Difference between revisions of "2015 AMC 10B Problems/Problem 13"
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==Solution 1== | ==Solution 1== | ||
− | We find the <math>x</math>-intercepts and the <math>y</math>-intercepts to find the intersections of the axes and the line. If <math>x=0</math>, then <math>y=12</math>. If <math>y</math> is <math>0</math>, then <math>x=5</math>. Our three vertices are <math>(0,0)</math>, <math>(5,0)</math>, and <math>(0,12)</math>. Two of our altitudes are <math>5</math> and <math>12</math>, and since it is a <math>5</math>-<math>12</math>-<math>13</math> right triangle, the hypotenuse is <math>13</math>. Since the area of the triangle is <math>30</math>, so our final altitude is <math>\frac{30(2)}{13}=\frac{60}{13}</math>. The sum of our altitudes is <math>\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}</math>. Note that there is no need to calculate the final answer after we know that the third altitude has length <math>\frac{60}{13}</math> since <math>E</math> is the only choice with a denominator of <math>13</math>. | + | We find the <math>x</math>-intercepts and the <math>y</math>-intercepts to find the intersections of the axes and the line. If <math>x=0</math>, then <math>y=12</math>. If <math>y</math> is <math>0</math>, then <math>x=5</math>. Our three vertices are <math>(0,0)</math>, <math>(5,0)</math>, and <math>(0,12)</math>. Two of our altitudes are <math>5</math> and <math>12</math>, and since it is a <math>5</math>-<math>12</math>-<math>13</math> right triangle, the hypotenuse is <math>13</math>. Since the area of the triangle is <math>30</math>, so our final altitude is <math>\frac{30(2)}{13}=\frac{60}{13}</math>. The sum of our altitudes is <math>\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}</math>. Note that there is no need to calculate the final answer after we know that the third altitude has length <math>\frac{60}{13}</math> since <math>E</math> is the only choice with a denominator of <math>13</math> and <math>13</math> is relatively prime to <math>5</math> and <math>12</math>. |
+ | |||
+ | ==Video Solution 1== | ||
+ | https://youtu.be/zFHtdWEhriQ | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Solution 2 (very similar to Solution 1)== | ==Solution 2 (very similar to Solution 1)== | ||
− | Noticing that the line has coefficients <math>12</math> and <math>5</math>, we can suspect that we have a <math>5</math>-<math>12</math>-<math>13</math> triangle on our hands. Because we want the sum of the altitudes, the third altitude that is not an axis must have a denominator of <math>13</math>. Since the other altitudes are integers, we choose the option with | + | Noticing that the line has coefficients <math>12</math> and <math>5</math>, we can suspect that we have a <math>5</math>-<math>12</math>-<math>13</math> triangle on our hands. Because we want the sum of the altitudes, the third altitude that is not an axis must have a denominator of <math>13</math>. Since the other altitudes are integers, we choose the option with <math>13</math> as the denominator, namely <math>E</math>. |
==See Also== | ==See Also== |
Latest revision as of 17:14, 2 August 2022
Contents
Problem
The line forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?
Solution 1
We find the -intercepts and the
-intercepts to find the intersections of the axes and the line. If
, then
. If
is
, then
. Our three vertices are
,
, and
. Two of our altitudes are
and
, and since it is a
-
-
right triangle, the hypotenuse is
. Since the area of the triangle is
, so our final altitude is
. The sum of our altitudes is
. Note that there is no need to calculate the final answer after we know that the third altitude has length
since
is the only choice with a denominator of
and
is relatively prime to
and
.
Video Solution 1
~Education, the Study of Everything
Solution 2 (very similar to Solution 1)
Noticing that the line has coefficients and
, we can suspect that we have a
-
-
triangle on our hands. Because we want the sum of the altitudes, the third altitude that is not an axis must have a denominator of
. Since the other altitudes are integers, we choose the option with
as the denominator, namely
.
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 |
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