Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I remember being presented a mathematical puzzle some years back that I still can't solve. The problem is defined as follows:

We have two points on a plane, and using only a compass, how do we find other two points, so that all four of them would be vertices of a square?

I'm not sure whether the first two points were supposed to be vertices of the same edge of a square or not, so solutions to both variants are welcome.

share|cite|improve this question
The trick is to figure out how to construct a pair of points which $\sqrt{2}$ apart... – Zhen Lin Mar 5 '12 at 7:46

Here is a description of one of the constructions:

Given the two points $\color{maroon}0$ and $\color{maroon}d$ that form one side of the square:

  1. Construct the five $\color{gray}{\text{gray}}$ circles of common radius $\color{maroon}{od}$ shown in the diagram and locate the points $\color{maroon}a$, $\color{maroon}b$, and $\color{maroon}c$.
  2. Construct the two $\color{maroon}{\text{maroon}}$ circles: one with center $\color{maroon}c$ and radius $\color{maroon}{cb}$, and one with center $\color{maroon}d$ and radius $\color{maroon}{ad}$. Note that $\color{maroon}{ad}$ and $\color{maroon}{cb}$ have the same length.
  3. Locate the point of intersection $\color{pink}e$ of the two maroon circles.
  4. Construct the $\color{pink}{\text{pink}}$ circles of radius $\color{pink}{oe}$ at centers $\color{maroon}c$ and $\color{maroon}d$.
  5. The point of intersection, $\color{pink}f$, of the pink circles is a vertex of the square.
  6. Draw the $\color{darkgreen}{darkgreen}$ circle centered at $\color{pink}f$ of radius $\color{maroon}{od}$.
  7. The point of intersection, $\color{darkgreen}g$, of the darkgreen circle with the gray circle centered at $\color{maroon}d$ is the final vertex of the square.

    enter image description here

Justification of step 5:

From step 1., $\color{maroon}{aboc}$ is a rhombus with common side length $ l(\color{maroon}{co})$. Since the point $\color{pink}e$ is equidistant to the points $\color{maroon}c$ and $\color{maroon}d$ , the segment $\color{pink}e\color{maroon}o$ is perpendicular to the segment $\color{maroon}{cd}$. Since $\color{pink}f$ is equidistant to the points $\color{maroon}c$ and $\color{maroon}d$, the points $\color{pink}e$, $\color{pink}f$, and $\color{maroon}o$ are colinear and the segment $of$ is perpendicular to the segment $\color{maroon}od$.

We need to show that $l(fo)=l(co)$. Proceeding with some abuse of notation:

Considering the rhombus $aboc$, we have $$ cb^2+ao^2=2(ac^2+co^2 ); $$ or, since $ao=co=ac$, $$ cb^2=3co^2. $$ Since $cb=ce$, we have $$\tag{1} ce^2=3co^2 $$
Considering now the right triangle $ceo$: $$\tag{2} ce^2=eo^2+co^2. $$ Combining equations $(1)$ and $(2)$: $$\tag{3} eo^2=2co^2. $$

Considering now the right triangle $cfo$: $$\tag{4} cf^2=fo^2+co^2 $$ since $cf=oe$: $$\tag{5} oe^2=fo^2+co^2 $$ From $(3)$ and $(5)$ now, we finally obtain $$ fo=co, $$ as desired.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.