2023 WSMO Accuracy Round Problems/Problem 6: Difference between revisions
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\angle(AXB) &= 180-\angle(AXD)\\ | \angle(AXB) &= 180-\angle(AXD)\\ | ||
&= 180-(180-\angle(XAD)-\angle(XDA))\\ | &= 180-(180-\angle(XAD)-\angle(XDA))\\ | ||
&= \angle(XAD)+\angle(XDA) | &= \angle(XAD)+\angle(XDA)\\ | ||
&= \angle(CAD)+\angle(BDA)\\ | &= \angle(CAD)+\angle(BDA)\\ | ||
&= \frac{\overparen{CD}}{2}+\frac{\overparen{AB}}{2}\\ | &= \frac{\overparen{CD}}{2}+\frac{\overparen{AB}}{2}\\ | ||
Revision as of 12:39, 13 September 2025
Problem
In quadrilateral 
there exists a point 
such that 
and 
Let 
be the foot of the perpendiculars from 
to 
to 
to and 
to 
If ![$[ABCD] = 20,$](http://latex.artofproblemsolving.com/0/b/1/0b1dd970eb4d965705bc1e4c66d3be7cd0ddf90f.png)
find ![$\left([KLMN]\right)^2.$](http://latex.artofproblemsolving.com/1/a/7/1a746823718edf1a819e486d1d951507611aa2f0.png)
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]](http://latex.artofproblemsolving.com/2/f/a/2fac749e4a770382db18ddd8c2223e5dee315025.png)
The existence of point implies that 
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{\overparen{CD}}{2}+\frac{\overparen{AB}}{2}\\
&= \frac{\angle(COD)}{2}+\frac{\angle(AOB)}{2}\\
&= \frac{120}{2} = 60^{\circ}.
\end{align*} (Error compiling LaTeX. Unknown error_msg)
So,
In the same manner, we have 
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*}](http://latex.artofproblemsolving.com/9/9/4/994af682253ec1e2f726b4bd673092e38a55d06b.png)
