# Inscribing an isosceles triangle in a circle

Hello it would be great if you could help with the following problem.

I had to solve it geometrically (no problem there) and then I thought I'd like to solve it by calculating the points as well. Well that wasn't so easy so here's the problem:

Given:

1. a circle $k: x^2+y^2=6^2$

2. a point $P = (4,-3.5)$

3. the distance $\overline{AB}=7$

Now find the isosceles triangle ABC where $A,B,C \in k$ and P lies on AB. There are two solutions for this problem. I already managed to use the parameters to get from 4 equations to 2 equations with 2 unknowns. But not even Mathematica wants to solve those. I can post them here but I didn't want to influence the answers with them yet.

Thanks for any help. :)

Edit:

Apparently you want me to provide my part of the solution so here we go. I'm lookking for points $A$ and $B$

Let them have the coordinates $A=(x_A,y_A)$ and $B=(x_B,y_B)$

I can use that to get the following 4 equations with 4 variables with the given points:

1. using the circle: $y_A^2=\pm\sqrt{6^2-x_A^2}$ and $y_B^2=\pm\sqrt{6^2-x_B^2}$

2. using a line through $A,B,P$: $y_A=\frac{(y_B+3.5)*(x_A-x_B)}{(x_B-4)}$

3. using the distance: $(x_B-x_A)^2+(y_B-y_A)^2=7^2$

plugging 1. into 2. and 3. I get 2 equations with two unknowns.

1. $\to$ 2. $y_A=\frac{(y_B+3.5)* \left( \sqrt{6^2-y_A^2}-\sqrt{6^2-y_B^2 }\right)}{\left(\sqrt{6^2-y_B^2} - 4\right)}$

and

1. $\to$ 3. $\left(\sqrt{6^2-y_B^2} - \sqrt{6^2-y_A^2} \right)^2 + (y_B-y_A)^2=7^2$

That's what I got and that doesn't seem like a sensible line of questioning.

• The point $(4,-3.5)$ doesn't seem to be on the circle. – zoli Mar 22 '17 at 15:48
• @zoli Fortunately not. Otherwise it would be difficult to find a chord through it with a length of 7. – Reinhard Meier Mar 22 '17 at 16:41
• @ReinhardMeier: Oh, yes. – zoli Mar 22 '17 at 16:44
• Is $\overline{AC}$ meant to be one of the two equal sides? – amd Mar 23 '17 at 0:17
• Either way, aren’t there at least four solutions? There are two chords of length 7 through $P$ and for each chord there are either two of four isosceles triangles that can be built on to (depending on the answer to my previous question). – amd Mar 23 '17 at 2:23

You can calculate everything that is constructable with compass and straightedge. So you can try to find a way to get your points $A$, $B$ and $C$ with compass and straightedge and then try to transform this into calculation.
Do a drawing at the side with a triangle $A'B'M'$, where $|A'B'|=7$, $|A'M'|=|B'M'|=6$. Now draw a circle with the radius $|MP|=\sqrt{28.75}$ around the center $M'$. This circle will intersect $A'B'$ at two points $P'_1$ and $P'_2$. The distance $|P'_iA'|$ is a candidate for the distance $|PA|$ and the distance $|P'_iB'|$ is a candidate for the distance $|PB|$. Draw a circle with radius $|P'_1A'|=|P'_2B'|$ around $P$ and a circle with radius $|P'_1B'|=|P'_2A'|$ around $P$. Each of these circles will intersect the original circle at two points. One point on the smaller circle around $P$ together with one point on the bigger circle around $P$ forms a chord through $P$ with length $7$. Once you have $A$ and $B$, it is easy to find $C$ such that you get an isosceles triangle.
Note: I have started to do some of the calculation, e.g. I have found $$|P'_1A'|=|P'_2B'| = \frac{7}{2} - \frac{3}{2}\sqrt{2} \\ |P'_1B'|=|P'_2A'| = \frac{7}{2} + \frac{3}{2}\sqrt{2}$$ But the number grew more and more ugly and finally I decided not to continue on this.