1993 AIME Problems/Problem 13
Problem
Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centered midway between the paths. At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let be the amount of time, in seconds, before Jenny and Kenny can see each other again. If
is written as a fraction in lowest terms, what is the sum of the numerator and denominator?
Contents
[hide]Solution
Solution 1
Consider the unit cicle of radius 50. Assume that they start at points and
Then at time
, they end up at points
and
The equation of the line connecting these points and the equation of the circle are
When they see each other again, the line connecting the two points will be tangent to the circle at the point
Since the radius is perpendicular to the tangent we get
or
Now substitute
into
and get
Now substitute this and
into
and solve for
to get
Finally, the sum of the numerator and denominator is
Solution 2
Let and
be Kenny's initial and final points respectively and define
and
similarly for Jenny. Let
be the center of the building. Also, let
be the intersection of
and
. Finaly, let
and
be the points of tangency of circle
to
and
respectively.
![[asy] size(8cm); defaultpen(linewidth(0.7)); pair A,B,C,D,P,Q,O,X; A=(0,0); B=(0,160); C=(200,0); D=(200,53.333); P=(100,0); Q=(123.529,94.118); O=(100,50); X=(300,0); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(O); dot(X); draw(A--B--X--cycle); draw(C--D); draw(P--O--Q); draw(O--X); draw(Circle(O,50)); label("$A$",A,SW); label("$B$",B,NNW); label("$C$",C,S); label("$D$",D,NE); label("$P$",P,S); label("$Q$",Q,NE); label("$O$",O,W); label("$X$",X,ESE); [/asy]](http://latex.artofproblemsolving.com/e/8/5/e854091993a5c5a12a513b87e411d392cda1b333.png)
From the problem statement, , and
. Since
,
.
Since ,
. So,
.
Since circle is tangent to
and
,
is the angle bisector of
.
Thus, .
Therefore, , and the answer is
.
Solution 3
Let be the time they walk. Then
and
.
Draw a line from point to
such that
is perpendicular to
. Further, draw a line passing through points
and
, so
is parallel to
and
and is midway between those two lines. Then
. Draw another line passing through point
and parallel to
, and call the point of intersection of this line with
as
. Then
.
We see that since they are corresponding angles, and thus by angle-angle similarity,
.
Then
And we obtain
so we have , and our answer is thus
.
Solution 4
We can use areas to find the answer. Since Jenny and Kenny are 200 feet apart, we know that they are side by side, and that the line connecting the two of them is tangent to the circular building.
We know that the radius of the circle is 50, and that ,
.
Illustration:
By areas, . Having right trapezoids,
. The other areas of right trapezoids can be calculated in the same way. We just need to find
in terms of
.
If we bring up to where the point J is, we have by the Pythagorean Theorem,
.
Now we have everything to solve for .
After isolating the radical, dividing by 50, and squaring, we obtain: .
Since Jenny walks feet at 1 foot/s, our answer is
.
Solution 5
Consider our diagram here, where Jenny goes from
to
and Kenny goes from
to
. Really what we are asking is the length of
, knowing that
and
are tangents and
is perpendicular to the parallel lines. Draw lines
and
as shown such that they are parallel to
and
and are tangent to the circle. Additionally, let
be the median of trapezoid
. Notice how
and
as well, so
and
are medians of trapezoids
and
, respectively. If we set
and
, then
so
and
.
Now we will find in two different ways. For the first way, notice that
has its inscribed circle, so
. This means that
. However, if we let
be the altitude from
to
, then notice how
. Since
, then by the Pythagorean Theorem, we also have that
. After that, then we have an equation in
, which is
Squaring both sides, it expands to
and since
is nonzero, then once we cancel
from both sides, we can divide
and get the new equation
Because Jenny travels this at
foot per second, then it will take
seconds, so our answer is
.
~~ethanzhang1001
Solution 6
Let the point where Jenny started be . Then let
the point Jenny is when she can see Kenny. Similarly, let
be where Kenny started, and let
the point Kenny is when she can see Jenny. Then the equation of the line passing through
is
. Now, note that the center of the building is at
, and the distance from
to
is
, so, using the distance from a point to a line formula, we get
which simplifies to
from clearing denominators, squaring to remove absolute value and radical, then cancelling and rearranging. But this is equivalent to
, so our answer is
.
See also
1993 AIME (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 | ||
All AIME Problems and Solutions |
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