A number of points is drawn inside a rectangle. The rectangle is divided into 210 triangles whose vertices coincide with the vertices of the rectangle and/or the points drawn inside the rectangle. Find the number of points drawn.

I have no idea how to interpret this question. My take is this: there are $210$ triangles that combines into a rectangle, then we find the total number of vertices that the triangles make inside the rectangle, excluding the vertices of the rectangle.

If my take is correct, then I absolutely have no idea how to solve the problem. I tried finding any pattern, but I couldn't. This is primarily because I don't know what counts as a 'vertex'.

A simple example is this: enter image description here

In the diagram on the left, there is only $1$ point, $A$, inside the rectangle made up of $4$ triangles, but in the diagram on the right, there are $2$ points, $B$ and $C$, inside the rectangle made up of the same number of triangles as the rectangle on the left.

How do you solve the problem? If it helps, the final answer should be $104$ points.

  • $\begingroup$ If there are $k$ triangles, the total sum of the angles of all the triangles is $180k$. These angles occur because of line segments meeting at the $n$ points inside the rectangle and the $4$ vertices of the rectangle. At each point in the interior, the angles formed at them add up to 360 and at the vertices they add up to 90. Can you relate the two angle sums and take it from there? $\endgroup$ Mar 10, 2023 at 17:32
  • 6
    $\begingroup$ I would guess that for the purposes of this problem the geometric triangles on your picture on the right aren't triangles. They're quadrilaterals that just happen to have one of their angles equal to 180 degrees. I agree you have no way of knowing this from the problem statement. $\endgroup$ Mar 10, 2023 at 17:45
  • $\begingroup$ @Umesh Shankar I don't see how the problem can be solved with your method... $\endgroup$
    – Jean Marie
    Mar 17, 2023 at 20:48

1 Answer 1


Consider a polyhedron that has 1 rectangle face glued to the "rectangle split into triangles" of your problem. Let $T = 210$ be the number of triangles. We have $1$ rectangular face and $T$ triangular ones. The number of edges of the polyhedron is $$E = \frac{3T + 4}{2} = \frac{3}{2}T + 2 $$ (each edge is counted twice as it joins $2$ faces)

The number of faces is $F = T+ 1$

The Euler characteristic formula gives $2 = V + F - E$ where $V$ is the number of vertices. This gives us $V = 2 + E - F = \frac{T}{2} + 3$ . If we subtract the $4$ vertices corresponding to the rectangle we get $V - 4 =\frac{T}{2} - 1$ interior points. With $T = 210$ this gives $104$ points.


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