Given a triangle $\Delta$ of unit area, how much area can always be covered by $k$ isosceles triangles contained in $\Delta$ and intersecting at most at their boundaries?

The answer is easy for $k=1$. Without error of my part, the worst case is given by a triangle with sides proportional to $1+\epsilon,2+\epsilon,3+\epsilon$ where one can cover roughly $2/3$ of the total area.

Is there an elegant way to compute the answer for $k=2,3,\dots$?


2 Answers 2


Three trivial observations (not an answer).

Let us denote by $a_k(\Delta)$ the maximal covered area in $\Delta$. Then

  • for any right triangle $a_2=1$,
  • for any acute triangle $a_3=1$,
  • for any triangle $a_4=1$.
  • $\begingroup$ Nice observations! $\endgroup$ Apr 29, 2010 at 7:13

I think I can completely find those triangles for which $a_{2}=1$.

We have two isosceles triangles forming a third triangle. Two of their angles must form a 180 degree angle. the two equal angles of both triangles cannot do this. So the vertex angle of one triangle must be adjacent to one of the angles of the second triangle. Let the second triangle have one angle $A$ and two angles B=$90-A$. Then if the angle of the first triangle is adjacent to $A$ then It must be of magnitude 180-$A$ and the other two angles of the triangle must be $A/2$ but we then must add $A/2$ two one of the remaining angles this the triangle with the following angles:

($A/2$, 90, $90-A/2$) this gives all right triangles which has already been noted in a separate answer.

If the angle of the first triangle is adjacent to $90-A/2$ then we get a new angle of $45-A/4$ we will have to add this quantity to either of the two angles giving two triangles with the following angles:

($45-A/4$, A, $135-3A/4$)

($45-A/4$, $90-A/2$, $45+3A/4$)

  • $\begingroup$ My question concerns the worst case. $\endgroup$ Apr 30, 2010 at 7:40

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