trapezoid inscribed in parabola y=x^2, area wanted

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trapezoid inscribed in parabola y=x^2, area wanted

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Source: 2011 Belarusian MO B 10.8

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#1 h Nov 18, 2020, 1:00 PM • 1 Y

Y by Mango247

On the parabola $y = x^2$ mark four points $A, B, C, D$ , so that the quadrangle $ABCD$ is a trapezoid ( $AD\parallel BC$ , $AD> BC$ ). Let $m, n$ be the distances from the intersection point of the diagonals of this trapezoid to the midpoints of its sides $AD, BC$ respectively. Calculate the area of the trapezoid $ABCD$ .

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#2 h Nov 18, 2020, 7:04 PM

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Let $A(a,a^2),B(b,b^2),C(c,c^2),D(d,d^2),a<b<c<d$
$AD\parallel BC \to a+d=b+c$

Let point of intersection of diagonals is $O$
$AC:y=(a+c)x-ac, BD:y=(b+d)x-bd \to O_x=\frac{ac-bd}{a+c-b-d}=\frac{ac-b(b+c-a)}{a+c-b-(b+c-a)}=\frac{(b+c)(a-b)}{2(a-b)}=\frac{b+c}{2}$
$O_y=\frac{(a+c)(b+c)}{2}-ac=\frac{ab+bc+c^2-ac}{2}=\frac{ab+cd}{2}$

$M$ is midpoint $AD$ and $M=(\frac{a+d}{2},\frac{a^2+d^2}{2})$
$m^2=OM^2= \frac{(a+d-b-c)^2}{4}+\frac{(a^2+d^2-ab-cd)^2}{4}=\frac{((a-d)(a-b))^2}{4} \to m =\frac{(d-a)(b-a)}{2}$
$n=\frac{(c-b)(b-a)}{2}$

$S=\frac{(a^2+d^2)(d-a)-(a^2+b^2)(b-a)-(b^2+c^2)(c-b)-(c^2+d^2)(d-c)}{2}=\frac{(c-a)(d-b)(b+d-a-c)}{2}=(c-a)^2(b-a)$
$m+n=\frac{(b-a)(d-a+c-b)}{2}=(b-a)(c-a)$
$m-n=\frac{(b-a)(d-a-c+b)}{2}=(b-a)^2$

$S=(c-a)^2(b-a)= \frac{(m+n)^2}{\sqrt{m-n}}$

This post has been edited 1 time. Last edited by RagvaloD, Nov 18, 2020, 7:06 PM

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