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Source: IMO 1968 A1

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1592 posts

Rushil

#1 h Nov 4, 2005, 6:58 AM • 2 Y

Y by Adventure10, Mango247

Find all triangles whose side lengths are consecutive integers, and one of whose angles is twice another.

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411 posts

campos

#2 h Nov 6, 2005, 3:22 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with such conditions, such that $A=2B$ . It's very well-known (or it can be deduced from the cosine law and the double-angle formula, or i think it's somewhere in the forum) that $a^2-b^2=bc$
Then, we just have to check the cases $(a,b,c)=(a,a-1,a-2), (a,b,c)=(a,a-2,a-1)$ , and $(a,b,c)=(a,a-1,a+1)$ .

$(a,b,c)=(a,a-1,a-2)$ gives $a^2-5a+3=0$ , but it has no integer roots.
$(a,b,c)=(a,a-1,a+1)$ gives $a^2-2a=0 \Rightarrow a=2 \Rightarrow a+b=3=c \Rightarrow\Leftarrow$
finally, $(a,b,c)=(a,a-2,a-1)$ gives $a^2-7a+6=0$ and since $a=1 \Rightarrow c=0$ , then we must have that $a=6$ .
then the triangle that satisfies such conditions is $(a,b,c)=(6,4,5)$ with $A=2B$

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sayantanchakraborty

505 posts

sayantanchakraborty

#3 h Dec 22, 2013, 4:15 AM • 2 Y

Y by Adventure10, Mango247

If <A=2<B, some simple manipulating with the sine rule gives $/a^2=b(b+c) /$ . Assuming the lengths of the sides to be $/ (a-1,a,a+1)/$ and considering all possible cases we get the solutions in (a,b,c) as
$/(1,2,3) and (4,5,6)/$ and obviously all its permutations.

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sayantanchakraborty

505 posts

sayantanchakraborty

#4 h Dec 22, 2013, 4:18 AM • 2 Y

Y by Adventure10, Mango247

Campos missed the solution $(1,2,3)$ .

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4401 posts

#5 h Dec 22, 2013, 5:12 AM • 2 Y

Y by Heisenberg09, Adventure10

sayantanchakraborty wrote:

Campos missed the solution $(1,2,3)$ .

That´s NOT a triangle, but a degenerated one!

Best regards,
sunken rock

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1498 posts

suli

#6 h May 17, 2015, 4:40 AM • 2 Y

Y by Adventure10, Mango247

Well technically a degenerate triangle is a triangle. The problem should be worded more clearly, or both solutions should be accepted as correct.

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501 posts

#8 h Nov 13, 2024, 8:13 PM

Y by

Bruh AMC 2024 #22

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