Geometry - Rectangle ABCD with inside point E. Find the least possible value for sum of interger distances from E to 4 vertices.

Question

The point E lies within the rectangle ABCD.

If the distances from the vertices to E are all distinct integers, what is the least possible value of AE + BE + CE + DE?

Answer 1

We want $$\left\{ \begin{align*} a^2&+c^2&=p^2& \hspace{4em} (1)\\ a^2&+d^2&=q^2 &\hspace{4em} (2)\\ b^2&+c^2&=r^2 &\hspace{4em} (3)\\ b^2&+d^2&=s^2 &\hspace{4em} (4) \end{align*} \right.$$

where $a+b$ and $c+d$ are the side lengths of the rectangle, and $p,q,r,s$ are the distances to the vertices. We want to find a way such that $p,q,r,s$ are distinct integers (i.e. let $\gcd(p,q,r,s)=1$).

\begin{align*} (1)-(2): && c^2-d^2&=p^2-q^2 &\hspace{4em} (5) \\ (3)-(4): && c^2-d^2&=r^2-s^2 &\hspace{4em} (6) \\ (5)-(6): && 0 & = p^2-q^2-r^2+s^2 \\ &&r^2-s^2 &=p^2-q^2 & \hspace{4em}(7) \end{align*}

A solution (and the solution containing the smallest numbers) for the diophantine equation $(7)$ is $$(p,q,r,s)=(8,4,7,1).$$ Therefore, the answer to your question is the sum of $p,q,r,s$, or $20$.

More info: https://oeis.org/A118882

Answer 2

Hint:

Recognize the relationship

$$AE^2+CE^2=BE^2+DE^2$$

which would well reduce the guess work.

Answer 3

Suppose we have such rectangle. Place it on a coordinate system so that the $E$ is on $(0,0)$ and $A = (-a, b); B= (-a,-c), C=(d,b); D=(d,-c)$ and we need for $\sqrt{a^2 + b^2}=m, \sqrt{a^2+c^2}=n, \sqrt{d^2 + b^2}=r, \sqrt{d^2+c^2}=s$ are all integers.

We must have $m^2 +s^2 = n^2 + r^2$ and that simply requires finding the smallest integer that is a sum of two distinct perfect squares in two possible ways that will have the smallest sum.

$m^2 + s^2 = n^2 + r^2 \implies m^2-n^2 = r^2 - s^2 \implies (m+n)(m-n)=(r+s)(r-s)$.

Now $m+n, m-n$ must be the same parity and and we could do $r+s=5, r-s=3$ and $m+n=15, m-n=1$ so $m=8;n=7;r=4;s=1$. This would be the smallest set of four distinct values for $m-n,r-s,r+s,m+n$. And so $s+r+n+m=20$.

And $8^2 + 1^2 = 7^2 + 4^2=65$ is a probably this smallest such integer.

(I'm not sure how to verify this other than trial and error. Oddly enough taking even values $m-n=2, r-s=4,r+s=6,m+n=12$ so $r=5;s=1;n=5;m=7$ gives us a smaller integer $7^2+1=5^2 + 5^2=50$ and as smaller sum of $18$ but the terms are not distinct.)

So we need $a^2 + b^2= 64; a^2+c^2=49; d^2+b^2=16; d^2 + c^2 =1$.

$d^2=1-c^2$ and so $a^2 + b^2 = 64; a^2+c^2=49; b^2-c^2=15$

$b^2 = 15+c^2$ and so $a^2+c^2=49$. So we need $0< c^2 <1$ the$1> d^2=1-c^2>0$, $b^2 = 15+c^2> 15$ and $a^2 = 49-c^2> 48$ are possible rectangles.