Bob the builder
Bob the builder has many sticks of length 3, 5, and 7. He wants to form triangles each of whose edges consists of exactly 1 stick. How many non-congruent triangles can he form with the sticks?
The answer is 9.
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7 solutions
Some things I think need to be cleared up about the problem.
Can Bob join sticks together to make the triangle's sides longer (like a 6 − 1 0 − 1 2 triangle, for example)? Do the endpoints of the sticks have to mark the vertices of the triangle?
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Then there may be infinitely many triangles
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But he doesn't have an infinite number of sticks.
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@Brad Morin – If you create a 3 − 5 − 7 triangle and then take the side of length 7 and slide it on the plane the triangle is in, you can create an infinite amount of triangles using only 1 of each stick.
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@Trevor B. – I don't think the sticks are meant to overlap or meet the other sticks at some points (not the endpoints), but no matter what, the question should mention that if can it be overlapped or the sticks can meet at some points of the other stick but not the endpoints.
We can't join two sticks, since we only need one stick in each side.
BUT ARE'NT 3-3-3 5-5-5 7-7-7 CONGRUENT SO FINALLY WE HAVE OLT 7? PLS CLARIFY
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They are similar but not congruent. For similarity to become congruence, the corresponding sides must be equal.
it was said tat triangles are non congruent 3-3-3 ;5-5-5; 7-7-7 are congruent
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3-3-3 and 5-5-5 are similar, but not congruent.
I bet you used google translate and misunderstood the meaning, I did it and couldn't figure any shit out of it.
A problem is flawed when it requires assumptions, even though the assumptions appear to be bizarre to others. I assumed he could more than one stick for a side. Yes, I was stumped, and I am embarrassed for the silly assumption.
thanx
i did it the same way but i think there should be an alternative solution in which we wont have to physically count the possibilities...for if the no. of sticks increase, it might pose a problem to count.
By using those unique lengths given in the question, we can make triplets...... and we will get 10 of them (counting all rotation and reflection):
BUT
We will leave the [3] [3] [7] <- triplet ( because sum of any two sides of a triangle is always greater than the third side and in this triplet, the rule has been violated)
Therefore, total of possible triangles are = \boxed{9}
( 3 , 3 , 3 ) , ( 5 , 5 , 5 ) , ( 7 , 7 , 7 ) , ( 3 , 3 , 5 ) , ( 3 , 3 , 7 ) , ( 5 , 5 , 3 ) , ( 5 , 5 , 7 ) , ( 7 , 7 , 3 ) , ( 7 , 7 , 5 ) , ( 3 , 5 , 7 ) These are all combinations and count to 10. But we cannot make a traingle with sides 3 , 3 , 7 as 3 + 3 < 7 [Triangular inequality is dissatisfied]. So the answer is 1 0 − 1 = 9
I solved it as 1 + 3p2 + 3 = 10 But i forgot what you said :-(
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1 + 3p2 + 3 = 10
True ... One scalene, six isosceles (five once you eliminate 3,3,7) and three equilateral.
But you can also do 5C3=10.
There are 1 0 different combinations of sticks. But remember that a side of a triangle is less than the sum of the other two, and greater than their difference. Look carefully. The triangle ( 3 , 3 , 7 ) isn't possible because 7 > 3 + 3 and it should be minor. So the answer is 1 0 − 1 = 9 .
possible triangles are 333,555,777,335,553,557,577,773and 357
disgusting solution
We use the triangle Inequality. We want to list all the possible triples ( a , b , c ) such that a + b > c , a + c > b , and b + c > a , and a , b , c ∈ { 3 , 5 , 7 } . We do not care about order. We get:
- (3, 3, 3)
- (5, 5, 5)
- (7, 7, 7)
- (3, 3, 5)
- (5, 5, 7)
- (7, 7, 5)
- (7, 7, 3)
- (3, 5, 5)
- (3, 5, 7)
Our answer is 9 .
We find the number of triples ( a , b , c ) (not ordered), such that a , b , c ∈ { 3 , 5 , 7 } , where a , b , c are the sides of a triangle. We note that the only triple with all different side-lengths is ( 3 , 5 , 7 ) . We check whether a triangle can be formed with these side-lengths using the Triangle Inequality and we find that ( a , b , c ) = ( 3 , 5 , 7 ) satisfies the inequality. We also note that we can have 3 equilateral triangles formed with ( a , b , c ) = ( 3 , 3 , 3 ) , ( 5 , 5 , 5 ) , ( 7 , 7 , 7 ) . Now we need to consider the case where the triangle is isosceles, i.e. 2 of the side lengths are the same and third one is different. There is 6 such triples: ( 3 , 3 , 5 ) , ( 3 , 3 , 7 ) , ( 5 , 5 , 3 ) , ( 5 , 5 , 7 ) , ( 7 , 7 , 5 ) , ( 7 , 7 , 3 ) . A quick check suggests that only the triple ( 3 , 3 , 7 ) doesn't satisfy the Triangle Inequality but the rest do. Therefore the number of unique triangles that can be formed is 1 + 3 + 5 = 9 .
Examine the unique triplets of numbers that can be created with 3 , 5 , and 7 . There are 1 0 of them. Any other triplet of numbers would be a rotation or reflection of these.
3 − 3 − 3 3 − 3 − 5 3 − 3 − 7 3 − 5 − 5 3 − 5 − 7 3 − 7 − 7 5 − 5 − 5 5 − 5 − 7 5 − 7 − 7 7 − 7 − 7
You might think that the answer is 1 0 , but remember that in a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. This eliminates 3 − 3 − 7 as a possible triangle, so the answer is 9