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Consider any point $P$ on the unit circle centred at the origin $O(0,0)$ in $\mathbb R^2$. Let $A$ be $(-2,0)$ and $B$ be $(2,0)$ be two point on the $x$-axis and $D$ be the sum of the two distances $AP$ and $BP$. Then $D$ is maximised when $P$ is at the top (or bottom) of the circle. This is easily proven using calculus.

Can anyone produce a simple direct geometric proof of this result. Calculus seems too heavy for this?

I must admit I was initially attracted to the option where $AP$ is a tangent to the circle, but it is inferior.

For those looking for problems, allowing A and B to be positioned randomly is also quite interesting and specialises to the above.

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  • $\begingroup$ Not purely geometric, but by the median length formula $PA^2+PB^2=10\,$. Then by the AM-RMS inequality $PA+PB$ is maximum when $PA=PB\,$. Btw, you should tag it with something geometry if that's what you are looking for. $\endgroup$
    – dxiv
    Commented Jun 14, 2021 at 7:48
  • $\begingroup$ Thanks for that, quite interesting and no calculus. Still seems very heavy for what seems an obvioius result. Maybe it is deeper than it looks. $\endgroup$
    – Mathman
    Commented Jun 14, 2021 at 8:16
  • $\begingroup$ Draw an ellipse with focii at $A (-2,0)$ and $B (2,0)$ and semi-minor axis being $1$. You can show that for every point on the circle $PA + PB \leq P'A + P'B$ where $P'$ is the point $(0,\pm1)$. $\endgroup$
    – Math Lover
    Commented Jun 14, 2021 at 8:33
  • $\begingroup$ Also by Cauchy-Schwartz inequality, you can show that $max (PA + PB)$ is $2 \sqrt5$. By AM-GM, you can show it too. $\endgroup$
    – Math Lover
    Commented Jun 14, 2021 at 8:37

2 Answers 2

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Let $P=(x,y)$ and $Q=(-x,-y)$ be points on the unit circle. Note that $PA+PB=PA+QA$. $P$ and $Q$ are the end-points of a diameter of the unit circle and $A$ is a point on the circle $x^2+y^2=4$. The problem is equivalent to maximizing $CR+DR$ where $C=(-1,0)$, $D=(1,0)$ and $R$ an arbitrary point on the circle $x^2+y^2=4$. Consider the family of ellipses with foci $C$ and $D$ which meets the circle $x^2+y^2=4$ at some point. The largest ellipse is the one that touches the circle $x^2+y^2=4$ at $(0,2)$ and $(0,-2)$.

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  • $\begingroup$ I really like the ellipse approach. It is very direct and pretty. Unfortunately here in Australia we have just removed conic sections from the high school courses, so the students will have no idea what a focus is. Still very nice! Thanks $\endgroup$
    – Mathman
    Commented Jun 16, 2021 at 10:15
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Attempt:

  1. An ellipse is the locus of points whose sum of distances to two fixed points, the loci, is constant.

  2. Consider an ellipse with foci $(-c, 0), (0,c)$, (here: $c=2$), major axis a, minor axis $b$, $b<a$; axes along $X,Y$ axes.

  3. Draw a circle centered at $(0,0)$ with radius $1$.

4)The sum of the distances from a point to the foci is $2a$;

and $a^2=c^2+b^2$.

5)Consider the cases:

A) $b >1; $

Then $a^2=c^2+b^2$;

The circle lies within the ellipse.

B) $b=1;$

The circle and ellipse touch at $(0,1)$ and $(0,-1)$.

$a^2=c^2+1=4+1$;

Sum of the lengths $2a=2\sqrt{5};$

C)$b<1$;

The ellipse and circle intersect.

Sum of lengths: $2a=2\sqrt{c^2+b^2}<2\sqrt{5}$.

6)It follows that the maximal sum of lengths $2\sqrt{5}$ is attained for

$b=1$, when circle and ellipse touch.

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