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The question came up while messing around with graph paper. I wanted to make an isosceles triangle where the length of one side and it's hight were both integers. The closest I could get was a base side of $8$ and a height of $7$. Which gives a length of the other two sides $\sqrt{65}$. It looks like the ratio of half the bast to the height is $\sqrt{3}$ if that helps.

Michael Albanese
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mark howe
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    If there exists an integer $p$ such that $p\sqrt{3}=q$ for some integer $q$, then $3p^2=q^2$. Now look at the prime factorization of the lhs and rhs of the previous equation – Amr Sep 28 '13 at 00:58
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    As a "practical" solution, you might try the number $15$. The product with $\sqrt{3}$ is about $25.98$. (The number was obtained by looking at the continued fraction expansion of $\sqrt{3}$.) – André Nicolas Sep 28 '13 at 01:15

4 Answers4

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No: This is, in fact, equivalent to showing that $\sqrt{3}$ is irrational.

To do this, suppose that $\sqrt{3} a = b$ for integers $a$ and $b$, which we may take without loss of generality to have no common factors. Then $3a^2 = b^2$, so $3$ is a divisor of $b^2$ and so also of $b$. So $9 | b^2 \implies 9 | 3a^2 \implies 3|a^2 \implies 3|a$. This contradicts that the integers are relatively prime.

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The title question asks to solve $x \sqrt{3} = y$ for integers $x$ and $y$. If $x \neq 0$, we can rearrange this to get $\sqrt{3} = y/x$: any solution to this equation thus prove $\sqrt{3}$ is a rational number. If we check $x=0$, we see that $x=y=0$ is a solution, and the only solution to the title question.

It's not clear what sort of triangle you're trying to draw: are you trying to make all side lengths integers along with the height?

In that case, there's an "easy" solution: if you draw in the altitude, you cut your isoceles triangle into two isomorphic right triangles, and you need all three sides of the right triangle to be an integer, or the side that is half of the base is allowed to be a half-integer.

If you take all sides of these right triangles to be integers, there is a classic solution: pick any integers $r,s,t$. Then you get a right triangle with sides

  • $2rst$
  • $r(s^2 - t^2)$
  • $r(s^2 + t^2)$

Furthermore, all solutions can be expressed in this way. The simplest of these so-called pythagorean triples is $(3,4,5)$. Thus, there is an isoceles triangle with side lengths $(5,5,6)$ and height 4 and another with side lengths $(5,5,8)$ and height 3.

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Yes. There is in fact one and only one integer $n$ such that $n\alpha$ is an integer for every number $\alpha$. However, the isosceles triangle that you produce with this number will have exactly one vertex.

Cameron Buie
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Here is my favorite proof that, if $n$ is a positive integer that is not a perfect square, then $\sqrt{n}$ is irrational. What I like about this proof is that it does not rely on any divisibility properties of the integers. In particular, it does not use unique factorization.

Since $n$ is not a perfect square, there is an integer $k$ such that $k^2 < n < (k+1)^2$, so that $k < \sqrt{n} < k+1$ or $0 < \sqrt{n}-k < 1$.

Suppose $\sqrt{n} = \dfrac{a}{b}$ where $a$ and $b$ are positive integers. Since $0 < \sqrt{n}-k < 1$, $0 < \dfrac{a}{b}-k < 1$ or $0 < a-kb < b$.

Then (watch closely - at no time do the fingers leave the hands!)

$\begin{align} \sqrt{n} &= \sqrt{n}\dfrac{\sqrt{n}-k}{\sqrt{n}-k}\\ &= \dfrac{n-k\sqrt{n}}{\sqrt{n}-k}\\ &= \dfrac{n-k\dfrac{a}{b}}{\dfrac{a}{b}-k}\\ &= \dfrac{bn-ka}{a-kb}\\ \end{align} $

Starting with the assum[tion that $\sqrt{n} = \dfrac{a}{b}$, we have found another fraction, $\dfrac{a_1}{b_1} = \dfrac{bn-ka}{a-kb} $ such that $\sqrt{n} = \dfrac{a_1}{b_1}$ and $b_1 < b$.

Continuing this process, we get an unending decreasing sequence of denominators, which is impossible. (This method of proof is called "infinite descent".)

Therefore, $\sqrt{n}$ is irrational.

Another way to argue is to suppose that $\sqrt{n} = \dfrac{a}{b}$ with $b$ as small as possible. But then we can find another fraction with smaller denominator equal to $\sqrt{n}$, which is a contradiction.

Note how the assumption that $n$ is not a perfect square is used in a very essential way in the proof.

A shorter version of this proof, which I think I saw in a post by Gower, is this:

Let $b$ be the smallest positive integer such that $b \sqrt{n}$ is an integer. Then, with $k = \lfloor \sqrt{n} \rfloor $ as above so that $0 < \sqrt{n}-k < 1$, $(b\sqrt{n}-bk)\sqrt{n} = bn-bk\sqrt{n} $ is also a positive integer and $b\sqrt{n}-bk$ is smaller that $b$, a contradiction.

marty cohen
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