I am self-studying Euclidean Geometry. One problem with studying alone is that when you’re stuck, you’re stuck. I can’t get past the following problem:

“To describe an equilateral triangle having given the distances of a point from each of its vertices.”

I know that the vertices must be on concentric circles with the given point as center. [Note: this is a problem on Euclid – Book I material and so nothing is known about the circle other than its definition]. It’s not obvious to me how one should proceed. So I looked at the special case in which the given distances were all equal to one another. Then the problem becomes one of constructing an equilateral triangle whose vertices are all on a given circle. This I was able to do. But when I considered the case when only two of the distances were equal, I got nowhere. I would almost have it, if only I could prove some little statement or other… But no, it was always just out of reach. I feel like Tantalus in Tartarus.

Any help would be most appreciated.


1 Answer 1


Let the distances be $r,s,t$.
Construct a triangle $ABC$ with sides $AB=r$, $AC=s$, $BC=t$ (exercise: show that $r,s,t$ are such that such a triangle exists).
Construct $D$ so that $ACD$ is an equilateral triangle with $D$ and $B$ on opposite sides of $AC$; let $BD=u$.
Construct $E$ so $BDE$ is an equilateral triangle with $A$ inside it.

Claim: $A$ is at distance $r,s,t$ from $B,D,E$, respectively (so $BDE$ is the triangle you want).

Proof of claim.
$AB=r$ by construction.
$AD=AC=s$ by construction of $D$.
So the only hard part is showing $AE=t$. It follows immediately if we can show triangles $ADE$ and $CDB$ are congruent, since that will imply $AE=CB=t$.

Looking at the two triangles, we have $AD=CD=s$ by construction of $D$, and $DE=DB=u$ by construction of $E$. $$\angle BDA+\angle CDB=\angle CDA=\pi/3=\angle BDE=\angle ADE+\angle BDA$$ so $\angle CDB=\angle ADE$. Thus, by side-angle-side, the triangles are congruent, and we're done.

  • $\begingroup$ Thank you! Now I can continue with my studies. $\endgroup$
    – BlakeDavis
    Jul 6, 2012 at 15:13

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