Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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==Solution 2== | ==Solution 2== | ||
− | From solution 1, we know that <math>AD = 12</math> and <math>DC = 9</math>. Since <math>\triangle ADC \sim \triangle DEC</math>, we can figure out that <math>DE = \frac{36}{5}</math>. We also know what <math>AC</math> is so we can figure what <math>AE</math> is: <math>AE = 15 - \frac{ | + | From solution 1, we know that <math>AD = 12</math> and <math>DC = 9</math>. Since <math>\triangle ADC \sim \triangle DEC</math>, we can figure out that <math>DE = \frac{36}{5}</math>. We also know what <math>AC</math> is so we can figure what <math>AE</math> is: <math>AE = 15 - \frac{27}{5} = \frac{48}{5}</math> . Quadrilateral <math>ABDF</math> is cyclic, implying that <math>\angle{B} + \angle{DFA}</math> = 180°. Therefore, <math>\angle{B} = 180 - \angle{DFA} = \angle{EFA}</math>, and triangles <math>\triangle AEF \sim \triangle ADB</math>. Solving the resulting proportion gives <math>EF = 4</math>. Therefore, <math>DF = ED - EF = \frac{16}{5}</math>. <math>m + n = 16 + 5 = 21</math> and our answer is <math>\boxed{\textbf{(B) } 21}</math>. |
<math>\triangle ADC \sim \triangle DEC</math> because of <math>AA \sim</math> . <math>\angle{ADC} = \angle{DEC} = 90°</math>. Lets say <math>\angle{ADE} = x</math>. So <math>\angle{EDC} = 90 - x</math> and <math>\angle{DEC} = 180 - 90 - (90 - x) = x</math> so <math>\angle{ADE} = \angle{DEC}</math> | <math>\triangle ADC \sim \triangle DEC</math> because of <math>AA \sim</math> . <math>\angle{ADC} = \angle{DEC} = 90°</math>. Lets say <math>\angle{ADE} = x</math>. So <math>\angle{EDC} = 90 - x</math> and <math>\angle{DEC} = 180 - 90 - (90 - x) = x</math> so <math>\angle{ADE} = \angle{DEC}</math> | ||
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+ | ~[https://artofproblemsolving.com/wiki/index.php/User:South South] | ||
==Solution 3== | ==Solution 3== | ||
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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>. | ||
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+ | ==Solution 5 (Coord Bash)== | ||
+ | If we draw the diagram like above, but make <math>BC</math> the base, we can set <math>B(0,0), C(14,0)</math> and <math>A(5,12)</math>, as <math>AD</math> can quickly be verified to be <math>12</math> by Pythagorean triples or similar triangles. Construct <math>X</math> on <math>BC</math> such that <math>EX \perp BC</math>. This implies <math>\triangle ADC \sim \triangle EXC</math> as <math>AD \parallel EX</math>, and <math> \angle ACD = \angle ECX </math>. Also construct <math>FY</math> such that <math>FY \perp BC</math>. | ||
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+ | Line <math>\overline {AC}</math> has a slope of <math>-\frac{4}{3}</math> by slope formula. Since <math>D = (5,0)</math> and <math>DE \perp AC,</math>, the equation of <math> DE = \frac{3}{4}x - \frac{15}{4}</math>. | ||
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+ | Furthermore, <math>F</math> can now be expressed as <math>(x,\frac{3}{4}x - \frac{15}{4}).</math> Since we know <math>AF \perp BF,</math> we can solve for <math>x</math> with the perpendicular slope formula like so: <cmath>\frac{\frac{3}{4}x - \frac{63}{4}}{x - 5} \cdot \frac{\frac{3}{4}x - \frac {15}{4}}{x} = -1</cmath> <cmath>\frac{x-21}{x-5} \cdot \frac{x-5}{x} = -\frac{16}{9}</cmath> <cmath>\frac{x-21}{x} = -\frac{16}{9}</cmath> <cmath> 9x = 189 -16x </cmath> <cmath>x = \frac{189}{25}</cmath> | ||
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+ | Plugging <math>\frac{189}{25}</math> into <math>F</math>, we get <math>F(\frac{189}{25},\frac{48}{25}).</math> Since <math>FY \perp DY</math>, we get that <math>\triangle FYD</math> has side lengths of <math>FY = \frac{48}{25}</math> | ||
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+ | and <math>DY = BY - BD = BY - 5 = \frac{64}{25}</math>. | ||
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+ | Clearly, <math>\triangle FYD</math> is a <math>3 - 4 - 5</math> pythagorean triple, so <math>DF = \frac{80}{25} = \frac{16}{5}</math>. | ||
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+ | <math>16 + 5 = m + n = \boxed{\textbf{(B) }21}</math>. | ||
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+ | ~JT0543164 | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 22:39, 29 July 2024
- 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
From solution 1, we know that and
. Since
, we can figure out that
. We also know what
is so we can figure what
is:
. Quadrilateral
is cyclic, implying that
= 180°. Therefore,
, and triangles
. Solving the resulting proportion gives
. Therefore,
.
and our answer is
.
because of
.
. Lets say
. So
and
so
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,
.
Solution 5 (Coord Bash)
If we draw the diagram like above, but make the base, we can set
and
, as
can quickly be verified to be
by Pythagorean triples or similar triangles. Construct
on
such that
. This implies
as
, and
. Also construct
such that
.
Line has a slope of
by slope formula. Since
and
, the equation of
.
Furthermore, can now be expressed as
Since we know
we can solve for
with the perpendicular slope formula like so:
Plugging into
, we get
Since
, we get that
has side lengths of
and .
Clearly, is a
pythagorean triple, so
.
.
~JT0543164
Video Solution
https://youtu.be/XZBKnobK-JU?t=3064
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.