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Inequality => square

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Source: INMO 1998 Problem 4

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1592 posts

Rushil

#1 h Oct 7, 2005, 5:12 AM • 2 Y

Y by Adventure10, Mango247

Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$ , prove that $ABCD$ is a square.

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3498 posts

#2 h Oct 7, 2005, 7:28 AM • 2 Y

Y by Adventure10, Mango247

Is $f(x)=\sin{x}$ concave on $[0, \pi]$ ? If so, then this problem can be proved by $l=2R\sin{\theta}$ then AM-GM then jensen.

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3660 posts

Arne

#3 h Oct 7, 2005, 7:32 AM • 5 Y

Y by AKS_9_54_61, srijonrick, Adventure10, Mango247, and 1 other user

Hm... I'd say that by Ptolemy we have $4 \geq AC \cdot BD = AB \cdot CD + AD \cdot BC \geq 2 \sqrt{AB \cdot BC \cdot CD \cdot DA} \geq 4$ (since $AC$ and $AB$ are not longer than a diameter of the circle).

So equality must hold everywhere. Hence $AC$ and $BD$ are diameters, $ABCD$ is a rectangle, and also $AB \cdot CD = AD \cdot BC$ which implies that $AB^2 = AD^2$ , so $ABCD$ is a square, and we're done.

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3498 posts

#4 h Oct 7, 2005, 7:54 AM • 2 Y

Y by Adventure10, Mango247

Oh...... I didn't think about using Ptolemy. Nice proof Arne.

Anyway, is my method correct?

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3660 posts

Arne

#5 h Oct 7, 2005, 12:34 PM • 2 Y

Y by Adventure10, Mango247

Yeah, I think so...

Could you write a full solution? Then it will be easier to judge

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3498 posts

#6 h Oct 7, 2005, 12:46 PM • 2 Y

Y by Adventure10, Mango247

OK.

Let $\angle{AOB}=2a$ , $\angle{BOC}=2b$ , $\angle{COD}=2c$ , $\angle{DOA}=2d$ . Then $a+b+c+d=180^o$ .
Since we also have: $AB=2R \sin{a}$ and etc, Hence:

$AB \cdot BC \cdot CD \cdot DA=16R^2 \cdot \sin{a} \sin{b} \sin{c} \sin{d}$
By AM-GM: $\sin{a} \sin{b} \sin{c} \sin{d} \leq (\dfrac{\sin{a}+\sin{b}+\sin{c}+\sin{d}}{4})^4$ .

Then jensen: $\dfrac{\sin{a}+\sin{b}+\sin{c}+\sin{d}}{4} \leq \sin{(\dfrac{a+b+c+d}{4})}=\dfrac{\sqrt{2}}{2}$ .

Thus $16R^2\cdot \sin{a} \sin{b} \sin{c} \sin{d} \leq 16 \cdot \frac14=4.$

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3660 posts

Arne

#7 h Oct 7, 2005, 12:58 PM • 2 Y

Y by Adventure10, Mango247

That looks fine to me!

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#8 h Sep 3, 2012, 1:08 AM • 1 Y

Y by Adventure10

using Ptolemy's theorem,we have $AB.DC+AD.BC=AC.BD\le4$ with equality iff $AC$ and $BD$ are diameters.
so , we have , by AM-GM , $AB.BC.CD.AD\le (\frac{AB.DC+AD.BC}{2})^2 \le(4/2)^2=4$ with equality iff $AB.DC=AD.BC$
so , from the given condition in the problem , we must have , $AB.BC.CD.DA=4.$
so , from the equality conditions and simple geometry , it is obvious that ABCD is a square.

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729 posts

SRIDEV

#9 h Oct 13, 2014, 10:55 PM • 2 Y

Y by Adventure10, Mango247

Dear @shobber ,

You wrote AB = 2RSina
Thus AB.BC.CD.DA should = 16R^4. Sina.Sinb.Sinc.Sind

But you wrote
AB.BC.CD.DA = 16R^2. Sina.Sinb.Sinc.Sind

Please do clarify, how R^4 becomes R^2 ?

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1566 posts

#10 h Dec 13, 2017, 3:52 PM • 2 Y

Y by Adventure10, Mango247

SRIDEV wrote:

Dear @shobber ,

You wrote AB = 2RSina
Thus AB.BC.CD.DA should = 16R^4. Sina.Sinb.Sinc.Sind

But you wrote
AB.BC.CD.DA = 16R^2. Sina.Sinb.Sinc.Sind

Please do clarify, how R^4 becomes R^2 ?

Don't forget that $R=1$

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55 posts

#12 h Mar 16, 2025, 10:52 AM

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some 20 years later :gleam:

:gleam:

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163 posts

lksb

#13 h Mar 16, 2025, 11:10 PM

Y by

shobber wrote:

Is $f(x)=\sin{x}$ concave on $[0, \pi]$ ? If so, then this problem can be proved by $l=2R\sin{\theta}$ then AM-GM then jensen.

take $(\sin(x))''$

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ohiorizzler1434

719 posts

ohiorizzler1434

#14 h Mar 16, 2025, 11:16 PM

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Bro! I can help prove that sin(x) is concave! The derivative of sin(x) is cos(x)! The derivative of cos(x) is -sin(x)! But because -sin(x) is below 0 from 0 to pi, we know that sin(x) is concave!

We can also prove it geometrically! Consider sin(a) and sin(b), which are the heights formed from a point to the x-axis on the unit circle, for 0<=a,b <= 180. Now, a linear combination of sin(a) and sin(b) represents the line between (a,sin(a)) and (b,sin(b)) on the graph of sin(x). However, sin(c) for c between a,b has higher value than any point on the line as can be seen on the circle! Thus sin(x) is concave from 0 to pi! Now that's rizz!

This post has been edited 1 time. Last edited by ohiorizzler1434, Mar 16, 2025, 11:18 PM

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