Difference between revisions of "2018 AMC 10A Problems/Problem 16"
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Elements2015 (talk | contribs) (→Solution) |
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==Solution== | ==Solution== | ||
| − | + | <asy> | |
pair A, B, C, E, P; | pair A, B, C, E, P; | ||
A=(-20, 0); | A=(-20, 0); | ||
| Line 19: | Line 19: | ||
draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
draw(B--P); | draw(B--P); | ||
| − | dot(" | + | dot("$A$", A, SW); |
| − | dot(" | + | dot("$B$", B, SE); |
| − | dot(" | + | dot("$C$", C, NE); |
| − | dot(" | + | dot("$P$", 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>. | 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>. | ||
Revision as of 18:49, 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("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$P$", P, S); [/asy]](http://latex.artofproblemsolving.com/b/d/9/bd9de3056615c94374c482055a7327326f575c53.png)
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. 
