Saturday, April 25, 2015

Putnam Optimization Hard Problem

Find the minimum of |sinx+cosx+tanx+cotx+secx+cscx|.

This is a horrible problem, you could have wasted an inordinate amount of your precious time finding how to, for example, express this given expression in terms of only one variable instead of the given six and then work from there because the methods that could be employed are not necessarily immediately obvious.  But that's life. If you want to acquire the skills of thinking flexibly and creatively, you need to solve complex and unfamiliar problems.

If you are a mathematics educator, then this kind of problem is what we should shower our students because we must acknowledge that the stronger performance of our students, the more it reflects our curricular emphasis and our schools’ efforts in developing problem solving skills in our students.

Now, back to our problem. There are at least two ways to tackle this problem effectively and I will lead you one after another on two separate blog posts:

Method I:

If we let x=a135, where aR, we see that we have:

sinx=sin(a135)=sinacosa2 and

cosx=cos(a135)=cosa+sina2 and adding these two yields

sinx+cosx=sinacosa2+cosa+sina2=2cosa

Similarly,

tanx=tan(a135)=sina+cosacosasina and

cotx=cot(a135)=cosasinasina+cosa and adding them up gives

tanx+cotx=sina+cosacosasina+cosasinasina+cosa=2cos2asin2a

Finally,

secx=1cosx=2sina+cosa and

cscx=1sinx=2sinacosa, adding the two gives

secx+cscx=2sina+cosa+2sinacosa=22cosasin2acos2a

|sinx+cosx+tanx+cotx+secx+cscx|

=|2cosa+2cos2asin2a+22cosasin2acos2a|

=|2cosa+2cos2asin2a22cosacos2asin2a|

=|2cosa+222cosacos2asin2a|

=|2cosa+222cosa2cos2a1|

=|2cosa+222cosa(2cosa1)(2cosa+1)|

=|2cosa+2(12cosa)(2cosa1)(2cosa+1)|

=|2cosa22cosa+1|

=|(2cosa+22cosa+1)|

=|2cosa+22cosa+1|

=|2cosa+1+22cosa+11|

Treat 2cosa>0, we can hence apply the AM-GM inequality to the following:

(2cosa+1)+22cosa+12(2cosa+1)(2(2cosa+1))=22

Since 2cosa+22cosa+1=221 does have solutions, we can conclude by now that

|sinx+cosx+tanx+cotx+secx+cscx| is 221.

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