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How
to Write a Solution
A Picture is Worth a Thousand Words
by
Richard Rusczyk
When you're writing a solution to a geometry problem, or any problem involving a picture, you should include the diagram. If you don't include the diagram, you often make the grader have to draw it for you. Even if the diagram is given in the problem, you should include it in your solution. If you make reader go looking somewhere else for a diagram, you are very likely to lose their attention.
Draw your diagram precisely. Use a geometry rendering program if you are typesetting your solution, or use a ruler and compass if you are writing your solution by hand.
Here's an example:
Here's a solution without a diagram.
We will show that AK = (AB + AC - BC)/2, and thus show that the length of AK is independent of D.
Since tangents from a point to a circle are equal, we have both DN = DL and DN' = DL'. Thus,
NN' = DN + DN' = DL + DL' = 2DL + LL'.
Similarly, we have
MM' = MK + KM' = KL + KL' = 2KL + LL'.
Since MM' = NN' by symmetry, we conclude that KL = DL. Hence,
MM' = NN' = DL + LL' + KL = KD.
We can compute NN', and hence KD, in terms of AD and the sides of the triangle:
DN
= (AB + AD + BD)/2 - AB
DN' = (AC + AD + CD)/2 - AC
Thus,
NN' = ND + DN' = AD + BC/2 - AC/2 - AB/2.
Since NN' = KD, we have
AD - KD = AK = (AC + AB - BC)/2,
as desired. Since A, B, and C are independent of D, we conclude that the length AK is independent of D.
Click here for a solution with a diagram.
In our solution above, we used the fact that the length of the segments from a vertex of a triangle (like vertex D of triangle ABD above) to the points of tangency of the inscribed circle with the sides of the triangle from that vertex (segments DN and DL above) equals half the perimeter of the triangle minus the opposite side of the triangle. Applying this principle to find length DN in triangle ABD gives us:
DN = (AB + AD + BD)/2 - AB
If you aren't familiar with this fact, try to prove it yourself (and write a nice solution). Every good geometer reaches for this fact as easily as she reaches for the Pythagorean Theorem.
