Difference between revisions of "2013 AMC 12B Problems/Problem 19"
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− | Line <math>\overline {AC}</math> has a slope of <math>-4 | + | 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> | ||
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> | 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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<math>16 + 5 = m + n = \boxed{\textbf{(B) }21}</math>. | <math>16 + 5 = m + n = \boxed{\textbf{(B) }21}</math>. | ||
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− | ==Video Solution== | + | ~JT0543164 |
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+ | ==Video Solution by Pi Academy== | ||
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+ | https://youtu.be/2wrOFzxYCcM?si=J-u-_5Yb8o3hXrAp | ||
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+ | ~ Pi Academy | ||
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+ | ==Video Solution 2== | ||
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+ | https://www.youtube.com/watch?v=X0YJfFiBy0g | ||
+ | |||
https://youtu.be/XZBKnobK-JU?t=3064 | https://youtu.be/XZBKnobK-JU?t=3064 | ||
Latest revision as of 21:24, 10 October 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
[hide]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 by Pi Academy
https://youtu.be/2wrOFzxYCcM?si=J-u-_5Yb8o3hXrAp
~ Pi Academy
Video Solution 2
https://www.youtube.com/watch?v=X0YJfFiBy0g
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.