Difference between revisions of "2018 AMC 10A Problems/Problem 16"
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==Solution== | ==Solution== | ||
− | + | [asy] | |
− | 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>. | + | 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("<math>A</math>", A, SW); | ||
+ | dot("<math>B</math>", B, SE); | ||
+ | dot("<math>C</math>", C, NE); | ||
+ | dot("<math>P</math>", P, S); | ||
+ | [/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>. | ||
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) | ||
==See Also== | ==See Also== |
Revision as of 19:47, 3 March 2018
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
[asy]
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, SW);
dot("
", B, SE);
dot("
", C, NE);
dot("
", P, S);
[/asy]
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)
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.