Difference between revisions of "2013 AMC 12B Problems/Problem 19"
(→Solution 1) |
(→Solution 1) |
||
Line 40: | Line 40: | ||
Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>, so <math>\triangle ABF \sim \triangle ADE</math> are similar. In addition, <math>\triangle ADE \sim \triangle ACD</math>. We can easily find <math>AD=12</math>, <math>BD = 5</math>, and <math>DC=9</math> using Pythagorean triples. | Since <math>\angle{AFB}=\angle{ADB}=90^{\circ}</math>, quadrilateral <math>ABDF</math> is cyclic. It follows that <math>\angle{ADE}=\angle{ABF}</math>, so <math>\triangle ABF \sim \triangle ADE</math> are similar. In addition, <math>\triangle ADE \sim \triangle ACD</math>. We can easily find <math>AD=12</math>, <math>BD = 5</math>, and <math>DC=9</math> using Pythagorean triples. | ||
− | So, the ratio of the longer leg to the hypotenuse of all three similar triangles is <math>\tfrac{12}{15} = \tfrac{4}{5}</math>, and the ratio of the shorter leg to the hypotenuse is <math>\tfrac{9}{15} = \tfrac{3}{5}</math>. It follows that <math>AF= | + | So, the ratio of the longer leg to the hypotenuse of all three similar triangles is <math>\tfrac{12}{15} = \tfrac{4}{5}</math>, and the ratio of the shorter leg to the hypotenuse is <math>\tfrac{9}{15} = \tfrac{3}{5}</math>. It follows that <math>AF=(\tfrac{4}{5}\cdot 13), BF=(\tfrac{3}{5}\cdot 13)</math>. |
− | Let <math>x=DF</math>. By [[Ptolemy's Theorem|Ptolemy's Theorem]], we have < | + | Let <math>x=DF</math>. By [[Ptolemy's Theorem|Ptolemy's Theorem]], we have <cmath>13x+\left(5\cdot 13\cdot \frac{4}{5}\right)= 12\cdot 13\cdot \frac{3}{5} \qquad \Leftrightarrow \qquad 13x+52=93.6.</cmath> Dividing by <math>13</math> we get <math>x+4=7.2\implies x=\frac{16}{5}</math> so our answer is <math>\boxed{\textbf{(B) }21}</math>. |
~Edits by BakedPotato66 | ~Edits by BakedPotato66 |
Revision as of 23:27, 16 October 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
?
Diagram
Solution 1
Since , quadrilateral
is cyclic. It follows that
, so
are similar. In addition,
. We can easily find
,
, and
using Pythagorean triples.
So, the ratio of the longer leg to the hypotenuse of all three similar triangles is , and the ratio of the shorter leg to the hypotenuse is
. It follows that
.
Let . By Ptolemy's Theorem, we have
Dividing by
we get
so our answer is
.
~Edits by BakedPotato66
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.