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Given distances from a point inside a rectangle to all the vertices, is it possible to find the sides of the rectangle, or is this analagous to the impossiblity of finding a third side of a triangle given only two of the triangle's sides?

Arne Erikson
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    Yes, draw perpendiculars from that point to the sides and giving each segment a variable, using Pythagoras you are basically having four equations and four unknowns. – cr001 Nov 11 '22 at 18:27
  • Say the distances are as follows. AP =4, BP=6, CP=5, DP= square root of 5. Could you please flesh out the details of how to find the sides? – Arne Erikson Nov 11 '22 at 18:37
  • I tried it out actually I was wrong plugging the numbers into wolfram alpha, looks like one of the equations is linearly dependent on the other three so indeed it seems impossible in general. – cr001 Nov 11 '22 at 18:44
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    $a^2+c^2=5, b^2+c^2=16, a^2+d^2=25, b^2+d^2=36$ is what you would enter into wolfram alpha, it actually outputs a family of solution in terms of $a$ as long as $a$ is within a certain range so every number is a real number. – cr001 Nov 11 '22 at 18:46

1 Answers1

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The side lengths of a rectangle cannot be determined just by the distances from an arbitrary point to the vertices.

For example, let $X$ be a point, and $A,B,C,D$ be a square with center $X$ and side length $2$. Then the distances from $X$ to each vertex would be $\sqrt{2}$. However, if we reduce the angles $AXB$ and $CXD$, while keeping the $\sqrt{2}$ distances the same (effectively making the rectangle longer and thinner), we are changing the side lengths of the rectangle.

S.L.
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