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Difference between revisions of "1968 IMO Problems/Problem 1"
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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.  
 
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.  
  
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==Solution==
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==Solution 1==
 
In triangle <math>ABC</math>, let <math>BC=a</math>, <math>AC=b</math>, <math>AB=c</math>, <math>\angle ABC=\alpha</math>, and <math>\angle BAC=2\alpha</math>. Using the [[Law of Sines]] gives that
 
In triangle <math>ABC</math>, let <math>BC=a</math>, <math>AC=b</math>, <math>AB=c</math>, <math>\angle ABC=\alpha</math>, and <math>\angle BAC=2\alpha</math>. Using the [[Law of Sines]] gives that
  
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{{alternate solutions}}
 
{{alternate solutions}}
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Alternate Solution:
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==Solution 2==
 
if in a triangle one angle is twice the other . Say in tr. ABC angle A=2angle B
 
if in a triangle one angle is twice the other . Say in tr. ABC angle A=2angle B
 
A=2B
 
A=2B
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b=4,c=5,a=6
 
b=4,c=5,a=6
 
putting similar cases we can show that all other solutions are non-integral
 
putting similar cases we can show that all other solutions are non-integral
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==Solution 3==
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NO TRIGONOMETRY!!!
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Let <math>a, b, c</math> be the side lengths of a triangle in which <math><C = 2<B.</math>
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Extend <math>AC</math> to <math>D</math> such that <math>CD = BC = a.</math> Then <math><CDB = <ACB/2 = <ABC</math>, so <math>ABC</math> and <math>ADB</math> are similar by AA Similarity. Hence, <math>c^2 = b(a+b)</math>. Then proceed as in Solution 2, as only algebraic manipulations are left.
  
 
==See Also==
 
==See Also==

Revision as of 22:01, 14 September 2014

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. $\blacksquare$


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Solution 2

if in a triangle one angle is twice the other . Say in tr. ABC angle A=2angle B A=2B which implies C=180-3B SinC=Sin3B Sin^2A = Sin^2 2B= 2sinBcosBSin2B = sinB(SinB + Sin3B) = SinB(SinB + SinC) Hence, a^2 = b(b+c) using the above relation we check for triangle with consecutive sides. Putting b as the smallest, (b+2)^2 = b^2 +b(b+1) (b-4)(b+1)=0 b=4,c=5,a=6 putting similar cases we can show that all other solutions are non-integral

Solution 3

NO TRIGONOMETRY!!!

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

Extend $AC$ to $D$ such that $CD = BC = a.$ Then $<CDB = <ACB/2 = <ABC$, 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.

See Also

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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