I squeeze, squeeze, and squeeze...

Geometry Level 4

Put three congruent triangles inside a unit square so that they don't overlap one another.

What is the maximum possible area of one of the triangles?

The answer is 0.28868.

1 solution

Jeremy Galvagni
Aug 15, 2018

No proof that this is the maximum possible triangle, but here's what I found.

The three triangles share a right angle so each is a 30-60-90 right triangle. The longer leg is $1$ , so the shorter leg is $\frac{1}{\sqrt{3}}$ .

The area is then $\frac{1}{2}\cdot \frac{1}{\sqrt{3}} \cdot 1 = \frac{1}{2\sqrt{3}} \approx \boxed{0.288675}$

Suppose you modify your problem to read like this:

Put FOUR congruent triangles inside a unit square so that they don't overlap one another.

What is the maximum possible area of one of the triangles ?

Does this answer look correct ?

Since we can place the triangles in a unit square like it is seen in this figure (Within the orange square)

https://i.stack.imgur.com/BYTn9.png

And we know that (1/2)^2 + (sqrt(3)/2)^2 = 1..

So one of these right triangles will have dimensions of (1/2) -> height and (sqrt(3)/2) -> base.

This means that the maximum area of one of these triangles is (1/2)*(1/2)(sqrt(3)/2) = sqrt(3)/8 = 0.2165

Vijay Simha - 2 years, 8 months ago

Yes, but the question asks for the maximum area. The one I showed is greater.

Jeremy Galvagni - 2 years, 8 months ago

If four CONGRUENT triangles are squeezed into a 1x1 square...

Can there be any triangle having sides a,b and c greater than a triangle having sides: 1/2, sqrt(3)/2 and 1 ?

Vijay Simha - 2 years, 8 months ago

@Vijay Simha – Did you have FOUR there before? If so I'm sorry I missed it. That changes the problem to be much simpler: Draw the two diagonals. Each triangle has area 1/4.

Jeremy Galvagni - 2 years, 8 months ago

@Jeremy Galvagni – Darn ! I agree

Vijay Simha - 2 years, 8 months ago

I squeeze, squeeze, and squeeze...

The answer is 0.28868.

1 solution

0 pending reports