Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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Points <math>A</math>, <math>B</math>, <math>D</math>, and <math>F</math> all lie on a circle whose diameter is <math>AB</math>. Let the point where the circle intersects <math>AC</math> be <math>G</math>. Using power of a point, we can write the following equation to solve for <math>AG</math>: <cmath>DC\cdot BC = CG\cdot AC</cmath> <cmath>9\cdot 14 = CG\cdot 15</cmath> <cmath>CG = 126/15</cmath> Using that, we can find <math>AG = \frac{99}{15}</math>, and using <math>AG</math>, we can find that <math>GE = 3</math>. | Points <math>A</math>, <math>B</math>, <math>D</math>, and <math>F</math> all lie on a circle whose diameter is <math>AB</math>. Let the point where the circle intersects <math>AC</math> be <math>G</math>. Using power of a point, we can write the following equation to solve for <math>AG</math>: <cmath>DC\cdot BC = CG\cdot AC</cmath> <cmath>9\cdot 14 = CG\cdot 15</cmath> <cmath>CG = 126/15</cmath> Using that, we can find <math>AG = \frac{99}{15}</math>, and using <math>AG</math>, we can find that <math>GE = 3</math>. | ||
− | We can use power of a point again to solve for <math>DF</math>: <cmath>FE\cdot DE = GE\cdot AE</cmath> <cmath>(\frac{36}{5} – DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}</cmath> <cmath>\frac{36}{5} – DF = 4</cmath> <cmath>DF = \frac{16}{5} = \frac{m}{n}</cmath> Thus, <math>m+n = 16+5 = 21</math> <math>\boxed{\textbf{(B)}}</math>. | + | We can use power of a point again to solve for <math>DF</math>: <cmath>FE\cdot DE = GE\cdot AE</cmath> <cmath>(\frac{36}{5} – DF)\cdot \frac{36}{5} = 3 \cdot \frac{48}{5}</cmath> <cmath>\frac{36}{5} – DF = 4</cmath> <cmath>DF = \frac{16}{5} = \frac{m}{n}</cmath> Thus, <math>m+n = 16+5 = 21</math> <math>\boxed{\textbf{(B)}}</math>. |
== See also == | == See also == |
Revision as of 20:15, 28 January 2021
- The following problem is from both the 2013 AMC 12B #19 and 2013 AMC 10B #23, so both problems redirect to this page.
Contents
Problem
In triangle ,
,
, and
. Distinct points
,
, and
lie on segments
,
, and
, respectively, such that
,
, and
. The length of segment
can be written as
, where
and
are relatively prime positive integers. What is
?
Solution 1
Since , quadrilateral
is cyclic. It follows that
. In addition, triangles
and
are similar, and triangles
and
are similar. We can easily find
,
, and
using pythagorean triples. So, the ratio of the hypotenuse to the longer leg of all three similar triangles is
, and the ratio of the hypotenuse to the shorter leg is
. It follows that
. By Ptolemy's Theorem, we have
where
. Dividing by
we get
so our answer is
.
Solution 2
Using the similar triangles in triangle gives
and
. Quadrilateral
is cyclic, implying that
= 180°. Therefore,
, and triangles
and
are similar. Solving the resulting proportion gives
. Therefore,
and our answer is
.
Solution 3
If we draw a diagram as given, but then add point on
such that
in order to use the Pythagorean theorem, we end up with similar triangles
and
. Thus,
and
, where
is the length of
. Using the Pythagorean theorem, we now get
and
can be found out noting that
is just
through base times height (since
, similar triangles gives
), and that
is just
. From there,
Now,
, and squaring and adding both sides and subtracting a 169 from both sides gives
, so
. Thus, the answer is
.
Solution 4 (Power of a Point)
First, we find ,
, and
via the Pythagorean Theorem or by using similar triangles. Next, because
is an altitude of triangle
,
. Using that, we can use the Pythagorean Theorem and similar triangles to find
and
.
Points ,
,
, and
all lie on a circle whose diameter is
. Let the point where the circle intersects
be
. Using power of a point, we can write the following equation to solve for
:
Using that, we can find
, and using
, we can find that
.
We can use power of a point again to solve for :
Thus,
.
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 |
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.