Difference between revisions of "2018 AMC 10A Problems/Problem 16"
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dot("$B$", B, SE); | dot("$B$", B, SE); | ||
dot("$C$", C, NE); | dot("$C$", C, NE); | ||
− | dot("$P$", P, S); | + | dot("$P$", P, S); |
draw(Circle((0,0), 21)); | draw(Circle((0,0), 21)); | ||
draw(Circle((0,0), 20)); | draw(Circle((0,0), 20)); |
Revision as of 23:45, 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.
It follows that we can draw circles of radii
and
that each contribute [b]two[/b] integer lengths 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.