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. | ||
− | ==Solution== | + | ==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}} | ||
− | + | ||
+ | ==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 | ||
+ | |||
+ | ==Solution 3== | ||
+ | NO TRIGONOMETRY!!! | ||
+ | |||
+ | Let <math>a, b, c</math> be the side lengths of a triangle in which <math><C = 2<B.</math> | ||
+ | |||
+ | 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
Contents
[hide]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 , let , , , , and . Using the Law of Sines gives that
Therefore . Using the Law of Cosines gives that
This can be simplified to . Since , , and are positive integers, . Note that if is between and , then is relatively prime to and , and cannot possibly divide . Therefore is either the least of the three consecutive integers or the greatest.
Assume that is the least of the three consecutive integers. Then either or , depending on if or . If , then is 1 or 2. couldn't be 1, for if it was then the triangle would be degenerate. If is 2, then , but and must be 3 and 4 in some order, which means that this triangle doesn't exist. therefore cannot divide , and so must divide . If then , so is 1, 2, or 4. Clearly cannot be 1 or 2, so must be 4. Therefore . This shows that and , and the triangle has sides that measure 4, 5, and 6.
Now assume that is the greatest of the three consecutive integers. Then either or , depending on if or . is absurd, so , and . Therefore is 1, 2, or 4. However, all of these cases are either degenerate or have been previously ruled out, so 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.
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 be the side lengths of a triangle in which
Extend to such that Then , so and are similar by AA Similarity. Hence, . 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 |