Triangle problem - solution

This is a solution to the triangle problem I posted here.

The picture is below; we are to find the area of the triangle.

Since AC is tangent to the circle at D and AB is tangent to the circle at E, then the distances AD and AE are equal. That is, AE must be 2.

I can't remember nor find a name for this theorem. It has to do with two tangents from the same point (and it is easy to prove using congruent triangles), saying that those distances are congruent.

Similarly, CD and CF are congruent -- both are 3. And also by the same reasoning, BE = BF (because BA and BC are tangets to the circle).

I will call BE and BF as x, it's easier to manipulate in an equation.

Now, we can solve x fairly easily by applying the Pythagorean theorem to the right triangle ABC.

(x + 2)2 + (x + 3)2 = (2 + 3)2

x2 + 4x + 4 + x2 + 6x + 9 = 25

2x2 + 10x + 13 = 25

2x2 + 10x − 12 = 0

x2 + 5x − 6 = 0

x = (−5 ± √( 25 − 4(−6))) / 2

x = (−5 ± 7) / 2

x = − 6 or x = 1.  Obviously we choose x = 1.

Then, we go on to solve the area of the triangle. The two legs of the triangle are thus 3 and 4, and so the area is 3 × 4 / 2 = 6 square units.

All done!


Cool said…
Wow! That looks very complicated. It's great to find resources like this online to help solve problems. Thank you!
Wally said…
It looks complicated BUT is very simple IF one knows about Special Right Triangles.

Since this right triangle has a hypotenuse of 5 the legs MUST BE 3 and 4.

A=6 sq units
David Chandler said…
I solved it with the same general technique, but I used a and b instead of specific lengths 2 and 3 along the hypotenuse. The answer is always ab. The algebra is monstrous in the middle, but in the end it all boils down to ab.

The solution is so elegant that I am led to think there must be a more elegant insight that would avoid the heavy algebra. If anyone finds such a solution I would like to see it.
Anonymous said…
Dear Maria,

Thank you for the information, really clever technique.

Dear Wally,

we can not predict the length of the right angle triangle be merely having the hypotenuse, we need another parameter (angle or another side)

best regards

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