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This question was asked in last year's "Mathcounts" competition.

What is the largest possible perimeter of a triangle whose sides have integer lengths and that can fit inside a circle of radius $20$ cm?

This is a question from a recent math contest. I was thinking that I should start by considering a circumcircle of the triangle with a circumradius $19<R<20$. (If that doesn't work I'd go smaller.)

I started by considering an equilateral triangle just because it would be easy to express $R$ in terms of the triangle side length, $s$. $R =\frac{s}{\sqrt 3}$.

Substituting into the above chain inequality yields $19<\frac{s}{\sqrt 3}<20$. So approximating, $33\leq s \leq 34$.

So I've got a ballpark size of my triangle, but I'm not sure how to proceed. This contest does allow the use of a calculator.

amWhy
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user1604149
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2 Answers2

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The largest possible perimeter of triangle with circumradius $20$ occurs when the triangle is equilateral and has sides of length $34.64$. So we know we can get a perimeter of $3\times 34=102$ and that we cannot get a perimeter of $3\times 34.64=103.9$

All you have to check is the circumradius of a $34,34,35$ triangle.

I get $19.82$ so $34,34,35$ is the triangle with the largest perimeter.

  • Is it easy to see that the largest perimeter of all triangles with the same circumradius $r$ is of an equilateral triangle with that circumradius (and therefore with the side $r\sqrt{3}$)? I am sure this is true btw, but just curious if the proof is easy. –  Feb 02 '22 at 01:13
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    Yes - fairly easy. I think the answer of https://math.stackexchange.com/questions/710796/maximum-perimeter-of-an-isosceles-triangle-inscribed-in-the-unit-circle?noredirect=1&lq=1 is a good one. –  Feb 02 '22 at 01:23
  • Yuck, trigonometry + derivatives. (I would prefer an elementary proof. But this is a personal preference.) –  Feb 02 '22 at 01:24
  • I think I vaguely remember a neat purely geometrical proof but I would need time to work it out again! –  Feb 02 '22 at 01:27
  • @StinkingBishop. Keeping an edge and the sum of the other two edges constant generates an ellipse containing the circumcircle. So the maximum has to occur when all sides are equal. –  Feb 02 '22 at 01:31
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Suppose:

R is radius of circle

S is the area of triangle

a, b and c are the measure of the sides

We may use this formula:

$$4RS=abc$$

R is known to be constant, If S is maximum abc will be maximum and it is possible only if $a=b=c$.Hence perimeter will be maximum if a=b=c and we can write:

$R=20$

$$S=\frac{\sqrt 3}4 a^2$$

$$4\times 20\times \frac{\sqrt 3}4 a^2=a^3$$

which gives:

$a\approx 34.6$

and perimeter is:

$$p=3\times 34.6\approx 104$$

Closest integers are :

$$(a, b, c)=(33, 34, 35)$$

sirous
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