Difference between revisions of "2018 AMC 10A Problems/Problem 16"
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Let the line segment be <math>BX</math>, with <math>X</math> on <math>AC</math>. As you move <math>X</math> along the hypotenuse from <math>A</math> to <math>P</math>, the length of <math>BX</math> strictly decreases, hitting all the integer values from <math>20, 19, \dots 15</math> (IVT). Similarly, moving <math>X</math> from <math>P</math> to <math>C</math> hits all the integer values from <math>15, 16, \dots, 21</math>. This is a total of <math>\boxed{(D) 13}</math> line segments. | Let the line segment be <math>BX</math>, with <math>X</math> on <math>AC</math>. As you move <math>X</math> along the hypotenuse from <math>A</math> to <math>P</math>, the length of <math>BX</math> strictly decreases, hitting all the integer values from <math>20, 19, \dots 15</math> (IVT). Similarly, moving <math>X</math> from <math>P</math> to <math>C</math> hits all the integer values from <math>15, 16, \dots, 21</math>. This is a total of <math>\boxed{(D) 13}</math> line segments. | ||
(asymptote diagram added by elements2015) | (asymptote diagram added by elements2015) | ||
+ | |||
+ | ==Solution 2 - Circles== | ||
+ | Note that if a circle with an integer radius <math>r</math> centered at vertex <math>B</math> intersects hypotenuse <math>\overline{AB}</math>, the lines drawn from <math>B</math> to the points of intersection are integer lengths. As in the previous solution, the shortest distance <math>14<\overline{BP}<15</math>. As a result, a circle of <math>14</math> will [b]not[/b] reach the hypotenuse and thus does not intersect it. We also know that a circle of radius <math>21</math> intersects the hypotenuse once and a circle of radius <math>\{15, 16, 17, 18, 19, 20 \}</math> 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(circle(B, 21)); | ||
+ | draw(circle(B, 20)); | ||
+ | draw(circle(B, 19)); | ||
+ | draw(circle(B, 18)); | ||
+ | draw(circle(B, 17)); | ||
+ | draw(circle(B, 16)); | ||
+ | draw(circle(B, 15)); | ||
+ | </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 </math>B<math> to </math>\overline{AC}<math> and one circle of radius </math>21$ 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 22:41, 1 February 2021
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
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(B, 21)); draw(circle(B, 20)); draw(circle(B, 19)); draw(circle(B, 18)); draw(circle(B, 17)); draw(circle(B, 16)); draw(circle(B, 15)); (Error making remote request. Unknown error_msg)
It follows that we can draw circles of radii and 20,B\overline{AC}21$ 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.