[solved] - I find this ugly...

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[solved] - I find this ugly...

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Source: flanders '88

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3615 posts

Peter

#1 h Aug 9, 2004, 8:26 PM • 2 Y

Y by Adventure10, Mango247

Be $R$ a positive real number. If $R, 1, R+\frac12$ are triangle sides, call $\theta$ the angle between $R$ and $R+\frac12$ (in rad).

Prove $2R\theta$ is between $1$ and $\pi$ .

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liyi

#2 h Aug 10, 2004, 1:59 PM • 2 Y

Y by Adventure10, Mango247

can we use derivatives?

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Peter

#3 h Aug 10, 2004, 2:21 PM • 2 Y

Y by Adventure10, Mango247

Bah, you can. Although I didn't mean it was THAT ugly.

There are solutions without using analytic geometry if that's what you meant

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#4 h Aug 28, 2004, 3:31 PM • 2 Y

Y by Adventure10, Mango247

Peter VDD wrote:

Be $R$ a positive real number. If $R, 1, R+\frac12$ are triangle sides, call $\theta$ the angle between $R$ and $R+\frac12$ (in rad).

Prove $2R\theta$ is between $1$ and $\pi$ .

I actually don't find it so ugly...

Here is my solution:

At first, clearly $0<\theta <\pi$ , since $\theta$ is an angle of a triangle. We must prove $1<2R\theta <\pi$ .

Since the numbers R, 1, $R+\frac12$ are the sides of a triangle, we must have $R+\left( R+\frac12\right) >1$ , what simplifies to 4R > 1. We denote u = 4R and thus have u > 1.

Now, the semiperimeter of our triangle is

$s=\frac{1}{2}\left( R+1+\left( R+\frac{1}{2}\right) \right) =R+\frac{3}{4}=\frac{1}{4}u+\frac{3}{4}=\frac{1}{4}\left( u+3\right)$

(since $R=\frac{1}{4}u$ , as u = 4R). After the half-angle formulas, we have

$\displaystyle \sin ^{2}\frac{\theta }{2}=\frac{\left( s-R\right) \left( s-\left( R+\frac{1}{2}\right) \right) }{R\left( R+\frac{1}{2}\right) }$ .

This quickly simplifies to

$\displaystyle \sin ^{2}\frac{\theta }{2}=\frac{3}{u\left( u+2\right) }$ .

Now we will show $1<2R\theta$ . This is obviously equivalent to

$\displaystyle \frac{\theta }{2}>\frac{1}{4R}$ ,

i. e. to

$\displaystyle \frac{\theta }{2}>\frac{1}{u}$ .

This, in turn, is equivalent to

$\displaystyle \sin ^{2}\frac{\theta }{2}>\sin ^{2}\frac{1}{u}$ ,

since the function $\sin ^2 x$ increases on $\displaystyle \left[ 0;\;\frac{\pi }{2}\right]$ , and both numbers $\displaystyle \frac{\theta }{2}$ and $\displaystyle \frac{1}{u}$ lie in this interval (since $\theta <\pi$ implies $\displaystyle \frac{\theta }{2}<\frac{\pi }{2}$ , and u > 1 implies $\displaystyle \frac{1}{u}<1<\frac{\pi }{2}$ ).

Now, in order to prove

$\displaystyle \sin ^{2}\frac{\theta }{2}>\sin ^{2}\frac{1}{u}$ ,

we will show

$\displaystyle \sin ^{2}\frac{\theta }{2}>\left( \frac{1}{u}\right) ^{2}$ ;

in fact, once this will be shown, from x > sin x for positive x we will immediately get $\displaystyle \sin ^{2}\frac{\theta }{2}>\left( \frac{1}{u}\right) ^{2}>\sin ^{2}\frac{1}{u}$ .

Since $\displaystyle \sin ^{2}\frac{\theta }{2}=\frac{3}{u\left( u+2\right) }$ , the inequality

$\displaystyle \sin ^{2}\frac{\theta }{2}>\left( \frac{1}{u}\right) ^{2}$

that we must prove simplifies to

$\displaystyle \frac{3}{u\left( u+2\right) }>\left( \frac{1}{u}\right) ^{2}$ .

This one, after multiplication with $u^{2}\left( u+2\right)$ , takes the form 3u > u + 2, or, in other words, 2u > 2, what is clearly true because u > 1. Thus we have proven the left bound $1<2R\theta$ .

Remains to establish the right bound $2R\theta <\pi$ . Again, this is equivalent to

$\displaystyle \frac{\theta }{2}<\frac{\pi }{4R}$ ,

i. e. to

$\displaystyle \frac{\theta }{2}<\frac{\pi }{u}$ .

