6000th Brilliant Problem Solved - Problem

Geometry Level 5

Find the largest constant m m and the smallest constant M M , such that given any 13 distinct real numbers , there always exists 2 numbers x x and y y , with x > y x > y , such that

m < x y 1 + x y < M . m < \dfrac {x-y}{1+xy} < M.

If the value of m + M m+M can be written as a + b c a+b\sqrt{c} , where a a , b b and c c are integers , and c c is square-free, find a + b + c a+b+c .


To commemorate my 6000th solved problem, which was this , I have decided to post this problem.


The answer is 4.

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1 solution

Sharky Kesa
May 26, 2016

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

Notice how the expression looks like the formula for the tangent of the difference of two angles. Indeed, if we substitute x = tan α x=\tan \alpha and y = tan β y=\tan \beta , where α , β [ 0 , 18 0 ] \alpha, \beta \in [0, 180^{\circ}] , we get that the expression is just tan ( α β ) \tan (\alpha - \beta) . Also, we have that α > β \alpha > \beta . Thus, since the expression must result in a non-negative value, we must have the tan ( α β ) > tan 0 = 0 \tan(\alpha - \beta) > \tan 0 = 0 .

We also have that α β \alpha - \beta can take any value but 0 0 . Thus, m = 0 m=0 . Also, we have that the tan \tan function has a period 18 0 180^{\circ} . Thus, by PHP, 2 of these 13 distinct reals must be within a 1 5 15^{\circ} interval bounded by 2 multiples of 15. Thus, the minimum maximum value of tan ( α β ) \tan(\alpha-\beta) must be tan 1 5 = 2 3 \tan 15^{\circ}=2-\sqrt{3} , which is given by the reals being equal to tan ( 15 n ) \tan (15n^{\circ}) . However, for this to be true, one of the real values must be tan 9 0 \tan 90^{\circ} , which is undefined. Thus, the supremum value must be tan 1 5 = 2 3 \tan 15^{\circ}=2-\sqrt{3} .

Therefore, the sum of the infimum and supremum value is 0 + 2 3 = 2 + ( 1 ) 3 0+2-\sqrt{3} = 2+(-1)\sqrt{3} . 2 + ( 1 ) + 3 = 4 2+(-1)+3=\boxed{4}

Congrats on solving 6000 problems. This was a nice problem as well.

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Thanks! Try my proof problem which I also posted as a commemoration of my 6000 solved problems.

Sharky Kesa - 5 years ago

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I've got a proof for that one. Since all the triangles are right angled, the inradius has a nice expression. Pugging the values gives the result.

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@A Former Brilliant Member Can you post that as your proof please? :P

Sharky Kesa - 5 years ago

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@Sharky Kesa I'll post after sometime. ;)

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