Difference between revisions of "2018 AMC 10A Problems/Problem 16"
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Right triangle <math>ABC</math> has leg lengths <math>AB=20</math> and <math>BC=21</math>. Including <math>\overline{AB}</math> and <math>\overline{BC}</math>, how many line segments with integer length can be drawn from vertex <math>B</math> to a point on hypotenuse <math>\overline{AC}</math>? | Right triangle <math>ABC</math> has leg lengths <math>AB=20</math> and <math>BC=21</math>. Including <math>\overline{AB}</math> and <math>\overline{BC}</math>, how many line segments with integer length can be drawn from vertex <math>B</math> to a point on hypotenuse <math>\overline{AC}</math>? | ||
Latest revision as of 18:14, 31 October 2024
Contents
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 1
![[asy] 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, NW); [/asy]](http://latex.artofproblemsolving.com/0/e/f/0ef5ecee03d86a1f795fcb2c11283dd1a2d4017d.png)
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 
distinct 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 not 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.
![[asy] 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(arc((0,0),21, 90, 180)); draw(arc((0,0),20, 90, 180)); draw(arc((0,0),19, 90, 180)); draw(arc((0,0),18, 90, 180)); draw(arc((0,0),17, 90, 180)); draw(arc((0,0),16, 90, 180)); draw(arc((0,0),15, 90, 180)); [/asy]](http://latex.artofproblemsolving.com/0/b/8/0b81d2900bccc7870e731abd49af60e93adcb37d.png)
It follows that we can draw circles of radii 
and 
that each contribute two integer lengths (since these circles intersect the hypotenuse twice) from to and one circle of radius that contributes only one such segment. Our answer is then ![\[6 \cdot 2 + 1 = 13 \implies \boxed{D}\]](http://latex.artofproblemsolving.com/5/1/e/51ecfade984a88a7670f07dec953eed79302951f.png)
~samrocksnature
Video Solution 1(The Beauty of Math)
~IceMatrix
Video Solution 2 by OmegaLearn
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. 
