Trigonometry Related Part II
( z + x ) ( z + y ) 2 x y + ( x + y ) ( x + z ) 2 y z + ( y + z ) ( y + x ) 3 z x
x , y and z are positive real numbers. If the minimum value of the expression above can be expressed as Y X for positive coprime integers X and Y , determine X + Y .
The answer is 8.
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1 solution
That's a great solution.
Note that instead of "Suppose a , b , c are three sides of a triangle" (where there's the possibility that the statement is not true), we should instead say that "We can observe that a , b , c are three sides of a triangle".
What program are you using to type this up? It would be better to just type it up here, so that we can make edits to your solution. You can refer to the math formatting guide to learn how to get started with Latex.
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Thank you. I used Micro Word to type and then captured the screen. I found this easier tbh. I took up latex writing few days ago but its so complicated and troublesome.
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Indeed, Latex takes a while to learn how it works, as you need to know parts of the code first, as opposed to Word's WWSIWYG.
Thanks for submitting a solution! I've been enjoying your problems. I especially like non-symmetric inequalities where one cannot simply "make all variables equal".
Did the same way, except that I ended up using a quadratic polynomial expression in sin ( 2 β ) . Still, got the same answer.
How do you know when you should apply a = y + z , b = z + x , c = x + y ?
On the last part, it seems like the value of cos((A - C)/2) is 1 because it disappeared. How is that? I'm stuck with that one.
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It's not that it "disappeared". What we applied is that cos 2 A − C ≤ 1 , along with sin 2 B ≥ 0 to conclude that sin 2 B cos 2 A − C ≤ sin 2 B .
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Oh, I got it! Since we have 7/2 - (cosA + cosC + 3/2cosB), we can minimized this one by finding the maximum value of (cosA + cosC + 3/2cosB). But still, I am not sure if this claim was right.
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@Reineir Duran – Yup that's right. If you want to minimize 1 0 − A , it's equivalent to maximizing the value of A .
Thanks for that quick reply sir Calvin. Now I finally understood that part. One more follow-up question: Why did she wrote <= 11/6? (again on the last part), instead of having >= 11/6 since we are looking for the minimum value.
Could you also verify for a maximum of 3? (If I am not mistaken.)
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Since the solution seems complicated, I wonder if the method can also determine for its maximum value. Why a down vote?
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I don't know, who did that??? And by this method we can determine the minimum value only. More interesting but much harder to figure out than conventional methods I guess.
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@Leah Smith – I responded to myself as I also didn't know who did it. You must be doing a lot of geometry or trigonometry and got the fantastic thing. So the method by some sort of coincidence can solve this question for a minimum in the way you described. I think you agree with me that a general way is also important to ensure ways we apply can have certain merit to achieve.
There are quite a few approaches but as I stated above, the trigonometric uses is applied here: