Difference between revisions of "2018 AMC 10A Problems/Problem 16"
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Line 48: | Line 48: | ||
dot("$C$", C, NE); | dot("$C$", C, NE); | ||
dot("$P$", P, S);] | dot("$P$", P, S);] | ||
− | draw(circle( | + | draw(circle((0,0), 21)); |
− | draw(circle( | + | draw(circle((0,0), 20)); |
− | draw(circle( | + | draw(circle((0,0), 19)); |
− | draw(circle( | + | draw(circle((0,0), 18)); |
− | draw(circle( | + | draw(circle((0,0), 17)); |
− | draw(circle( | + | draw(circle((0,0), 16)); |
− | draw(circle( | + | draw(circle((0,0), 15)); |
</asy> | </asy> | ||
− | It follows that we can draw circles of radii <math>15, 16, 17, 18, 19,</math> and 20,<math> that each contribute [b]two[/b] integer lengths from < | + | It follows that we can draw circles of radii <math>15, 16, 17, 18, 19,</math> and <math>20,</math> that each contribute [b]two[/b] integer lengths from <math>B</math> to <math>\overline{AC}</math> and one circle of radius <math>21</math> that contributes only one such segment. Our answer is then <cmath>6 \cdot 2 + 1 = 13 \implies \boxed{D}</cmath> ~samrocksnature |
==Video Solution 1== | ==Video Solution 1== |
Revision as of 22:42, 1 February 2021
Contents
[hide]Problem
Right triangle has leg lengths and . Including and , how many line segments with integer length can be drawn from vertex to a point on hypotenuse ?
Solution
As the problem has no diagram, we draw a diagram. The hypotenuse has length . Let be the foot of the altitude from to . Note that is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for , which is between and .
Let the line segment be , with on . As you move along the hypotenuse from to , the length of strictly decreases, hitting all the integer values from (IVT). Similarly, moving from to hits all the integer values from . This is a total of line segments. (asymptote diagram added by elements2015)
Solution 2 - Circles
Note that if a circle with an integer radius centered at vertex intersects hypotenuse , the lines drawn from to the points of intersection are integer lengths. As in the previous solution, the shortest distance . As a result, a circle of will [b]not[/b] reach the hypotenuse and thus does not intersect it. We also know that a circle of radius intersects the hypotenuse once and a circle of radius intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.
unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$P$", P, S);] draw(circle((0,0), 21)); draw(circle((0,0), 20)); draw(circle((0,0), 19)); draw(circle((0,0), 18)); draw(circle((0,0), 17)); draw(circle((0,0), 16)); draw(circle((0,0), 15)); (Error making remote request. Unknown error_msg)
It follows that we can draw circles of radii and that each contribute [b]two[/b] integer lengths from to and one circle of radius that contributes only one such segment. Our answer is then ~samrocksnature
Video Solution 1
~IceMatrix
Video Solution 2
https://youtu.be/4_x1sgcQCp4?t=3790
~ pi_is_3.14
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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.