2018 AMC 10A Problems/Problem 16: Difference between revisions
No edit summary |
|||
| Line 14: | Line 14: | ||
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>. 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{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>. 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{13}</math> line segments. | ||
==See Also== | |||
{{AMC10 box|year=2018|ab=A|num-b=9|num-a=11}} | |||
{{MAA Notice}} | |||
Revision as of 16:38, 8 February 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
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
. Similarly, moving
from
to
hits all the integer values from
. This is a total of
line segments.
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America.