Question

Here is a question my teacher once said was unsolvable.
I have a circle with radius R with a line going directly through the middle of it horizontally. At the two points where the line intersects the circle, there is an isosceles triangle formed by two other lines crossing these intersect points. The triangle's third point does not touch or cross border of the circle, however the sides that construct the triangle continue until they do cross the boundary. The length of this excess line segment is Z. Solve this problem for X where X is the height of the triangle in the circle.

Answer 1

Be nice if I could see a picture of that

Answer 2

one moment then.

Answer 3

Yeah, same. Could you attach one?

Answer 4

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Answer 5

not quite :)

Answer 6

lol, nice Triforce!

Answer 7

link to the rescue :)

Answer 8

This problem is my favorite.

Answer 9

It looks solvable

Answer 10

The proof I gave my teacher back then was correct, and got me out of a final. I am posting it here as a brain teaser for those that like this kind of thing.

Answer 11

I'm not convinced that this is difficult

Answer 12

I solved it using a fourth degree polynomial back then.

Answer 13

If i label the hypotenuse as H, then the following rule definitely applies
(x+r)(r-x) = hz

Answer 14

Okay, first attempt was incorrect (I foiled wrong.) Let's try this again.

Consider the right triangles formed by dividing the isosceles triangle by x. Each of these right triangles has side lengths r, x, and a third side we shall refer to as b. Consider the chords b + z and r + x + (r-z). By the intersecting chord theorem, bz = (r+x)(r-x) = r^2 - x^2, or b = (r^2 - x^2)/z.

By the Pythagorean Theorem, x^2 + r^2 = b^2, or x^2 = b^2 - r^2. Then x^2 = ((r^2 - x^2)/z)^2 - r^2.

Answer 15

Multiplying this out gives us x^2 = (r^4 - 2r^2x^2 + x^4)/z^2 - r^2, or \[1/z^2 * x^4 - (2r^2/z^2 - 1)x^2 + (r^4 - r^2)/z^2 = 0\]. Solve for the roots, pick the real positive one, and you're done!

Answer 16

I still don't understand the problem statement (see attached).

Answer 17

solving for AF(red)

Answer 18

Blacksteel is closest. I actually got this in terms of x = ? using another proof that the fourth degree polynomial could be treated as a second degree polynomial given the properties of the problem.

Answer 19

If you complete the right triangle from where Z touches the circle to the diameter and label that completion as V and label the hypotenuse which is the square root of R^2 + X^2 as K, then

V^2 = (2R)^2 -(K+Z)^2
K^2 = V^2 + Z^2 and
K^2 = R^2 + X^2

From this you can get a statement in R,X and Z.

Answer 20

Well, yes, you clearly can. Since the product only has terms in even powers, you can rewrite the whole thing as a second-degree polynomial squared, and since you're looking for roots the solutions to the second degree polynomial and the second degree polynomial squared will be the same.

Answer 21

AF = 1/2R*sin(angleABF)

Answer 22

Blacksteel, how do I mark you as haven given the correct answer?

Answer 23

Yes, but we don't know the angle ABF

Answer 24

Just hit the Good Answer button next to my post

Answer 25

teh good answer button doesnt appear in all browsers; it will usually show up if you refresh the browser by hitting the f5 key :)

Answer 26

Yup, had to refresh.

Answer 27

Estudier, in your example the hypotenuse of that right triangle would be sqrt(x^2 + r^2) + z.

Answer 28

Sorry, I had visitors....

I don't understand, which right triangle?

One is r^2 +x^2, the other is v^2 z^2.

Answer 29

this website keeps freezing on me

Answer 30

I'm not sure i understand. You know that z is the distance from the vertex of the isosceles triangle to the edge of the circle, not the entire segment connecting the two intersections with the circle, right?

Answer 31

Is diagram right?

Answer 32

Still can't see what I have wrong, someone look at the diagram, please?

Answer 33

Estudier's Basis:
V^2 = (2R)^2 -(K+Z)^2
K^2 = V^2 + Z^2
K^2 = R^2 + X^2

Me:
V^2 = K^2 - Z^2
V^2 = R^2 + X^2 - Z^2
V^2 = (2R)^2 - (sqrt(R^2 + x^2) + Z)^2
(2R)^2 - (sqrt(R^2 + x^2) + Z)^2 = R^2 + X^2 - Z^2
3R^2 - X^2 + Z^2 = (sqrt(R^2 + x^2) + Z)^2
3R^2 - X^2 + Z^2 = R^2 + x^2 + 2Zsqrt(R^2 + x^2) + Z^2
2R^2 - 2X^2 = 2Zsqrt(R^2 + x^2)
R^2Z^(-1) - X^2Z^(-1) = sqrt(R^2 + x^2)
R^4Z^(-2) - 2X^2R^2Z^(-2) + X^4Z^(-2) = R^2 + x^2
R^4 - 2X^2R^2 + X^4 = R^2Z^2 + x^2Z^2
X^4 - 2X^2R^2 - x^2Z^2 = R^2Z^2 - R^4
X^4 - (2R^2 - Z^2)X^2 = (Z^2 - R^2)R^2

This final solution is exactly the same as Blacksteel's solution, the system of equations just needed some tuning. Blacksteel used chords instead of identifying the larger triangle to build an association between Z and R to find X.