Difference between revisions of "1968 IMO Problems/Problem 1"
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==Solution 2== | ==Solution 2== | ||
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In a given triangle <math>ABC</math>, let <math>A=2B</math>, <math>\implies C=180-3B</math>, and <math>\sin C=\sin 3B</math>. | In a given triangle <math>ABC</math>, let <math>A=2B</math>, <math>\implies C=180-3B</math>, and <math>\sin C=\sin 3B</math>. | ||
− | Then < | + | Then |
− | Hence, <math>(*)</math> <math>a^2 = b(b+c)</math> ( | + | |
+ | <math>\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)</math> | ||
+ | |||
+ | Hence, | ||
+ | |||
+ | <math>a^2 = b(b + c)\ (*)</math> | ||
+ | |||
+ | Indeed, we know from the [[Law of Sines]] that | ||
+ | |||
+ | <math>\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}</math>. | ||
+ | |||
+ | Denoting this ratio by <math>r</math>, we have <math>\sin A = ra, \sin B = rb, \sin C = rc</math>. | ||
+ | Substituting in <math>\sin ^2 A = \sin B(\sin B + \sin C)</math> and simplifying by <math>r^2</math>, | ||
+ | we get <math>(*)</math>. | ||
+ | |||
+ | At this point, notice that <math>(*)</math> is equivalent to the equality | ||
+ | <math>a^2c = b(a^2 + c^2 - b^2)</math> from Solution 1. Indeed, the latter | ||
+ | can be rewritten as <math>a^2(c - b) = b(c + b)(c - b)</math>, and we know | ||
+ | that <math>c \ne b</math>. So we could simply quote the fact proven in | ||
+ | Solution 1 that if <math>a, b, c</math> are consecutive integers and | ||
+ | <math>a^2 = b(c + b)</math>, then <math>b = 4, c = 5, a = 6</math> is the only solution. | ||
+ | |||
+ | For the sake of completeness, and for fun, we give a slightly | ||
+ | different proof here. | ||
+ | |||
+ | We have six possibilities, depending on how the three consecutive | ||
+ | numbers are ordered. The six possibilities are: | ||
+ | |||
+ | 1: <math>a = b - 2, c = b - 1, b</math> | ||
+ | |||
+ | 2: <math>c = b - 2, a = b - 1, b</math> | ||
+ | |||
+ | 3: <math>a = b - 1, b, c = b + 1</math> | ||
+ | |||
+ | 4: <math>c = b - 1, b, a = b + 1</math> | ||
+ | |||
+ | 5: <math>b, a = b + 1, c = b + 2</math> | ||
+ | |||
+ | 6: <math>b, c = b + 1, a = b + 2</math> | ||
+ | |||
+ | For each case, we substitute <math>a, c</math> in <math>(*)</math>, get an equation | ||
+ | in <math>b</math>, solve it, and get all the possible solutions. As a | ||
+ | shortcut, notice that (*) implies that <math>b|a^2</math>. If <math>a, b</math> | ||
+ | are consecutive integers, then they are relatively prime, so | ||
+ | <math>b|a^2</math> can not be true unless <math>b = 1</math>. In this case the | ||
+ | triangle would have sides <math>1, 2, 3</math>, which is impossible. | ||
+ | This eliminates cases 2, 3, 4 and 5. | ||
+ | |||
+ | In case 1, <math>(*)</math> becomes | ||
+ | |||
+ | <math>(b - 2)^2 = b(b + b - 1)</math>, or <math>b^2 + 3b - 4 = 0</math>. | ||
+ | |||
+ | This has solutions <math>1, -4</math>. The value <math>b = -4</math> is impossible. | ||
+ | The value <math>b = 1</math> yields <math>a = -1, c = 0</math>, which is impossible. | ||
+ | |||
+ | In case 6, <math>(*)</math> becomes | ||
+ | |||
+ | <math>(b + 2)^2 = b(b + b + 1)</math>, or <math>b^2 - 3b - 4 = 0</math>. | ||
+ | |||
+ | |||
+ | |||
+ | |||
If <math>b</math> is the shortest side, <math>(b+2)^2 = b^2 +b(b+1)</math> | If <math>b</math> is the shortest side, <math>(b+2)^2 = b^2 +b(b+1)</math> | ||
<math>\implies (b-4)(b+1)=0</math>, | <math>\implies (b-4)(b+1)=0</math>, |
Revision as of 17:56, 21 August 2024
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
In a given triangle , let , , and . Then
Hence,
Indeed, we know from the Law of Sines that
.
Denoting this ratio by , we have . Substituting in and simplifying by , we get .
At this point, notice that is equivalent to the equality from Solution 1. Indeed, the latter can be rewritten as , and we know that . So we could simply quote the fact proven in Solution 1 that if are consecutive integers and , then is the only solution.
For the sake of completeness, and for fun, we give a slightly different proof here.
We have six possibilities, depending on how the three consecutive numbers are ordered. The six possibilities are:
1:
2:
3:
4:
5:
6:
For each case, we substitute in , get an equation in , solve it, and get all the possible solutions. As a shortcut, notice that (*) implies that . If are consecutive integers, then they are relatively prime, so can not be true unless . In this case the triangle would have sides , which is impossible. This eliminates cases 2, 3, 4 and 5.
In case 1, becomes
, or .
This has solutions . The value is impossible. The value yields , which is impossible.
In case 6, becomes
, or .
If is the shortest side,
,
,
No other permutation of , and in terms of size gives integral values to [show]. So there is only one such triangle.
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 |