqrt{3}</math>, then we get <cmath>\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}</cmath> Solving the system of equations, we get <math>a = \sqrt{6}</math> and <math>b = \sqrt{2}</math>. Alternatively, you can square all the possible solutions until you find one that is equal to <math>8 - 4\sqrt{3}. \newline A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math> | + | We solve the nested root. We want to turn <math>8 - 4\sqrt{3}</math> into the square of something. If we have <math>(a - b) ^ 2 = 8 - 4\sqrt{3}</math>, then we get <cmath>\begin{cases} a^2 + b^2 = 8 \\ ab = 2\sqrt{3} \end{cases}</cmath> Solving the system of equations, we get <math>a = \sqrt{6}</math> and <math>b = \sqrt{2}</math>. Alternatively, you can square all the possible solutions until you find one that is equal to <math>8 - 4\sqrt{3}.</math> <math>\newline A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math> |
~ZericH | ~ZericH | ||
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Like Solution 1, we can use the Pythagorean theorem to find <math>P'Q'</math> and <math>Q'R'</math>, which are <math>2\sqrt2</math> and <math>2\sqrt6</math> respectively. Since the only answer choice that has <math>\sqrt2</math> and <math>\sqrt6</math> is <math>D</math>, we can make an educated guess that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer. | Like Solution 1, we can use the Pythagorean theorem to find <math>P'Q'</math> and <math>Q'R'</math>, which are <math>2\sqrt2</math> and <math>2\sqrt6</math> respectively. Since the only answer choice that has <math>\sqrt2</math> and <math>\sqrt6</math> is <math>D</math>, we can make an educated guess that <math>\boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}}</math> is the answer. | ||
+ | |||
+ | |||
+ | ==Solution 7 (Quickest and Easiest Using Trapezoids)== | ||
+ | |||
+ | The fastest way to find the area of <math>[PQR]</math> is by breaking it into shapes with easy-to-calculate areas. We can see that the area of the whole shape <math>[PQRR'P']</math> is made up of trapezoids <math>[PP'QQ']</math> and <math>[Q'QRR']</math>. So, <math>[PQR]=[PP'QQ']+[Q'QRR']-[PRR'P']</math>. | ||
+ | |||
+ | We already know the lengths of <math>PP'</math> and <math>QQ'</math> being <math>1</math> and <math>2</math> respectively (they are the radiuses of the circles). The "height" of the trapezoid is <math>P'Q'=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}</math> which can be found from the Pythagorean Theorem. So, the area of trapezoid <math>[PP'QQ']=\frac{1+2}2*{(2\sqrt{2})}=3\sqrt{2}</math>. Similarly, we can find the area of trapezoid <math>[Q'QRR']</math>. <math>Q'R'=5\sqrt{6}</math>, so <math>[Q'QRR']=\frac{2+3}2*{(2\sqrt{6})}=5\sqrt{2}</math>. | ||
+ | |||
+ | Now we can put the pieces together. <math>[PQR]=[PP'QQ']+[Q'QRR']-[PRR'P']=3\sqrt{2}+5\sqrt{6}-[PRR'P']=</math>. The area of <math>PRR'P'</math> is the long rectangle <math>[PR"R'P]'+ [PRR"]=4\sqrt{2}+4\sqrt{6}</math>. <math>R"</math> is the point on <math>RR'</math> in which <math>PR"</math> is perpendicular to <math>RR'</math>. | ||
+ | |||
+ | So, <math>[PQR]=3\sqrt{2}+5\sqrt{6}-(4\sqrt{2}+4\sqrt{6})=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}</math> | ||
+ | |||
+ | ~hwan | ||
==Video Solutions== | ==Video Solutions== | ||
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https://www.youtube.com/watch?v=UanfIBpDTh8&ab_channel=ArtofProblemSolving | https://www.youtube.com/watch?v=UanfIBpDTh8&ab_channel=ArtofProblemSolving | ||
− | ~Art of Problem Solving | + | ~by the official Art of Problem Solving channel on YouTube |
==See Also== | ==See Also== |
Latest revision as of 23:55, 24 August 2024
Contents
Problem
Circles with centers and
, having radii
and
, respectively, lie on the same side of line
and are tangent to
at
and
, respectively, with
between
and
. The circle with center
is externally tangent to each of the other two circles. What is the area of triangle
?
Solution 1
Notice that we can find in two different ways:
and
Since we want
we use the latter method, so we have
. Additionally,
. Therefore,
. Similarly,
. We can calculate
easily because
.
.
Plugging into first equation, the two sums of areas, so therefore
Solution 2
Use the Shoelace Theorem.
Let the center of the first circle of radius 1 be at .
Draw the trapezoid and using the Pythagorean Theorem, we get that
so the center of the second circle of radius 2 is at
.
Draw the trapezoid and using the Pythagorean Theorem, we get that
so the center of the third circle of radius 3 is at
.
Now, we may use the Shoelace Theorem!
Solution 3
and
because they are the sum of two radii.
and
, the difference of the radii. Using pythagorean theorem, we find that
and
are
and
,
.
Draw a perpendicular from to line
, then we can use the Pythagorean theorem to find
.
. We get
To make our calculations easier, let . The semi-perimeter of our triangle is
. Symbolize the area of the triangle with
. Using Heron's formula, we have
We can remove the outer root of a.
We solve the nested root. We want to turn into the square of something. If we have
, then we get
Solving the system of equations, we get
and
. Alternatively, you can square all the possible solutions until you find one that is equal to
~ZericH
Solution 4
The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that
can be calculated in two ways:
and
. Solving, we get:
- ColtsFan10, diagram partially borrowed from Solution 1
Solution 5 (Heron’s)
We can use the Pythagorean theorem to find that the lengths are . If we apply Heron’s, we know that it must be the sum (or difference) of two or more square roots, by instinct. This means that
is the answer.
Solution 6 (Educated Guess)
Like Solution 1, we can use the Pythagorean theorem to find and
, which are
and
respectively. Since the only answer choice that has
and
is
, we can make an educated guess that
is the answer.
Solution 7 (Quickest and Easiest Using Trapezoids)
The fastest way to find the area of is by breaking it into shapes with easy-to-calculate areas. We can see that the area of the whole shape
is made up of trapezoids
and
. So,
.
We already know the lengths of and
being
and
respectively (they are the radiuses of the circles). The "height" of the trapezoid is
which can be found from the Pythagorean Theorem. So, the area of trapezoid
. Similarly, we can find the area of trapezoid
.
, so
.
Now we can put the pieces together. . The area of
is the long rectangle
.
is the point on
in which
is perpendicular to
.
So,
~hwan
Video Solutions
https://www.youtube.com/watch?v=sWurLJqf02Y ~by Punxsutawney Phil
https://www.youtube.com/watch?v=UanfIBpDTh8&ab_channel=ArtofProblemSolving ~by the official Art of Problem Solving channel on YouTube
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
2016 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.