2013 AMC 12B Problems/Problem 19
- 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.