Difference between revisions of "2016 AMC 10A Problems/Problem 19"
(→Solution 2) |
|||
Line 24: | Line 24: | ||
</asy> | </asy> | ||
− | Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2} | + | Since <math>\triangle APD \sim \triangle EPB,</math> <math>\frac{DP}{PB}=\frac{AD}{BE}=3.</math> Similarly, <math>\frac{DQ}{QB}=\frac{3}{2}</math>. This means that <math>{DQ}=\frac{3\cdot BD}{5}</math>. As <math>\triangle ADP</math> and <math>\triangle BEP</math> are similar, we see that <math>\frac{PD}{PB}=\frac{3}{1}</math>. Thus <math>PB=\frac{BD}{4}</math>. Therefore, <math>r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,</math> so <math>r+s+t=\boxed{\textbf{(E) }20.}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 17:23, 19 February 2016
Contents
[hide]Problem
In rectangle
and
. Point
between
and
, and point
between
and
are such that
. Segments
and
intersect
at
and
, respectively. The ratio
can be written as
where the greatest common factor of
and
is
What is
?
Solution 1
Since
Similarly,
. This means that
. As
and
are similar, we see that
. Thus
. Therefore,
so
Solution 2
Coordinate Bash:
We can set coordinates for the points. and
. The line
's equation is
, line
's equation is
, and line
's equation is
. Adding the equations of lines
and
, we find that the coordinates of
is
. Furthermore we find that the coordinates
is
. Using the Pythagorean Theorem, the length of
is
, and the length of
=
The length of
. Then
The ratio
Then
and
is
and
, respectively. The problem tells us to find
, so
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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.