Difference between revisions of "2016 AMC 10A Problems/Problem 21"
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<math>= \sqrt{6}-\sqrt{2}</math> <math>\fbox{D}</math>. | <math>= \sqrt{6}-\sqrt{2}</math> <math>\fbox{D}</math>. | ||
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+ | ==Solution 3== | ||
+ | <math>PQ = 3</math> and <math>QR = 5</math> because they are the sum of two radii. <math>QQ' - PP' = 1</math> and <math>RR' - QQ' = 1</math>, the difference of the radii. Using pythagorean theorem, we find that <math>P'Q'</math> and <math>Q'R'</math> are <math>\sqrt{8}</math> and <math>\sqrt{24}</math>, <math>P'R' = \sqrt{8} + \sqrt{24}</math>. | ||
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+ | Draw a perpendicular from <math>P</math> to line <math>RR'</math>, then we can use the Pythagorean theorem to find <math>PR</math>. <math>RR' - PP' = 2</math>. We get <cmath>PR^2 = (\sqrt{8} + \sqrt{24})^2 + 4 = 36 + 16\sqrt{3} \Rightarrow PR = \sqrt{36 + 16\sqrt{3}} = 2\sqrt{9 + 4\sqrt{3}}</cmath> | ||
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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> | ||
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+ | 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>. <cmath>A = \sqrt{8 - 4\sqrt{3}} = \sqrt{(\sqrt{6} - \sqrt{2})^2} = \sqrt{6} - \sqrt{2} \rightarrow \fbox{D}</cmath> | ||
+ | ~ZericH | ||
==See Also== | ==See Also== |
Revision as of 11:27, 12 February 2019
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 ?
Contents
Solution 1
Notice that we can find in two different ways: and , so
. Additionally, . Therefore, . Similarly, . We can calculate easily because . .
Plugging into first equation, the two sums of areas, .
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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
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 |
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