Constant = m_1+m_2

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Constant = m_1+m_2

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Source: APMO 2003

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#1 h Mar 5, 2006, 5:23 AM • 2 Y

Y by Adventure10, Mango247

Suppose $ABCD$ is a square piece of cardboard with side length $a$ . On a plane are two parallel lines $\ell_1$ and $\ell_2$ , which are also $a$ units apart. The square $ABCD$ is placed on the plane so that sides $AB$ and $AD$ intersect $\ell_1$ at $E$ and $F$ respectively. Also, sides $CB$ and $CD$ intersect $\ell_2$ at $G$ and $H$ respectively. Let the perimeters of $\triangle AEF$ and $\triangle CGH$ be $m_1$ and $m_2$ respectively.

Prove that no matter how the square was placed, $m_1+m_2$ remains constant.

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#2 h Mar 11, 2006, 8:01 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Let p be a line perpendicular to the lines $l_1 \parallel l_2$ intersecting these 2 parallel lines at $P_1, P_2$ , so that their distance is $d = P_1P_2$ . Let (O) be a circle with sufficiently large radius $r$ centered at an arbitrary point O on the segment $P_1P_2$ . Let ABCD be a square with the side $a = r \sqrt 2$ inscribed in this circle intersecting both the lines $l_1, l_2$ in the way described in the problem. Let the angle $\phi = \angle P_1OA$ be fixed and let the square circumcenter O move on the perpendicular p. The triangles $\triangle AEF \sim \triangle CHG$ remain similar for all positions of the circumcenter O and the sum of their A- and C-altitudes is constant, $h_a + h_c = 2r \cos \phi - d$ , and the sum $m_1 + m_2$ of their perimeters (proportional to their altitudes) is then also constant. Consequently, we only have to consider the square circumcenter O being the midpoint of the segment $P_1P_2$ , when the triangles $\triangle AEF \cong \triangle CHG$ are congruent, i.e., their perimeters $m_1 = m_2$ are equal, and we have to show that the perimeter $m_1$ of the triangle $\triangle AEF$ is constant for any rotation angle $\phi = \angle P_1OA$ iff a = d.

Denote e = AF, f = AE, h = EF the sides of the right triangle $\triangle AEF$ . Let $A' \in EF$ be the foot of the bisector AO of the right angle $\angle FAE$ . Using $h^2 = e^2 + f^2$ , the angle bisector length b = AA' is

$b = \sqrt{ef\left(1 - \frac{h^2}{(e + f)^2}\right)} = \frac{ef \sqrt 2}{e + f}$

WLOG, assume $\angle AEF < \angle AFE$ . Then $\angle EFA = \phi + 45^\circ$ and

$\frac f e = \tan \angle EFA = \frac{1 + \tan \phi}{1 - \tan \phi}$

Solving for e, f, h

$e = \frac{b \sqrt 2}{1 + \tan \phi},\ \ f = \frac{b \sqrt 2}{1 - \tan \phi}$

$h = \sqrt{e^2 + f^2} = \frac{2b \sqrt{1 + \tan^2 \phi}}{1 - \tan^2 \phi} = \frac{2b \cos \phi}{\cos^2 \phi - \sin^2 \phi}$

The perimeter $m_1$ of the triangle $\triangle AEF$ is then

$m_1 = e + f + h = \frac{2b \sqrt 2}{1 - \tan^2 \phi} + \frac{2b \cos \phi}{\cos^2 \phi - \sin^2 \phi} =$

$= \frac{2b \sqrt 2 \cos^2 \phi + 2b \cos \phi}{\cos^2 \phi - \sin^2 \phi} = \frac{2b \cos \phi (\sqrt 2 \cos \phi + 1)}{2 \cos^2 \phi - 1} =$

$= \frac{2b \cos \phi}{\sqrt 2 \cos \phi - 1}$

On the other hand, we have

$b = AA' = OA - OA' = r - \frac{d}{2 \cos \phi} = d\ \frac{\frac{2r}{d} \cos \phi - 1}{2 \cos \phi}$

$m_1 = d\ \frac{\frac{2r}{d} \cos \phi - 1}{\sqrt 2 \cos \phi - 1}$

It is clear that the perimeter $m_1 = d$ is constant, iff $\frac{2r}{d} = \sqrt 2$ or using $a = r \sqrt 2$ , iff $a = d$ .

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#3 h Mar 11, 2006, 8:17 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Just a little bit of thinking before replying will, no doubt, produce in an olympiad

caliber person the following observation:

the ex-circles opposite the right angles of the triangles in question, having

perimeters of m1 and m2, have centers located in the same point on the

diagonal of the square connecting these right angles.

From this point the problem is just a pis cake.

T.Y.

M.T.

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#4 h Nov 23, 2010, 10:57 PM • 2 Y

Y by Adventure10, Mango247

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO2003Problem2

by Vo Duc Dien

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#5 h Nov 25, 2010, 6:13 PM • 2 Y

Y by Adventure10, Mango247

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO2003Problem2

Vo Duc Dien

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#6 h Jul 11, 2023, 6:28 AM

Y by

WLOG the square is a unit square. Employ Cartesian coordinates. Let $A=(0,0),B=(0,1),C=(1,1),D=(1,0).$ Then, suppose that $E=(0,b)$ and $F=(a,0)$ (sorry for using the variable $a$ again, it just means an arbitrary real between 0 and 1 now). Then, let line $\ell_2$ intersect line $AB$ again at $P$ . We then have $P=(0,b+\frac{\sqrt{a^2+b^2}}{a})$ by e.g. dropping a perpendicular from $E$ to $\ell_2$ and using similar triangles. Thus, the equation of line $\ell_2$ is $y=-\frac{b}{a}x+b+\frac{\sqrt{a^2+b^2}}{a}.$ Thus, intersecting this with line $y=1$ , we have $G=(\frac{ab-a+\sqrt{a^2+b^2}}{b},1).$ This means that $GC=1-\frac{ab-a+\sqrt{a^2+b^2}}{b}=\frac{a+b-ab-\sqrt{a^2+b^2}}{b}.$ Furthermore, note that $\triangle CGH\sim \triangle AFE$ due to the parallel lines, so the perimeter of $\triangle CGH$ is also equal to $CG(\frac{a+b+\sqrt{a^2+b^2}}{a})$ $=(\frac{a+b-ab-\sqrt{a^2+b^2}}{b})(\frac{a+b+\sqrt{a^2+b^2}}{a})$ $=\frac{ab+b^2+b\sqrt{a^2+b^2}-a^2b-ab^2-ab\sqrt{a^2+b^2}+a^2+ab+a\sqrt{a^2+b^2}-a\sqrt{a^2+b^2}-b\sqrt{a^2+b^2}-a^2-b^2}{ab}$ $=\frac{ab-a^2b-ab^2-ab\sqrt{a^2+b^2}+ab}{ab}$ $=2-a-b-\sqrt{a^2+b^2}.$ Since the perimeter of $\triangle AFE$ is $a+b+\sqrt{a^2+b^2}$ , the sum of the two perimeters is just 2, so we are done.

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