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Square Roots as Triangle Sidelengths

Main post link -> https://www.artofproblemsolving.com/Forum/viewtopic.php?f=151&t=502059

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 \( a, b, c \).

Note by Russelle Guadalupe
3 years, 11 months ago

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The 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 \((\frac{\Pi}{2})^c=90^o.\) Arnab Animesh Das · 3 years, 11 months ago

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@Arnab Animesh Das How did you find the sets? Are these 8 the only ones satifying the conditions? Ton De Moree · 3 years, 11 months ago

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@Ton De Moree 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. Calvin Lin Staff · 3 years, 11 months ago

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@Ton De Moree 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. Arnab Animesh Das · 3 years, 11 months ago

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@Arnab Animesh Das This does not explain why there are no other solutions where one of the numbers is not 16. Calvin Lin Staff · 3 years, 11 months ago

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@Arnab Animesh Das By transforming the equations to yield an equation: \[ab+bc+ca=256+\frac{1}{16}abc\] you can gradually worm your way to showing that one of $a,\,b,\,c$ is divisible by 16. It takes a while, but eventually you exhaust all the possibilities, and they all lead to this conclusion. Igor Kotrasiński · 3 years, 11 months ago

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@Igor Kotrasiński Yes, with positive integer values, you have finitely many cases to check, and can use divisibility arguments to restrict the cases.

What if the equations were for positive real values instead?

As a side note, there are more solutions than just the 8 listed sets. Calvin Lin Staff · 3 years, 11 months ago

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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. Calvin Lin Staff · 3 years, 11 months ago

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is answer 90 degrees? a = 4 , b = 12 , c =16 ? Rahul Nahata · 3 years, 11 months ago

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@Rahul Nahata That's a degenerate triangle, as c = a+b Ivan Stošić · 3 years, 11 months ago

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@Ivan Stošić No - the triangle has side lengths \(\sqrt a, \sqrt b, \sqrt c,\) not \(a, b, c.\) Michael Tang · 3 years, 11 months ago

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@Michael Tang I think it's funny that people upvoted your comment but downvoted the original answer :P Sotiri Komissopoulos · 3 years, 11 months ago

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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 \[256=16a+16b-ab+\sqrt{ab(16-a)(16-b)}\] \[256+ab-16a-16b=\sqrt{ab(16-a)(16-b)}\] \[(16-a)(16-b)=\sqrt{ab(16-a)(16-b)}\] \[(16-a)(16-b)=ab\] \[256-16a-16b+ab=ab\] \[16=a+b\] Luis Rivera · 3 years, 11 months ago

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@Luis Rivera Obviously we could do the same for the other sides, I just chose \(c\) because I liked it :) Luis Rivera · 3 years, 11 months ago

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Hint:Heron formula Luis Rivera · 3 years, 11 months ago

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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 Ali Khan · 3 years, 11 months ago

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@Ali Khan 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 Ali Khan · 3 years, 11 months ago

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@Ali Khan in triangle 8 8 16, the triangle cannot be made Ali Khan · 3 years, 11 months ago

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@Ali Khan 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. Jon Erik Gonzaga · 3 years, 11 months ago

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1,15,16. Angles will be calculated by cosine rule Ali Khan · 3 years, 11 months ago

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