Difference between revisions of "2018 AMC 10A Problems/Problem 16"
Ishankhare (talk | contribs) (Created page with "Right triangle <math>ABC</math> has leg lengths <math>AB=20</math> and <math>BC=21</math>. Including <math>\overline{AB}</math> and <math>\overline{BC}</math>, how many line s...") |
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\textbf{(E) }15 \qquad | \textbf{(E) }15 \qquad | ||
</math> | </math> | ||
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+ | ==Solution== | ||
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+ | 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>. | ||
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+ | 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. |
Revision as of 14:11, 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.