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Pythagorean Theorem

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Source: Indian RMO 1993 Problem 4

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1592 posts

Rushil

#1 h Oct 23, 2005, 3:04 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a rectangle with $AB = a$ and $BC = b$ . Suppose $r_1$ is the radius of the circle passing through $A$ and $B$ touching $CD$ ; and similarly $r_2$ is the radius of the circle passing through $B$ and $C$ and touching $AD$ . Show that $r_1 + r_2 \geq \frac{5}{8} ( a + b) .$

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#2 h Oct 24, 2005, 2:02 AM • 1 Y

Y by Adventure10

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varun

#3 h Oct 24, 2005, 3:04 AM • 2 Y

Y by Adventure10, Mango247

Altheman, I did the same thing as you.
Note that we are correct just that the process was not adequate toobtain so strong inequality.
we proved $r_1 + r_2 \ge \frac{1}{2}(a+b)$ which is true for given case.

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270 posts

varun

#4 h Oct 24, 2005, 7:23 AM • 2 Y

Y by Adventure10, Mango247

As solved above by Altheman,

$r_1=\frac{a^2}{8b}+\frac{b}{2}=\frac{a^3+4ab^2}{8ab}$
$r_2=\frac{b^3+4ba^2}{8ab}$
Thus,
$r_1+r_2=\frac{a^3+b^3+4ab(a+b)}{8ab}=\frac{(a+b)(a^2+b^2-ab)+4ab(a+b)}{8ab}$
= $\frac{(a+b)(a^2+b^2+3ab)}{8ab}\ge\frac{(a+b)(2ab+3ab)}{8ab}=\frac{(a+b)(5ab)}{8ab}=\frac{5}{8}(a+b)$

$r_1+r_2\ge\frac{5}{8}(a+b)$
Thus, proved.

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1660 posts

#5 h Oct 20, 2017, 6:47 PM • 2 Y

Y by Adventure10, Mango247

I have just one another proof by geometry. Should I post it? This proof is by my friend, not by me.

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#6 h Apr 11, 2019, 11:02 AM • 1 Y

Y by Adventure10

We know:
$r_1=\frac{a^2}{8b}+\frac{b}{2}$ and $r_2=\frac{b^2}{8a}+\frac{a}{2}$

Adding these two up, we get:
$r_1 + r_2 = \frac{1}{8}(\frac{a^2}{b} + \frac{b^2}{a} + 4(a + b))$
All that remains to show is $\frac{a^2}{b} + \frac{b^2}{a} \ge a + b$ $\iff a^3 + b^3 \ge (a+b)ab$ $\iff a^2 - ab + b^2 \ge ab$ $\iff (a-b)^2 \ge 0$ which is indeed true :-D

:-D

.

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ATGY

#7 h Jul 2, 2021, 11:51 AM

Y by

Let the circle with radius $r_1$ have center $O$ . Let the point where the circle touches $CD$ be $F$ and extend $FO$ to meet $AB$ at point $E$ . Clearly, $E$ is the midpoint of $AB$ as $OA$ and $OB$ are both radii of the circle (meaning that $\triangle{OAB}$ is isosceles).

Since $OF = r_1$ , $OE = b - r_1$ . As $E$ is the midpoint of $AB$ , $AE = \frac{a}{2}$ . Using the Pythagorean Theorem on $\triangle{OEA}$ , we have:
${r_1}^2 = (b - r_1)^2 + \frac{a^2}{4}$ $2br_1 = b^2 + \frac{a^2}{4}$ $r_1 = \frac{a^2}{8b} + \frac{b}{2}$ Similarly, by symmetry:
$r_2 = \frac{b^2}{8a} + \frac{a}{2}$
Adding $r_1$ and $r_2$ , we have:
$\frac{a^3 + b^3}{8ab} + \frac{a + b}{2} = \frac{(a + b)(a^2 - ab + b^2)}{8ab} + \frac{a + b}{2}$ $= (a + b)\left(\frac{(a - b)^2 + ab}{8ab} + \frac{1}{2}\right)$ $= (a + b)\left(\frac{(a - b)^2}{8ab} + \frac{5}{8}\right) > \frac{5}{8}(a + b)$
We are done $\blacksquare$

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#8 h Oct 30, 2024, 9:07 AM

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AOPSMasters wrote:

I have just one another proof by geometry. Should I post it? This proof is by my friend, not by me.

Yes, pleasee

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