Difference between revisions of "2017 AMC 8 Problems/Problem 16"
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− | Since point <math>D</math> is on line <math>BC</math>, it will split it into <math>CD</math> and <math>DB</math>. Let <math>CD = 5 - x</math> and <math>DB = x</math>. Triangle <math>CAD</math> has side lengths <math>3, 5 - x, AD</math> and triangle <math>DAB</math> has side lengths <math>x, 4, AD</math>. Since both perimeters are equal, we have the equation <math>3 + 5 - x + AD = 4 + x + AD</math>. Eliminating <math>AD</math> and solving the resulting linear equation gives <math>x = 2</math>. Draw a perpendicular from point <math>D</math> to <math>AB</math>. Call the point of intersection <math>F</math>. Because angle <math>ABC</math> is common to both triangles <math>DBF</math> and <math>ABC</math>, and both are right triangles, both are similar. The hypotenuse of triangle <math>DBF</math> is 2, so the altitude must be <math>6/5</math> Because <math>DBF</math> and <math>ABD</math> share the same altitude, the height of <math>ABD</math> therefore must be <math>6/5</math>. The base of <math>ABD</math> is 4, so <math> | + | Since point <math>D</math> is on line <math>BC</math>, it will split it into <math>CD</math> and <math>DB</math>. Let <math>CD = 5 - x</math> and <math>DB = x</math>. Triangle <math>CAD</math> has side lengths <math>3, 5 - x, AD</math> and triangle <math>DAB</math> has side lengths <math>x, 4, AD</math>. Since both perimeters are equal, we have the equation <math>3 + 5 - x + AD = 4 + x + AD</math>. Eliminating <math>AD</math> and solving the resulting linear equation gives <math>x = 2</math>. Draw a perpendicular from point <math>D</math> to <math>AB</math>. Call the point of intersection <math>F</math>. Because angle <math>ABC</math> is common to both triangles <math>DBF</math> and <math>ABC</math>, and both are right triangles, both are similar. The hypotenuse of triangle <math>DBF</math> is 2, so the altitude must be <math>6/5</math> Because <math>DBF</math> and <math>ABD</math> share the same altitude, the height of <math>ABD</math> therefore must be <math>6/5</math>. The base of <math>ABD</math> is 4, so <math>[ABD] = \frac{1}{2} * 4 * \frac{6}{5} = \frac{12}{5} \boxed{\textbf{(D) } \frac{12}{5}}</math> |
==See Also== | ==See Also== |
Revision as of 16:31, 22 November 2017
Contents
Problem 16
In the figure below, choose point on
so that
and
have equal perimeters. What is the area of
?
Solution
We know that the perimeters of the two small triangles are and
. Setting both equal and using
, we have
and
. Now, we simply have to find the area of
. Since
, we must have
. Combining this with the fact that
, we get
Solution 2
Since point is on line
, it will split it into
and
. Let
and
. Triangle
has side lengths
and triangle
has side lengths
. Since both perimeters are equal, we have the equation
. Eliminating
and solving the resulting linear equation gives
. Draw a perpendicular from point
to
. Call the point of intersection
. Because angle
is common to both triangles
and
, and both are right triangles, both are similar. The hypotenuse of triangle
is 2, so the altitude must be
Because
and
share the same altitude, the height of
therefore must be
. The base of
is 4, so
See Also
2017 AMC 8 (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 AJHSME/AMC 8 Problems and Solutions |
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