Friday Puzzler -- Tangent

Re: Friday Puzzler -- Tangent

Henry Higginbotham

Starting at the lower left, label the four corners of the square as A, B, C, and D. Radius point is E. Intersection of the diagonal line with A-B is F.

From F, a horizontal line intersects C-D at point G. Line FG = 1.

Angle DCE = tan 0.5 = 26.565°

Line CE is the bisector of FCD, thus FCD = 53.130°, leaving CFG as 26.870°.

Cos 26.870 = 0.8000, so line FC = 1/0.8 = 1.25.

My math background is limited compared to some others here, so this presentation is clumsy. IF the answer and reasoning is correct, I'd appreciate a hint or two on how to clean things up.

Re: Friday Puzzler -- Tangent

Larry Barrett

Henry

Alex will jump in here pretty soon, but in my opinion I think you have the correct answer (1.25"), but I think your reasoning is a little off, or maybe there is just a simpler way to look at it.

(BTW, I have been looking at this off and on and did not see how to proceed at all, so hats off to you).

Here is your reasoning, and interspersed is my suggestion:

Starting at the lower left, label the four corners of the square as A, B, C, and D. Radius point is E. Intersection of the diagonal line with A-B is F.

From F, a horizontal line intersects C-D at point G. Line FG = 1.

Angle DCE = tan 0.5 = 26.565°

Line CE is the bisector of FCD, thus FCD = 53.130°, leaving CFG as 26.870°.

It is not obvious to me why CE is a bisector of angle FCD, although I think you are correct that angle FCD is 53.130°. And, a minor error - if FCD is 53.13° , then angle CFG is (90° - 53.13°) = 37.87°. However, cos 37.87° = .8, as you say.

Instead of your bisector argument, lable the point where the semicircle radius intersects line FC as point X. Now look at triangles CXE and CDE. They are both right triangles. Sides EX and ED are equal (.5"), Line CE is the hypotamus for both. Therefore sides CX and CD are equal and the triangles are congruent. Therefore angle XCE (which is the same as FCE) and angle DCE are equal. And this is what you concluded from your bisector argument.

So angle FCD is indeed 2x26.565° = 53.13°, and therefore angle GFC is (90° - 53.13°) = 37.87° and cos 37.87° = .8.

And cos GFC = 1/(length of the red line) = .8 so the red line = 1/.8 = 1.25, as you say.

Nice Job

Alex Y

And correct answer, Henry. I follow your and Larry's reasoning and I agree. I haven't checked the trig numbers, but you got the right answer..

Here is a solution with no trig. Using your and Larry's labels:

As Larry pointed out, triangles CXE and CDE are similar. From that, we know that length of CX is the same as CD, or 1.

Similarly, XFE is similar to AFE, so AF = FX. Let's call that length "z".

Consider the triangle FBC. Its hypotenuse is 1+z and its sides are of lengths 1 and 1-z. Pythagoras tells us that

(1+z)^2 = (1-z)^2 + 1^2.

1+2z+z^2 = 1-2z+z^2 + 1

4z=1

z=.25,

so CF = 1.25