2009 AMC 10B Problems/Problem 22
Contents
Problem
A cubical cake with edge length inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where
is the midpoint of a top edge. The piece whose top is triangle
contains
cubic inches of cake and
square inches of icing. What is
?
Solution
Let's label the points as in the picture above. Let be the area of
. Then the volume of the corresponding piece is
. This cake piece has icing on the top and on the vertical side that contains the edge
. Hence the total area with icing is
. Thus the answer to our problem is
, and all we have to do now is to determine
.
Solution 1
Introduce a coordinate system where ,
and
.
In this coordinate system we have , and the line
has the equation
.
As the line is orthogonal to
, it must have the equation
for some suitable constant
. As this line contains the point
, we have
.
Substituting into
, we get
, and then
.
We can note that in
is the height from
onto
, hence its area is
, and therefore the answer is
.
Solution 2
Extend to intersect
at
:
It is now obvious that is the midpoint of
. (Imagine rotating the square
by
clockwise around its center. This rotation will map the segment
to a segment that is orthogonal to
, contains
and contains the midpoint of
.)
From we can compute that
.
Observe that and
have the same angles and therefore they are similar. The ratio of their sides is
.
Hence we have , and
.
Knowing this, we can compute the area of as
.
Finally, we compute and conclude that the answer is
- You could also notice that the two triangles in the original figure are similar.
Solution 3
Use trigonometry.
The length of and
is
and
respectively. So
, and
.
From the right-angled triangle , the hypotenuse,
So
, and
Knowing this, . So we proceed as follows:
So the answer is .
Note that we didn't use a calculator, but we used trigonometric identities
Solution 4 (Pythagorean Theorem only)
Since and
, we know that
. If we let
, then
. Now, by the Pythagorean Theorem, we have:
Expanding and rearranging the second equation gives:
Since , we have that:
Knowing , we can solve for the height
:
Therefore, the area of triangle is
. Since the solution to the problem is
, the answer is
.
Solution 4b
QM=\sqrt{1^{2}+2^{2}}=sqrt{5}
since \angle PMQ = 90 =
\angle MQR
therefore \angle MQR
and because \angle QNR = 90
therefore and
are similar
therefore \frac {QR}{QM}\ = \frac {RN}{PQ}\ = \frac {QN}{PM}\
\frac {2}{\sqrt{5}\}\ = \frac {RN}{2}\ = \frac {QN}{1}\
RN = \frac {4}{\sqrt{5}\}\, QN = \frac {2}{\sqrt{5}\}\
Therefore, the area of triangle is
. Since the solution to the problem is
, the answer is
.
See Also
2009 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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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