Difference between revisions of "2016 AMC 10A Problems/Problem 21"
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Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>? | Circles with centers <math>P, Q</math> and <math>R</math>, having radii <math>1, 2</math> and <math>3</math>, respectively, lie on the same side of line <math>l</math> and are tangent to <math>l</math> at <math>P', Q'</math> and <math>R'</math>, respectively, with <math>Q'</math> between <math>P'</math> and <math>R'</math>. The circle with center <math>Q</math> is externally tangent to each of the other two circles. What is the area of triangle <math>PQR</math>? | ||
Revision as of 23:56, 15 February 2019
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
, so
. Additionally,
. Therefore,
. Similarly,
. We can calculate
easily because
.
.
Plugging into first equation, the two sums of areas, .
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.