Triangle Inequality Extension

Algebra Level 2

Suppose a , b , a, b, and c c are positive integers.

If a 2 + b 2 < c 2 , a^2 + b^2 < c^2 , is it necessarily true that a + b < c ? a + b < c ?

Yes No

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4 solutions

Consider a (non-degenerate) triangle with integer side lengths a , b , c a,b,c and opposing angles A , B , C A,B,C . Now if the triangle is obtuse with C > 9 0 \angle C \gt 90^{\circ} then cos ( C ) < 0 \cos(C) \lt 0 , and so by the cosine law

c 2 = a 2 + b 2 2 a b cos ( C ) > a 2 + b 2 c^{2} = a^{2} + b^{2} - 2ab\cos(C) \gt a^{2} + b^{2} .

But by the triangle inequality a + b > c a + b \gt c , so the given statement is not true, i.e., if a 2 + b 2 < c 2 a^{2} + b^{2} \lt c^{2} it is not necessarily the case that a + b < c a + b \lt c .

Nice way of relating it to the triangle inequality!

Calvin Lin Staff - 4 years, 7 months ago
Siva Budaraju
Apr 22, 2017

NO.

As a counter-example, take 2,3,4. 2+3>4 but 2^2+3^2<4^2.

I couldn't think of a general explanation so I will give an example: (0.5^2)+(0.6^2) = 0.25 + 0.36 = 0.61< 1^2 = 1, but 0.5+0.6 = 1.1>1.

Therefore it is not always true that a^2 + b^2 < c^2.

While your argument can be easily modified, the question states that the variables are positive integers.

Lolly Lau - 4 years, 5 months ago
Munem Shahriar
Aug 29, 2018

If a = b = 1 a = b = 1 and c = 2 c = 2 then 1 2 + 1 2 < 2 2 1^2 + 1^2 < 2^2 and 1 + 1 2 1 + 1 \not< 2 . Hence the answer is no.

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