Difference between revisions of "2018 AMC 10A Problems/Problem 16"
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− | ==Solution== | + | ==Solution 1== |
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<asy> | <asy> | ||
unitsize(4); | unitsize(4); | ||
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dot("$B$", B, SE); | dot("$B$", B, SE); | ||
dot("$C$", C, NE); | dot("$C$", C, NE); | ||
− | dot("$P$", P, | + | dot("$P$", P, NW); |
</asy> | </asy> | ||
+ | |||
As the problem has no diagram, we draw a diagram. The hypotenuse has length <math>29</math>. Let <math>P</math> be the foot of the altitude from <math>B</math> to <math>AC</math>. Note that <math>BP</math> is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for <math>BP=\dfrac{20\cdot 21}{29}</math>, which is between <math>14</math> and <math>15</math>. | As the problem has no diagram, we draw a diagram. The hypotenuse has length <math>29</math>. Let <math>P</math> be the foot of the altitude from <math>B</math> to <math>AC</math>. Note that <math>BP</math> is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for <math>BP=\dfrac{20\cdot 21}{29}</math>, which is between <math>14</math> and <math>15</math>. | ||
− | 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> distinct line segments. |
(asymptote diagram added by elements2015) | (asymptote diagram added by elements2015) | ||
==Solution 2 - Circles== | ==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 < | + | 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. |
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<asy> | <asy> | ||
unitsize(4); | unitsize(4); | ||
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draw(arc((0,0),15, 90, 180)); | draw(arc((0,0),15, 90, 180)); | ||
</asy> | </asy> | ||
− | 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 | + | |
+ | 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 (since these circles intersect the hypotenuse twice) 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== |
Latest revision as of 13:47, 10 August 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 1
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.
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 ~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.