1968 IMO Problems/Problem 1

Problem

Prove that there is one and only one triangle whose side lengths are consecutive integers, and one of whose angles is twice as large as another.

Solution 1

In triangle $ABC$ , let $BC=a$ , $AC=b$ , $AB=c$ , $\angle ABC=\alpha$ , and $\angle BAC=2\alpha$ . Using the Law of Sines gives that

$\frac{b}{\sin{\alpha}}=\frac{a}{\sin{2\alpha}}\Rightarrow \frac{\sin{2\alpha}}{\sin{\alpha}}=2\cos{\alpha}=\frac{a}{b}$

Therefore $\cos{\alpha}=\frac{a}{2b}$ . Using the Law of Cosines gives that

$\cos{\alpha}=\frac{a^2+c^2-b^2}{2ac}=\frac{a}{2b}$

This can be simplified to $a^2c=b(a^2+c^2-b^2)$ . Since $a$ , $b$ , and $c$ are positive integers, $b|a^2c$ . Note that if $b$ is between $a$ and $c$ , then $b$ is relatively prime to $a$ and $c$ , and $b$ cannot possibly divide $a^2c$ . Therefore $b$ is either the least of the three consecutive integers or the greatest.

Assume that $b$ is the least of the three consecutive integers. Then either $b|b+2$ or $b|(b+2)^2$ , depending on if $a=b+2$ or $c=b+2$ . If $b|b+2$ , then $b$ is 1 or 2. $b$ couldn't be 1, for if it was then the triangle would be degenerate. If $b$ is 2, then $b(a^2+c^2-b^2)=42=a^2c$ , but $a$ and $c$ must be 3 and 4 in some order, which means that this triangle doesn't exist. therefore $b$ cannot divide $b+2$ , and so $b$ must divide $(b+2)^2$ . If $b|(b+2)^2$ then $b|(b+2)^2-b^2-4b=4$ , so $b$ is 1, 2, or 4. Clearly $b$ cannot be 1 or 2, so $b$ must be 4. Therefore $b(a^2+c^2-b^2)=180=a^2c$ . This shows that $a=6$ and $c=5$ , and the triangle has sides that measure 4, 5, and 6.

Now assume that $b$ is the greatest of the three consecutive integers. Then either $b|b-2$ or $b|(b-2)^2$ , depending on if $a=b-2$ or $c=b-2$ . $b|b-2$ is absurd, so $b|(b-2)^2$ , and $b|(b-2)^2-b^2+4b=4$ . Therefore $b$ is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so $b$ cannot be the greatest of the three consecutive integers. This shows that there is exactly one triangle with this property - and it has side lengths of 4, 5, and 6. $\blacksquare$

Solution 2

(Note: this proof is an expansion by pf02 of an outline of a solution posted here before.)

In a given triangle $ABC$ , let $A=2B$ , $\implies C=180-3B$ , and $\sin C=\sin 3B$ . Then

$\sin ^2 A = \sin ^2 2B = 2 \sin B \cos B \sin 2B = \sin B(\sin B + \sin 3B) = \sin B(\sin B + \sin C)$

Hence,

$a^2 = b(b + c)\ (*)$

Indeed, we know from the Law of Sines that

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ .

Denote this ratio by $r$ ; we have $\sin A = ra, \sin B = rb, \sin C = rc$ . Substitute in $\sin ^2 A = \sin B(\sin B + \sin C)$ and simplify by $r^2$ . We get $(*)$ .

At this point, notice that $(*)$ is equivalent to the equality $a^2c = b(a^2 + c^2 - b^2)$ from Solution 1. Indeed, the latter can be rewritten as $a^2(c - b) = b(c + b)(c - b)$ , and we know that $c \ne b$ . So we could simply quote the fact (proven in Solution 1) that if $a, b, c$ are consecutive integers and $a^2 = b(c + b)$ , then $b = 4, c = 5, a = 6$ is the only solution which could be the sides of a triangle.

For the sake of completeness, and for fun, I give a slightly different proof here.

We have six possibilities, depending on how the three consecutive numbers are ordered. The six possibilities are:

1: $a = b - 2, c = b - 1, b$

2: $c = b - 2, a = b - 1, b$

3: $a = b - 1, b, c = b + 1$

4: $c = b - 1, b, a = b + 1$

5: $b, a = b + 1, c = b + 2$

6: $b, c = b + 1, a = b + 2$

For each case, we could substitute $a, c$ in $(*)$ , get an equation in $b$ , solve it, and get all the possible solutions. As a shortcut, notice that (*) implies that $b|a^2$ . If $a, b$ are consecutive integers, then they are relatively prime, so $b|a^2$ can not be true unless $b = 1$ . In this case the triangle would have sides $1, 2, 3$ , which is impossible. This eliminates cases 2, 3, 4 and 5.

In case 1, $(*)$ becomes

$(b - 2)^2 = b(b + b - 1)$ , or $b^2 + 3b - 4 = 0$ .

This has solutions $1, -4$ . The value $b = -4$ is impossible. The value $b = 1$ yields $a = -1, c = 0$ , which is impossible.

In case 6, $(*)$ becomes

$(b + 2)^2 = b(b + b + 1)$ , or $b^2 - 3b - 4 = 0$ .

The solutions are $-1, 4$ . The value $b = -1$ is impossible. Thus, we get the unique triangle $a = 6, b = 4, c = 5$ .

Solution 3

NO TRIGONOMETRY!!!

Let $a, b, c$ be the side lengths of a triangle in which $\angle C = 2\angle B.$

Extend $AC$ to $D$ such that $CD = BC = a.$ Then $\angle CDB = \frac{\angle ACB}{2} = \angle ABC$ , so $ABC$ and $ADB$ are similar by AA Similarity. Hence, $c^2 = b(a+b)$ . Then proceed as in Solution 2, as only algebraic manipulations are left.

Solution 4

Note: Adding this 4th solution is justified by the fact that it is extremely straightforward, and by the fact that in fact, it does not use the hypothesis that the sides of the triangle are integers. We only need the fact that the sides differ by $1$ (i.e. they are $x, x + 1, x + 2$ for some $x$ ), and the condition on the angles (that one is twice the other).

So, let us start by assuming that the two angles are $\alpha, 2\alpha$ and the sides are $x, x + 1, x + 2$ .

1968 IMO (Problems) • Resources
Preceded by First Problem	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 2
All IMO Problems and Solutions

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