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#1 h Mar 31, 2016, 1:31 PM • 1 Y

Y by Adventure10

Let $AB$ and $CD$ be two parallel chordes on hyperbola $y=1/x$ . Lines $AC$ and $BD$ intersect axis $Oy$ at points $A_1$ and $D_1$ respectively, and axis $Ox$ - at points $C_1$ and $B_1$ respectively. Prove that the area of $\triangle A_1OC_1$ equals the area of $\triangle D_1OB_1$

This post has been edited 1 time. Last edited by proximo, Mar 31, 2016, 1:43 PM

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#2 h Mar 31, 2016, 1:35 PM • 2 Y

Y by Adventure10, Mango247

chords?
$\text{[math]}$
Edit: so im assuming that the points are ABCD in clockwise order, A at the "top"

This post has been edited 1 time. Last edited by rafayaashary1, Mar 31, 2016, 1:41 PM
Reason: configuration

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#3 h Mar 31, 2016, 1:43 PM • 1 Y

Y by Adventure10

rafayaashary1 wrote:

chords?
$\text{[math]}$
Edit: so im assuming that the points are ABCD in clockwise order, A at the "top"

I think it doesn't matter

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#4 h Apr 4, 2016, 11:11 PM • 2 Y

Y by Adventure10, Mango247

For two points $X_1(x_1,1/x_1),X_2(x_2,1/x_2)$ on hyperbola equality of chord $X_1X_2$ is
$y=\frac{x_1+x_2-x}{x_1x_2}=-\frac{1}{x_1x_2}x+\frac{x_1+x_2}{x_1x_2}$
Let $A,B,C,D$ have coordinates $(a,1/a),(b,1/b),(c,1/c)(d,1/d)$
$AB$ and $CD$ are parallel, so $\frac{1}{ab}=\frac{1}{cd}$ => $ab=cd$
$A_1,A,C,C_1$ lies on $y=\frac{a+c-x}{ac}$
Coordinates of $A_1$ are $(0,\frac{a+c}{ac})$ , coordinates of $C_1$ are $(a+c,0)$ . Area of $\triangle A_1OC_1$ is $\frac{(a+c)^2}{2ac}$
Similarly , area of $\triangle D_1OB_1$ is $\frac{(b+d)^2}{2bd}$
$\frac{(a+c)^2}{2ac}-\frac{(b+d)^2}{2bd}=\frac{a^2bd+2abcd+c^2bd-b^2ac-2abcd-d^2ac}{2abcd}=\frac{ab(ad-ac)+cd(cb-ad)}{2abcd}=\frac{(ab-cd)(cb-ad)}{2abcd}=0$ . So areas are equals.

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