A square covers the rectangle

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0 replies

jlacosta

Apr 2, 2025

0 replies

Inspired by 2025 Nepal

sqing 1

N 4 minutes ago by sqing

Source: Own

Let $a, b, c$ be positive reals such that $a+b +c+abc = 4$ . Prove that
$\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+ 1}\leq\frac{3}{2}(2 - abc)$ $\frac{1}{ab+1} + \frac{1}{bc+1} + \frac{1}{ca + 1}\leq\frac{3}{2}(2 - abc)$

1 reply

+2 w

sqing

7 minutes ago

sqing

4 minutes ago

Thanks u!

Ruji2018252 2

N 11 minutes ago by lbh_qys

Let $a^2+b^2+c^2-2a-4b-4c=7(a,b,c\in\mathbb{R})$
Find minimum $T=2a+3b+6c$

2 replies

1 viewing

Ruji2018252

Yesterday at 5:52 PM

lbh_qys

11 minutes ago

Inspired by Ruji2018252

sqing 0

21 minutes ago

Source: Own

Let $a,b,c$ be reals such that $a^2+b^2+c^2-2a-4b-4c=7.$ Prove that
$-4\leq 2a+b+2c\leq 20$ $5-4\sqrt 3\leq a+b+c\leq 5+4\sqrt 3$ $11-4\sqrt {14}\leq a+2b+3c\leq 11+4\sqrt {14}$

0 replies

sqing

21 minutes ago

0 replies

Isos Trap

MithsApprentice 38

N 30 minutes ago by eg4334

Source: USAMO 1999 Problem 6

Let $ABCD$ be an isosceles trapezoid with $AB \parallel CD$ . The inscribed circle $\omega$ of triangle $BCD$ meets $CD$ at $E$ . Let $F$ be a point on the (internal) angle bisector of $\angle DAC$ such that $EF \perp CD$ . Let the circumscribed circle of triangle $ACF$ meet line $CD$ at $C$ and $G$ . Prove that the triangle $AFG$ is isosceles.

38 replies

MithsApprentice

Oct 3, 2005

eg4334

30 minutes ago

No more topics!

A square covers the rectangle

orl 3

N Nov 5, 2022 by parmenides51

Source: CGMO 2005, Problem 8

Given an $a \times b$ rectangle with $a > b > 0,$ determine the minimum side of a square that covers the rectangle. (A square covers the rectangle if each point in the rectangle lies inside the square.)

3 replies

orl

Dec 28, 2008

parmenides51

Nov 5, 2022

A square covers the rectangle

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Source: CGMO 2005, Problem 8

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orl

3647 posts

orl

#1 h Dec 28, 2008, 11:42 PM • 2 Y

Y by Adventure10, Mango247

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SCP

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SCP

#2 h Aug 28, 2012, 1:04 PM • 1 Y

Y by Adventure10

It has to be the least perimeter of the square possibly.

solution $\pm$

first you show all $4$ points of the rectangle are on the boundary of the square.

then you have the angle between rectangle and square $0$ or $45$ degrees.

so $min (a, \frac{\sqrt2} 2 (a+b) )$

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mathgirl199

656 posts

mathgirl199

#3 h Jul 2, 2020, 1:08 AM • 4 Y

Y by v4913, centslordm, Mango247, Mango247

Potential Solution?

Let $R$ denote the rectangle, and let $S$ denote the square with minimum length that covers $R$ . Let $s$ denote the length of a side of $S$ . We claim that $R$ is inscribed in $S$ . We also claim that $R$ can only be inscribed in the two ways shown below. Let $S_1$ and $S_2$ denote the squares shown on the left-hand and right-hand sides, respectively. For $S_2$ , the sides of $R$ are parallel to the diagonals of $S2$ . We can see that $s = a$ if $S = S_1$ .

https://lh3.googleusercontent.com/8IBseHsKPR3t0qdwjmS_YCmzdjC7tevE6s-KzyEQKCojYni3RwjDSLTX1sSWPXAqziqrSA=s170

