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2023 WSMO Accuracy Round Problems/Problem 6: Difference between revisions

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Created page with "==Problem== In quadrilateral <math>ABCD,</math> there exists a point <math>O</math> such that <math>AO = BO = CO = DO</math> and <math>\angle(AOB)+\angle(COD) = 120^{\circ}.<..."
 
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==Solution==
==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>

Revision as of 12:26, 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]

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