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Difference between revisions of "2023 WSMO Accuracy Round Problems/Problem 6"
 
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Line 46: Line 46:
 
draw(k--l--m--n--cycle,green);
 
draw(k--l--m--n--cycle,green);
 
</asy>
 
</asy>
 +
 +
The existence of point <math>O</math> implies that <math>ABCD</math> is a cyclic quadrilateral. Now, we have
 +
<cmath>\begin{align*}
 +
\angle(AXB) &= 180-\angle(AXD)\\
 +
&= 180-(180-\angle(XAD)-\angle(XDA))\\
 +
&= \angle(XAD)+\angle(XDA)\\
 +
&= \angle(CAD)+\angle(BDA)\\
 +
&= \frac{\overarc{CD}}{2}+\frac{\overarc{AB}}{2}\\
 +
&= \frac{\angle(COD)}{2}+\frac{\angle(AOB)}{2}\\
 +
&= \frac{120}{2} = 60^{\circ}.
 +
\end{align*}</cmath>
 +
So,
 +
<cmath>\begin{align*}
 +
KM &= KX+XM\\
 +
&= AX\cos60^{\circ}+XC\cos^{\circ}\\
 +
&= \frac{AX}{2}+\frac{XC}{2}\\
 +
&= \frac{AC}{2}.
 +
\end{align*}</cmath>
 +
In the same manner, we have <math>LN = \frac{BD}{2}.</math> We have
 +
<cmath>\begin{align*}
 +
[KLMN] &= \frac{1}{2}(KM)(LN)\sin\angle(LXK)\\
 +
&= \frac{1}{2}\left(\frac{AC}{2}\right)\left(\frac{BD}{2}\right)\sin\angle(AXB)\\
 +
&= \frac{1}{4}\left(\frac{1}{2}(AC)(BD)\sin\angle(AXB)\right)\\
 +
&= \frac{[ABCD]}{4}\implies\\
 +
[KLMN] &= \frac{20}{4} = 5\implies\\
 +
([KLMN]) &= 5^2 = \boxed{25}.
 +
\end{align*}</cmath>
 +
 +
~pinkpig

Latest revision as of 13:42, 13 September 2025

Problem

In quadrilateral $ABCD,$ there exists a point $O$ such that $AO = BO = CO = DO$ and $\angle(AOB)+\angle(COD) = 120^{\circ}.$ Let $K,L,M,N$ be the foot of the perpendiculars from $A$ to $BD,$ $B$ to $AC,$ $C$ to $BD,$ and $D$ to $AC.$ If $[ABCD] = 20,$ find $\left([KLMN]\right)^2.$

Solution

[asy] import geometry; unitsize(5cm); pair a = dir(110); pair b = dir(160); pair c = dir(310); pair d = dir(20); pair x = intersectionpoint(a--c,b--d); pair k = foot(a,b,d); pair l = foot(b,a,c); pair m = foot(c,b,d); pair n = foot(d,a,c); draw(a--c); draw(b--d); draw(a--k,dotted+red); draw(b--l,dotted+red); draw(c--m,dotted+red); draw(d--n,dotted+red); label("$A$",a,N); label("$B$",b,W); label("$C$",c,SE); label("$D$",d,E); label("$X$",x,NE); label("$K$",k,SW); label("$L$",l,NE); label("$M$",m,NE); label("$N$",n,SW); draw(rightanglemark(a,k,b,2),dotted); draw(rightanglemark(b,l,a,2),dotted); draw(rightanglemark(c,m,d,2),dotted); draw(rightanglemark(d,n,c,2),dotted); draw(Circle((0,0), 1),black); draw(a--b--c--d--cycle,blue); draw(k--l--m--n--cycle,green); [/asy]

The existence of point $O$ implies that $ABCD$ is a cyclic quadrilateral. Now, we have \begin{align*} \angle(AXB) &= 180-\angle(AXD)\\ &= 180-(180-\angle(XAD)-\angle(XDA))\\ &= \angle(XAD)+\angle(XDA)\\ &= \angle(CAD)+\angle(BDA)\\ &= \frac{\overarc{CD}}{2}+\frac{\overarc{AB}}{2}\\ &= \frac{\angle(COD)}{2}+\frac{\angle(AOB)}{2}\\ &= \frac{120}{2} = 60^{\circ}. \end{align*} So, \begin{align*} KM &= KX+XM\\ &= AX\cos60^{\circ}+XC\cos^{\circ}\\ &= \frac{AX}{2}+\frac{XC}{2}\\ &= \frac{AC}{2}. \end{align*} In the same manner, we have $LN = \frac{BD}{2}.$ We have \begin{align*} [KLMN] &= \frac{1}{2}(KM)(LN)\sin\angle(LXK)\\ &= \frac{1}{2}\left(\frac{AC}{2}\right)\left(\frac{BD}{2}\right)\sin\angle(AXB)\\ &= \frac{1}{4}\left(\frac{1}{2}(AC)(BD)\sin\angle(AXB)\right)\\ &= \frac{[ABCD]}{4}\implies\\ [KLMN] &= \frac{20}{4} = 5\implies\\ ([KLMN]) &= 5^2 = \boxed{25}. \end{align*}

~pinkpig

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