Solution to Problem 64
Congratulations to this week’s winners
Nathan Pauli, Anna McCullough, Tracy Thatcher
Further correct solutions were also received from Silvain Vinassac, Al Zimmermann, Burkart Venzke,
Philippe Fondanaiche, William Webb.
The easiest solution was presented by Nathan Pauli.
It has the advantage of showing
that the fact that the triangle is a right triangle is irrelevant since that fact is never
used in the solution!
[IMAGE]
Consider the diagram on the right. The fence,
f
,
is in blue; the units for the
lengths of the sides are in hundreds of feet. The area of the entire field is 6 square
units, so the fence is to subdivide the field into parts of 3 square units each.
Look at
the point
C
which is
b
units from the corner labeled
A
.
The fence goes to a point
labeled
B
which is
c
units from
A
.
The angle
CAB
is labeled q; this gives tan(q) =
3/4, sin(q) = 3/5, and cos(q) = 4/5.
Now, the area of triangle
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 Spring '10
 AlbertoDelgado
 Combinatorics, Pythagorean Theorem, Right triangle, Hypotenuse, triangle, Fence, Nathan Pauli

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