Now, this is trivial for $u\leq 2$ (since in this case we just have $\displaystyle \frac{\theta }{2}<\frac{\pi }{2}\leq \frac{\pi }{u}$ ); thus, in the following, we will only consider the case u > 2. Then, again, both numbers $\displaystyle \frac{\theta }{2}$ and $\displaystyle \frac{\pi }{u}$ lie in the interval $\displaystyle \left[ 0;\;\frac{\pi }{2}\right]$ (in fact, for $\displaystyle \frac{\theta }{2}$ we have already showed this, and for $\displaystyle \frac{\pi }{u}$ it's clear because $\displaystyle \frac{\pi }{u}<\frac{\pi }{2}$ follows from u > 2). Hence, instead of proving

$\displaystyle \frac{\theta }{2}<\frac{\pi }{u}$ ,

it is enough to show that

$\displaystyle \sin ^{2}\frac{\theta }{2}<\sin ^{2}\frac{\pi }{u}$ ,

i. e. to show that

$\displaystyle \frac{3}{u\left( u+2\right) }<\sin ^{2}\frac{\pi }{u}$

(here we have used our formula $\displaystyle \sin ^{2}\frac{\theta }{2}=\frac{3}{u\left( u+2\right) }$ again).

To prove the inequality $\displaystyle \frac{3}{u\left( u+2\right) }<\sin ^{2}\frac{\pi }{u}$ , we take its reciprocal:

$\displaystyle \frac{u\left( u+2\right) }{3}>\csc ^{2}\frac{\pi }{u}$ .

Now we subtract 1 on both sides:

$\displaystyle \frac{u\left( u+2\right) }{3}-1>\csc ^{2}\frac{\pi }{u}-1$ .

But $\displaystyle \frac{u\left( u+2\right) }{3}-1=\frac{\left( u-1\right) \left( u+3\right) }{3}$ and $\displaystyle \csc ^{2}\frac{\pi }{u}-1=\cot ^{2}\frac{\pi }{u}$ . Thus, we have to prove

$\displaystyle \frac{\left( u-1\right) \left( u+3\right) }{3}>\cot ^{2}\frac{\pi }{u}$ .

Again, we take the reciprocal (we can do it since both sides of our inequality are positive, the left one since u > 1 and the right one since $\displaystyle \frac{\pi }{u}<\frac{\pi }{2}$ ):

$\displaystyle \frac{3}{\left( u-1\right) \left( u+3\right) }<\tan ^{2}\frac{\pi }{u}$ .

In order to verify this, it is enough to show that

$\displaystyle \frac{3}{\left( u-1\right) \left( u+3\right) }<\left( \frac{3}{u}\right) ^{2}$ ;

in fact, once this will be shown, we will immediately get

$\displaystyle \frac{3}{\left( u-1\right) \left( u+3\right) }<\left( \frac{3}{u}\right) ^{2}<\left( \frac{\pi }{u}\right) ^{2}<\tan ^{2}\frac{\pi }{u}$ ,

where we have used $3<\pi$ and x < tan x for all $\displaystyle x\in \left[ 0;\;\frac{\pi }{2}\right]$ . So we have to show that

$\displaystyle \frac{3}{\left( u-1\right) \left( u+3\right) }<\left( \frac{3}{u}\right) ^{2}$ .

After multiplication with $u^{2}\left( u-1\right) \left( u+3\right)$ , this takes the form $3u^{2}<9\left( u-1\right) \left( u+3\right)$ . Dividing by 3 and simplifying, we can rewrite this inequality as $2u^2+6u-9>0$ . But the roots of the quadratic trinomial $2u^2+6u-9$ are $\displaystyle -\frac{3}{2}+\frac{3}{2}\sqrt{3}$ and $\displaystyle -\frac{3}{2}-\frac{3}{2}\sqrt{3}$ , both < 2. Since u > 2, we thus have $2u^2+6u-9>0$ . And this completes the proof of the right bound.

This ends the solution of the problem. But what makes me a bit mistrustful is the wealth of traps and possibilities to make some dumb errors in this solution. I have fixed some before posting, but I am still not at all sure I have no mistakes left now...

Darij

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Peter

#5 h Aug 28, 2004, 9:34 PM • 2 Y

Y by Adventure10, Mango247

darij grinberg wrote:

This ends the solution of the problem. But what makes me a bit mistrustful is the wealth of traps and possibilities to make some dumb errors in this solution. I have fixed some before posting, but I am still not at all sure I have no mistakes left now...
Darij

Hell yeah, such a solution!

I'll let you know if I find a mistake, but it can take a while before I'm finished with such a long answer.

Anyway the official sols (there were 3 I guess) all were less than half a page

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#6 h Nov 23, 2005, 11:08 PM • 1 Y

Y by Adventure10

darij grinberg wrote:

This ends the solution of the problem. But what makes me a bit mistrustful is the wealth of traps and possibilities to make some dumb errors in this solution. I have fixed some before posting, but I am still not at all sure I have no mistakes left now...
Darij

Darij,

Time is ripe to fix your solution to this problem a bit. It must be clear that

when the difference between two sides of a triangle is fixed, then the vertex

moves on hyperbola (in this case focal distance 2c=1 and 2a=1/2).

The upper bound is obvious, the lower can be worked out rather quickly.

I leave you the honors to clean up your own stuff, so that the gatekeeper

of 'Posterity' will undoubtedly induct you in there too.

Thank you.

M.T.

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