2016 AMC 12A Problems/Problem 15
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 Solution 1 by Pi Academy
https://youtu.be/tMVGfNSwLq0?si=TL8gtFQr3xC1Hg1R
~ Pi Academy
Video Solution 2
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 |
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