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Suppose that \(a, b,c\) are positive integers such that \[\begin{aligned} a+b+c &= 32\\ \frac{b+c-a}{bc}+\frac{c+a-b}{ca}+\frac{a+b-c}{ab} &= \frac{1}{4} \end{aligned}\]

Does there exist a triangle whose sidelengths are \(\sqrt{a}, \sqrt{b}\) and \(\sqrt{c}\)? If there is, find its largest angle.

Edit by Calvin: As mentioned, solve this question for positive real values of \(a, b, c\) so that you do not have finitely many triples to check.

Remember that the triangle which we are concerned about has side lengths \( \sqrt{a}, \sqrt{b}, \sqrt{c} \) and not \( a, b, c \).

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TopNewestThe distinct triplets \((a,b,c)\) satisfying the conditions are 8 sets \([16,1\rightarrow8,15\rightarrow8]\), where \(a=b+c\), Hence by

Pythagoras theorem, we can conclude that all the triangles are right-angled, since, \[(\sqrt{b})^2+(\sqrt{c})^2=(\sqrt{a})^2\]. As in a right-angled triangle, the largest angle has to be the right-angle. Therefore, the largest angle is \((\frac{\Pi}{2})^c=90^o.\)Log in to reply

How did you find the sets? Are these 8 the only ones satifying the conditions?

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This is an interesting question, though the restriction of "positive integers" implies that there are only finitely many cases to check.

It would be more interesting if the values were positive reals instead. If there are other real solutions, that could imply that Diophantine techniques need to be applied, but I don't see any direct approach.

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It's not fully mathematical, but its how I found it. Actually, at first I saw,

Rahul N.'s post., which forced me into thinking that if \(c=a+b\) was a general condition. Then, I saw that it led to nowhere. Then, I did a specific checking, I put \(c=16\), in the second equation, which gave me \[(a+b-16)^2=0\], which gives \(a+b=\{16,16\}\), Surprisingly, it automatically satisfies the first condition (since \(c=16\) was assumed),and it is true for all reals (and not only integers).Then, I took the distinct sets into account. N.B - As it is symmetrical about all \(a,b,c\), so putting \(a=16\text{ or }b=16\), would also lead to the same, which is obvious.P.S - I tried to solve in a more concrete way. But always somewhere, I had to assume \(c=16\) to solve. Hoping somebody would post a more logical solution, clearing up, why take \(c=16\) in the first place, and why not any other number. But, I know it has something to do with the \(\frac{1}{4}\) present in the \(R.H.S\) of the second equation.Log in to reply

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What if the equations were for positive real values instead?

As a side note, there are more solutions than just the 8 listed sets.

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This problem intrigued me (see my other comment). The condition of "positive integers" can be replaced with "positive reals", which makes this an algebra question instead (and hence affects the approaches that you would use).

There are infinitely many sets of positive reals which satisfy the conditions. Find them. For each of these sets, determine the largest angle.

Hint:Homogenize the equation.Log in to reply

is answer 90 degrees? a = 4 , b = 12 , c =16 ?

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That's a degenerate triangle, as c = a+b

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No - the triangle has side lengths \(\sqrt a, \sqrt b, \sqrt c,\) not \(a, b, c.\)

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Multiply evrything by abc, you get \[2ab+2bc+2ac-a^2-b^2-c^2=\frac{abc}{4}\] Then factor it as \[(\sqrt{a}+\sqrt{b}+\sqrt{c})(-\sqrt{a}+\sqrt{b}+\sqrt{c}) (\sqrt{a}-\sqrt{b}+\sqrt{c}) (\sqrt{a}+\sqrt{b}-\sqrt{c})=\frac{abc}{4}\] Divide everything by 16 \[\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\frac{-\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\frac{\sqrt{a}-\sqrt{b}+\sqrt{c}}{2} \frac{\sqrt{a}+\sqrt{b}-\sqrt{c}}{2}=\frac{abc}{64}\] Split them \[\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}-\sqrt{a})(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}-\sqrt{b})(\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{x}-\sqrt{a})=\frac{abc}{64}\] put \[p=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{2}\], we get \[\sqrt{p(p-\sqrt{a})(p-\sqrt{b})(p-\sqrt{c})}=\frac{\sqrt{abc}}{8}\] Similarly, we could do the same for the other sides, I just chose \(c\) cause I liked it So the area of the triangle \((\sqrt{a},\sqrt{b},\sqrt{c})\) is \(\frac{\sqrt{abc}}{8}\) So the altitude relative to the side \(\sqrt{c}\) is \(\frac{\sqrt{ab}}{4}\) Therefore, from pythagoras we get \[\sqrt{c}=\sqrt{a-\frac{ab}{16}}+\sqrt{b-\frac{ab}{16}}\] And also that \[a,b<16\] Squaring \[c=a+b-\frac{ab}{8}+2\sqrt{ab(1-\frac{a}{16})(1-\frac{b}{16})}\] Summing \(a+b\) to complete the 32 \[32=2a+2b-\frac{ab}{8}+2\sqrt{ab(1-\frac{a}{16})(1-\frac{b}{16})}\] Multiplying by 8 \[256=16a+16b-ab+\sqrt{ab(16-a)(16-b)}\] \[256+ab-16a-16b=\sqrt{ab(16-a)(16-b)}\] \[(16-a)(16-b)=\sqrt{ab(16-a)(16-b)}\] \[(16-a)(16-b)=ab\] \[256-16a-16b+ab=ab\] \[16=a+b\]

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Obviously we could do the same for the other sides, I just chose \(c\) because I liked it :)

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Hint:Heron formula

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(1 15 16), (2 14 16), (3 13 16), (4 12 16), (5 11 16), (6 10 16), (7 9 16), (8 8 16) all these side lengths satisfy the two equations and can form sides of triangles. Used MSExcel to find them

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oh. i found a, b c but side lengths are square roots of a b c. Still a b c candidates are (1 15 16), (2 14 16), (3 13 16), (4 12 16), (5 11 16), (6 10 16), (7 9 16). (6 10 16) or (7 9 16) can provide largest angle

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in triangle 8 8 16, the triangle cannot be made

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Uhh, those triples form a degenerate triangle, as the longest side (which is 16) is equal to the sum of the other two sides for all of the triples you presented.

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1,15,16. Angles will be calculated by cosine rule

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