It is easy to see that $s=\frac{\sqrt{2}(a+b)}{2}$ if $S=S_2$ . Taking the minimum value of $(a,\frac{\sqrt{2}(a+b)}{2})$ , we conclude that

$s= \left\{ \begin{array}{ll} a & a<(\sqrt{2}+1)b \\ \frac{\sqrt{2}(a+b)}{2} & a\geq (\sqrt{2}+1)b \\ \end{array} \right.$

Now we prove our claim that these are only two ways. Let $R = ABCD$ and $S = XYZW$ . WLOG, we place $XY$ horizontally. By the minimality of $S$ , we can assume that at least one vertex, say $A$ , of $R$ lies on one side of $S$ , say $WX$ (see the left diagram). If neither $B$ nor $D$ lies on the sides of $S$ , we can then slide $R$ down vertically so that one of them, say $B$ , lies on side $XY$ (see the middle figure). If neither $C$ or $D$ lies on the sides of $S$ , then we can apply an enlargement, centered at $X$ with scale less than $1$ , to $S$ such that the image of $S$ still covers $R$ .
This violates the minimality of $S$ . Therefore, at least one of $C$ and $D$ lies on the sides of $S$ , that is, three consecutive vertices of $R$ lie on the sides of $S$ (see the right-hand side figure). WLOG assume that they are $A, B, C$ . If any of these three vertices coincide with any of the vertices of $S$ , then we clearly have $S = S_1$ . Hence we may assume that $A , B, C$ are on sides $WX$ , $XY$ and $YZ$ , respectively. By symmetry, we may also assume that $AB =a>b=BC$ .

https://lh3.googleusercontent.com/lX6YAPjUs7xeaT73NSstZRW_31vZ4-Rui5zM7ZRqOvhw-W_DrjH3AHEOdhmGxa9R7-d9Mw=s170

Note

All rectangular-like shapes are indeed rectangles

If $D$ does not lie on line segment $ZW$ , then we can slide $R$ up so both $B$ and $D$ lie in the interior of $S$ (see the left-hand slide) Let $O$ be the center of $R$ . We can then rotate $R$ around $O$ with a small angle so that all four vertices lie inside $S$ (see the middle figure shown below). It is easy to see that we can use a smaller square to cover $R$ (by applying an enlargement centered at $O$ with a scale less that $1$ ) , violating the minimality of $S$ . This means our assumption is wrong, and D must lie on side $ZW$ (see the righthand side figure shown below) , which is the case when $S = S_2$ .

https://lh3.googleusercontent.com/j6nT3PMMp8rueVTHgUEE_B-m-ZHc99H1uG2L4e6-E_Qfi_AKxCUBaUYFn7HOnP8_2NUp=s170

Note

All rectangular-like shapes are indeed rectangles

We finish our proof that if $S = S_2$ , then the sides of $R$ are parallel to the diagonals of $S_2$ . By symmetry, it suffices to show that $AX = XB$ . It is not difficult to see that $\angle XAB = \angle YBC = \angle ZCD= \angle WDA$ . So triangles $ABX$ , $BCY$ , $CDZ$ and $DAW$ are similar. Set $AX = ax$ and $XB = ay$ . Then $BY = bx$ and $CY = by$ . Also, $DW = bx$ and $WA = by$ , which gives $by + ax = WA + AX = WX = XY = XB + BY =ay+bx$ , implying that $(a-b)x= (a-b)y$ , or $x=y$ , as desired $\blacksquare$

This post has been edited 1 time. Last edited by mathgirl199, Jul 2, 2020, 1:09 AM

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parmenides51

30630 posts

parmenides51

#4 h Nov 5, 2022, 7:59 AM

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discussed also here and here

This post has been edited 2 times. Last edited by parmenides51, Nov 5, 2022, 8:49 AM

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