4

Triangle ABC

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.

Limestone
  • 2,658
  • 1
    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. – Math Lover Feb 04 '21 at 17:48
  • @MathLover Thanks for your response! Yes, it can be but i didn't expect it. Could you please elaborate it a bit? – Limestone Feb 04 '21 at 17:55
  • 1
    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. – Math Lover Feb 04 '21 at 17:59
  • 1
    Here is something I found that may be of interest - https://math.stackexchange.com/questions/831352/locus-using-euclidean-geometry – Math Lover Feb 04 '21 at 18:05
  • 1
    @MathLover Thank you very much. I was searching for similar problems but couldn't find one. This answers my question. – Limestone Feb 04 '21 at 18:09
  • 1
    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 '21 at 01:48
  • Sorry i didn't know about that, I'll delete this one now – Qui Gonn Jinn May 21 '21 at 16:16

3 Answers3

4

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 $\sqrt{2ab\cos C}$.

jlammy
  • 9,424
  • 1
    Yes using vectors or barycentric coordinates, it is straightforward. – Math Lover Feb 04 '21 at 18:07
  • Very nice answer. For obtuse triangles, there'll be no point P. – cosmo5 Feb 04 '21 at 18:14
  • @MathLover how would you approach the problem with barycentric coordinates? – Dr. Mathva Feb 04 '21 at 19:34
  • Thanks for your response. But i am an elementary student so i haven't studied vectors yet. – Limestone Feb 05 '21 at 02:39
  • 1
    @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. – Math Lover Feb 05 '21 at 08:34
  • @Dr.Mathva I was thinking of using the centroid (center of mass) and use the fact that $\vec{GA} + \vec{GB} + \vec{GC} = 0$ and then represent $\vec{AP}$ etc, in terms of $\vec{GP}$ and $\vec{GA}$.. so nothing more than the affine property of the triangle. But I posted another answer given OP has not studied vectors yet. – Math Lover Feb 05 '21 at 08:46
  • 1
    @MathLover I simplified it as well. – Limestone Feb 05 '21 at 09:30
2

enter image description here

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 $a^2 + b^2 \geq c^2$.

Math Lover
  • 52,200
1

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.

bubba
  • 44,617