Difference between revisions of "1992 AIME Problems/Problem 9"
Kevinr9828 (talk | contribs) (→Solution 2) |
Kevinr9828 (talk | contribs) (→Solution 1) |
||
Line 12: | Line 12: | ||
Then <math>h^2 +z^2 =70^2</math> and <math>(73-z)^2 + h^2 =50^2</math> so <math>h =\frac{\sqrt{44710959}}{146}</math> | Then <math>h^2 +z^2 =70^2</math> and <math>(73-z)^2 + h^2 =50^2</math> so <math>h =\frac{\sqrt{44710959}}{146}</math> | ||
− | now substitute this into <math>(1)</math> to get <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>. | + | now substitute this into <math>(1)</math> to get <math>x= \frac{11753}{219} = \frac{161}{3}</math> and <math>m+n = 164</math>. |
+ | |||
+ | you don;t have to use trig nor angles A and B ..From similar triangles, | ||
+ | h/r = 70/x and h/r = 50/ (92-x) | ||
+ | |||
+ | this implies that 70/x =50/(92-x) so x = 161/3 | ||
== Solution 2 == | == Solution 2 == |
Revision as of 15:53, 23 June 2008
Contents
Problem
Trapezoid has sides
,
,
, and
, with
parallel to
. A circle with center
on
is drawn tangent to
and
. Given that
, where
and
are relatively prime positive integers, find
.
Solution 1
Let be the base of the trapezoid and consider angles
and
. Let
and let
equal the height of the trapezoid. Let
equal the radius of the circle.
Then
and
Let be the distance along
from
to where the perp from
meets
.
Then and
so
now substitute this into
to get
and
.
you don;t have to use trig nor angles A and B ..From similar triangles,
h/r = 70/x and h/r = 50/ (92-x)
this implies that 70/x =50/(92-x) so x = 161/3
Solution 2
From above,
and
. Adding these equations yields
. Thus,
, and
.
from solution 1 we get from 1 that h/r = 70/x and h/r = 50/ (92-x)
this implies that 70/x =50/(92-x) so x = 161/3
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |