# Find the locus of this point P

Find the locus of all points inside $$\triangle ABC$$ such that $$PA^{2}+PB^{2}=PC^{2}$$.

At first, i tried finding a right angled triangle and then tried to go on applying Pythagorean Theorem and finding other trivia but that didn't seem to work. I think the locus is most likely to be some line segment rather than an arc of some circle as that seems to be quite unrelated seen from an Euclidean perspective. Applying Apollonius' Theorem and Stewart's Theorem on some triangles might be the key to this problem although i could not find such triangles. I don't know whether this can be solved by trigonometry or not but since this is a problem from a chapter on euclidean geometry, i am sure that there is some clever way to look at this problem.

• Nice question. I think the locus is a circle. Say using coordinate geometry, we placed two vertices on $x-$axis (with one on the origin) and the third one in first quadrant, it shows the locus is a circle. Feb 4, 2021 at 17:48
• Say we took coordinates $(0,0), (a,0), (b,c)$ as vertices of a triangle. Then locus of point $P(x,y)$ will be $x^2+y^2 + (x-a)^2+y^2 = (x-b)^2+(x-c)^2$ and if we simplify it will lead to equation of a circle. Feb 4, 2021 at 17:59
• Here is something I found that may be of interest - math.stackexchange.com/questions/831352/… Feb 4, 2021 at 18:05
• @MathLover Thank you very much. I was searching for similar problems but couldn't find one. This answers my question. Feb 4, 2021 at 18:09
• It shouldn't be a line, since when $\triangle ABC$ is equilateral, then the locus is an arc of a circle of points subtending on $AB$ an angle of $150$ degrees.. This case can be solved by rotating the triangle around $C$ by $60$ degrees.
– plop
May 21, 2021 at 1:48

Use vectors with $$C$$ at the origin. Then $$|\mathbf{PA}|^2+|\mathbf{PB}|^2=|\mathbf{PC}|^2\implies|\mathbf P|^2-2(\mathbf A+\mathbf B)\cdot\mathbf P+|\mathbf A|^2+|\mathbf B|^2=\mathbf 0.$$ Then if $$\mathbf D=\mathbf A+\mathbf B$$, this rearranges to $$|\mathbf P-\mathbf D|^2=2\mathbf A\cdot\mathbf B.$$ So as long as $$\angle BCA<90^{\circ}$$, the locus will be a circle with centre $$D$$ radius $$\sqrt{2ab\cos C}$$.

• Yes using vectors or barycentric coordinates, it is straightforward. Feb 4, 2021 at 18:07
• Very nice answer. For obtuse triangles, there'll be no point P. Feb 4, 2021 at 18:14
• @MathLover how would you approach the problem with barycentric coordinates? Feb 4, 2021 at 19:34
• @DRSKMOBINULHAQUE I was going through the answer in the link I sent. I think that answer can be simplified quite a bit. I have added an answer. Feb 5, 2021 at 8:34
• @MathLover I simplified it as well. Feb 5, 2021 at 9:30

We extend median $$CM$$ such that $$CM = MD$$. Then please note that $$MP$$ is median of $$\triangle CPD$$ and $$\triangle APB$$. Applying Apollonius's theorem,

$$PA^2 + PB^2 = 2(MP^2 + AM^2)$$, $$PC^2 + PD^2 = 2(MP^2 + CM^2)$$

Subtracting second from first,

$$PA^2 + PB^2 - PC^2 = 0 = PD^2 + 2 (AM^2 - CM^2)$$

i.e $$PD^2 + 2(\frac{c^2}{4} - \frac{1}{2}(a^2 + b^2) + \frac{c^2}{4}) = 0$$

That leads to $$PD = \sqrt{a^2+b^2-c^2} \$$. So the locus of point $$P$$ satisfying $$PA^2 + PB^2 = PC^2$$ is a circular arc with point $$D$$ being the center and radius $$\sqrt{a^2+b^2-c^2}$$. We also note that this is possible only if $$a^2 + b^2 \geq c^2$$.

• I did the exact same thing! Feb 5, 2021 at 9:22
• Good to know! :) Feb 5, 2021 at 9:31
• Thank you for staying engaged! Feb 5, 2021 at 9:34

Hint: choose a coordinate system in which $$A=(0,0)$$ and $$B=(1,0)$$. Then $$C$$ will have some fixed coordinates, say $$C=(p,q)$$. Then, if $$P=(x,y)$$, we have $$PA^2 = x^2 + y^2$$, $$PB^2 = (x-1)^2 + y^2$$, and so on. You’ll end up with an equation that you should be able to identify.