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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.

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  • $\begingroup$ The point $(4,-3.5)$ doesn't seem to be on the circle. $\endgroup$ – zoli Mar 22 '17 at 15:48
  • $\begingroup$ @zoli Fortunately not. Otherwise it would be difficult to find a chord through it with a length of 7. $\endgroup$ – Reinhard Meier Mar 22 '17 at 16:41
  • $\begingroup$ @ReinhardMeier: Oh, yes. $\endgroup$ – zoli Mar 22 '17 at 16:44
  • $\begingroup$ Is $\overline{AC}$ meant to be one of the two equal sides? $\endgroup$ – amd Mar 23 '17 at 0:17
  • $\begingroup$ 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). $\endgroup$ – amd Mar 23 '17 at 2:23
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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.

This is one possible approach:

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.

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  • $\begingroup$ thanks for the answer. Ii edited my question to show my try for a solution. can you tell me why that didn't work? thanks $\endgroup$ – Notgiven Mar 24 '17 at 9:08
  • $\begingroup$ @Notgiven There is no particular step in your calculation where I could point to and say this is where you did something wrong. It is not unusual that some attempts to solve a mathematical problem turn out to be dead ends. Sometimes it is a matter of luck, sometimes a matter of experience, sometimes a matter of intuition and sometimes a matter of effort to find an adequate way to solve a problem. By back luck, you have chosen an approach that did not take you anywhere. You could have started with a completely different approach, e.g. trying to avoid the square roots as long as possible. $\endgroup$ – Reinhard Meier Mar 24 '17 at 11:12

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