1
$\begingroup$

The side of a triangle inscribed in a given circle subtends angles $a, b,$ and $y$ at the center. The minimum value of the arithmetic mean mean of $ \cos (a+ \frac{\pi}{2}), \cos(b+\frac{\pi}{2})$ and $\cos(y+\frac{\pi}{2}) $ is . . . ?

One thing I noticed here is that either $a + b+y = 2 \pi $ a+B+y = pi

or $a+y = b$ enter image description here.

I used a diagram. I don't know if there is a more rigorous way to prove that. The problem can be solved if$ a + b + y = 2 \pi $ but what if it is the other case?

My book solve this problem using AM GM inequality but all these terms are negative so how can that be valid here , also they forgot the other case???

@Drmathva helped me solved it $ with a+b+c = 2 \pi$ case but what about the other case???

$\endgroup$
  • $\begingroup$ Could you please provide a picture? I don't really understand what you mean in the first sentence... $\endgroup$ – Dr. Mathva May 5 at 19:02
  • $\begingroup$ @Dr.Mathva I am on mobile,. I currently can't . This app will not allow me. I will still try. $\endgroup$ – user541396 May 5 at 19:04
  • $\begingroup$ drive.google.com/file/d/14xSZ2KXuAL4sS5cThTLUyF1etkrh-B4a/…. $\endgroup$ – user541396 May 5 at 19:09
  • $\begingroup$ Oh, see. So you want to minimize $$\sum\cos\big(a+\frac\pi2\big)$$? $\endgroup$ – Dr. Mathva May 5 at 19:12
  • $\begingroup$ @Dr.Mathva right! $\endgroup$ – user541396 May 5 at 19:14
1
$\begingroup$

Case 1: $a+b+c=2\pi$

Observe, first of all, that we can expand $$\cos\big(a+\frac\pi2\big)=-\sin a$$

Thus, we want to maximize $$S=\sin a+\sin b+\sin c$$ under the constraint $a+b+c=2\pi$. Since we want to maximize $S$, we want all of it terms to be positive (this is not as rigorous as it might sound, so prove it!). Thus we can assume that $a,b,c\in (0,\pi)$. In that interval $$f(x)=\sin x\implies f''(x)=-\sin x<0$$ We now apply Jense's inequality $$\sin\bigg(\frac{a+b+c}{3}\bigg)=\frac{\sqrt 3}2\ge\frac{\sin a+\sin b+\sin c}3\iff S\le\frac{3\sqrt 3}2$$ Equality holds iff $a=b=c=\frac{2\pi}3$


Addendum

The second case doesn't really exist

enter image description here In fact, we didn't consider the angle subtended by $\color{fuchsia}c$ correctly! And we are done!


Addendum 2

Please observe that the proof given by your book is wrong!

Consider for instance $f(x):=x+3$. With the same argument given by your book, $f(x)_{min}$ in $(0, \infty)$ is achieved when $x=3$. But this is wrong!

Of course $$x+3\ge 2\cdot \sqrt{3x}$$

enter image description here

and they are equal when $x=3$. But this doesn't imply that $x=3$ gives the minimum value for $f$...

$\endgroup$
  • $\begingroup$ Isn't cos negative in second quadrant? $\endgroup$ – user541396 May 5 at 19:20
  • $\begingroup$ Upps... You're right $\endgroup$ – Dr. Mathva May 5 at 19:22
  • $\begingroup$ In your answer you made an assumption that triangle is distributed . But what if it is on the one side of diameter , drive.google.com/file/d/1T1YXVj3eerymrlegCNiV56v1lYDQfpUA/… $\endgroup$ – user541396 May 5 at 19:33
  • $\begingroup$ Done!! @swarnim $\endgroup$ – Dr. Mathva May 7 at 17:16
  • $\begingroup$ How can one be sure Jensen's inequality will not make the same mistake as am-gm??? $\endgroup$ – user541396 May 15 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.