Can you solve this triangle puzzle
[From: ] [author: ] [Date: 11-12-23] [Hit: ]
which does not yield any integersolutions for c.For a = 65, (91 - b) = 91 - (182 - 52 - c) = c - 26, so using (A), we get7 * 3 * 3 * 2 = (c - 26) (91 - c), which yields two integer solutions:c = 89 and c = 28.......
7 * 3 * 3 = (c - 52) (91 - c), which does not yield any integer
solutions for c.
For a = 52, (91 - b) = 91 - (182 - 52 - c) = c - 39, so using (A), we get
7 * 3 * 2 * 2 = (c - 39) (91 - c), which does not yield any integer
solutions for c.
For a = 65, (91 - b) = 91 - (182 - 52 - c) = c - 26, so using (A), we get
7 * 3 * 3 * 2 = (c - 26) (91 - c), which yields two integer solutions:
c = 89 and c = 28. For the former, b = 28; in the latter case, b =
89. So, this yields the sides {65, 28, 89}, which is NOT a right triangle.
For a = 78, (91 - b) = 91 - (182 - 78 - c) = c - 13, so using (A), we get
7 * 3 * 3 * 2 * 2 = (c - 13) (91 - c), which yields no integer
solution for c.
Thus the two possible solutions are: {13, 84, 85} (right triangle) and
{65, 28, 89} (not a right triangle).
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the right triangle sides measure 13, 84, and 85
i'll get back to you on the other one... 28, 65, 89 is a possibility, but i need to verify that it works, before i can say for sure.
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yes
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You're giving out Christmas gifts Rita! An easy problem.
By Heron's triangle area formula {http://en.wikipedia.org/wiki/Heron%27s_f… },
with sides "a", "b", and "c":
546² = (182/2) × (182/2 - a) × (182/2 - b) × (182/2 - c)
⇒ (2)²(3)²(7)(13) = (91 - a) (91 - b) (91 - c)
Also: (91 - a) + (91 - b) + (91 - c) = 3×91 - (a+b+c) = 3×91 - 182 = 91
So it only requires a partition of (2)²(3)²(7)(13) into three (positive) factors which add up to 91. Since each factor must be less than 91, the primes (7) and (13) must be in different factors. Also, since 91 is odd and all three factors cannot be odd (the prime (2) must appear twice), exactly two factors are even. That gives:
14u, 26v, 1w or 14u, 13v, 2w or 7u, 26v, 2w
where (uvw)=3². Since 13 divides 91, consider the first two triplets mod 13:
1u + 0 + 1w or 1u + 0 + 2w
Those are easily seen NOT to be multiples of 13 for (u,w)=(1,1) or (1,3) or (3,1) or (3,3) or (1,9) or (9,1). Hence they can give no solutions.
Consider the last triplet mod 3: 1u+2v+2w≡1 requiring (u,v,w) = (1,3,3) or (9,1,1)
Giving the two solutions:
(91-a)=7, (91-b)=78, (91-c)=6 ⇒ (a,b,c) = (84, 13, 85)
(91-a)=63, (91-b)=26, (91-c)=2 ⇒ (a,b,c) = (28, 65, 89)
Note that 84²+13²=85² while 28²+65²≠89²
so the first solution is a right triangle while the second is not.
Edit: I see that Deepak beat me to it with a complete solution. ("To BE..." also has an answer, though a bit short on details). But mine might be a little simpler so I'll let it stand.
