Equal lengths in cyclic quadrilateral

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Equal lengths in cyclic quadrilateral

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Source: All-Russian MO 2024 9.4

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#1 h Apr 22, 2024, 10:31 PM • 4 Y

Y by GeoKing, Rounak_iitr, trolled767, Amirreza.J

In cyclic quadrilateral $ABCD$ , $\angle A+ \angle D=\frac{\pi}{2}$ . $AC$ intersects $BD$ at ${E}$ . A line ${l}$ cuts segment $AB, CD, AE, DE$ at $X, Y, Z, T$ respectively. If $AZ=CE$ and $BE=DT$ , prove that the diameter of the circumcircle of $\triangle EZT$ equals $XY$ .

This post has been edited 2 times. Last edited by LoloChen, Jun 8, 2024, 11:12 AM

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#2 h Apr 22, 2024, 10:37 PM • 1 Y

Y by GeoKing

Let the circumcircle of $ABCD$ be $\odot O$ and rays $AB, DC$ intersect at ${F}$ . Call the midpoints of $AC, BD, ZT$ be $P, Q, M$ .
Obviously $ZP=PE$ and $TQ=QE$ , so ${O}$ is the circumcenter of $\triangle EZT$ , so $OM \perp ZT$ . By Klement Theorem $XM=YM$ . Notice that $\angle AFD=\frac{\pi}{2}$ , we only need to prove $FM=EO$ . We will actually prove that $FEOM$ is a parallelogram.

Claim 1: $FE \parallel OM$ .
Since ${O}$ is the circumcenter of $\triangle EZT$ , we only need to prove that $EO$ and $EF$ are isogonal lines of $\angle ZET$ . But actually $O, F$ are isogonal conjugates of $\triangle EAD$ .

Claim 2: $FE = OM$ .
By angle chase $\angle PFQ+\angle PMQ=\angle PFQ+\angle PEQ=\pi$ , so ${F}$ is on $\odot PMQ$ , the NPC of $\triangle EZT$ . Combining $FE \perp ZT$ , we get $FE = OM$ . Done.

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#3 h Jun 7, 2024, 1:01 PM • 2 Y

Y by pokpokben, soryn

Let $AB$ and $CD$ intersect at $F$ . Apply Menalus' Theorem twice on triangle $EZT$ with respect to lines $AB$ and $CD$ $\frac{EB}{TB}\cdot\frac{TX}{ZX}\cdot \frac{ZA}{EA}=1,\;\;\;\; \frac{ED}{TD}\cdot\frac{TY}{ZY}\cdot \frac{ZC}{EC}=1$ Dividing out yields $XZ=TY$ . Let $M$ , $N$ , and $O$ be the midpoints of segments $AC$ , $BD$ , and $XY$ , respectively. Notice that $\angle FMO=\angle AEB+\angle FMC=90^{\circ}$ $\angle FNO=\angle DEC+\angle BNF=90^{\circ}$ Thus $FMON$ is cyclic with diameter $FO$ . Finally, $OF=\frac{MN}{\sin(\angle MON)}\Rightarrow XY=\frac{ZT}{\sin(\angle ZET)}$

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#4 h Jun 7, 2024, 1:09 PM • 2 Y

Y by NO_SQUARES, LoloChen

I believe there is a typo in the problem, it should be $BE=DT$ not $BZ=DE$ . (In the grade 10 one it is written as $BE=DT$ ).

This post has been edited 1 time. Last edited by sami1618, Jun 7, 2024, 2:14 PM

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#5 h Today at 5:05 AM

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Here is my solution (literally combined two above).

$O$ is the circumcenter of $\triangle ZET$ . Since $O$ lies on bisector of $AC$ , it lies on bisector of $ZE$ . Similarly $O$ lies on perpendicular bisector of $ET$ , and we're done.

Claim is that $XZ=TY$ . By the Menelau's theorem on the $\triangle ZET$ and the line $AB$ , we will get that: $\frac{BE}{BT}* \frac{XT}{XZ} *\frac{AZ}{AE}=1$ . Similarly on the $\triangle ZET$ and the line $CD$ , we get that $\frac{CE}{CZ}* \frac{YZ}{YT} * \frac{DT}{DB}=1$ , from here we can easily conclude the claim.

Let $M, P, H$ be the midpoints of $ZT, ZE, ET$ . Then we have that $\angle AFP=\angle PAF=\angle BDF$ $\implies$ $FP \perp BD$ . Similarly we get that $FH \perp AC$ . So $FPMH$ is parallelogram with diameter $FM$ .

Since $\angle FPC=2*\angle FAC=2*\angle BDC=\angle BHF$ , and $PEHM$ is parallelogram by the parallelogram isogonalaty lemma in $\triangle PFH$ we can say that $FM, FE$ are isogonal. Since $FM$ contains circumcenter $FE$ is the altitude. It means that $XY \perp FE$ . In $ZET$ we have that $EO, EF$ are isogonal since $EO$ contains circumcenter and $EF$ is altitude.

Now some easy angle chase gives us that $OMFE$ is parallelogram. So $OE=MF= \frac{XY}{2}$ .

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