Let $ABCD$ be a rectangle. Construct equilateral triangles $BCX$ and $DCY$, in such a way that both of these triangles share some of their interior points with some interior points of the rectangle. Line $AX$ intersects line $CD$ on $P$, and line $AY$ intersects line $BC$ on $Q$. Prove that triangle $APQ$ is equilateral.
Problem
Source: Ibero American 2012
Tags: geometry, parallelogram, geometry proposed
Concyclicboy
03.10.2012 02:27
1° $\triangle ADY \cong\triangle XBA$ then $AX=AY$ and $\angle PAQ=60$. 2° $X$ and $Y$ are midpoints of $AP$ and $AQ$ respectively, because $X$ and $Y$ are on the perpendicular bisectors of $BC$ and $CD$ respectively, and $AP=AQ$.
sunken rock
16.11.2012 21:12
Remark: $\Delta AXY$ is equilateral if $ABCD$ is a parallelogram. Best regards, sunken rock
Math_Is_Fun_101
28.09.2021 17:37
Note that $X$ and $Y$ lie on the perpendicular bisectors of $BC$ and $AB$ respectively. Since $ABCD$ is a rectangle, $X$ and $Y$ lie on the perpendicular bisectors of $AD$ and $AB$ respectively. Now, from the equilateral triangles note we have $\angle DCX = \angle XCY = \angle YCB = 30^\circ$, so $CX$ and $CY$ are perpendicular bisectors of $DY$ and $BX$ respectively. These give $XA=XD=XY=BY=AY$, so $AXY$ is equilateral. Finally, note that dropping the perpendiculars from $X$ and $Y$ to $BC$ and $CD$ respectively shows $X$ and $Y$ are midpoints of $AP$ and $AQ$ respectively, so $APQ$ is equilateral as well.
Blastoor
28.09.2021 23:13
Notice that $\triangle AYD=\triangle DXC$. From symmetry we get $AY=AX$, from $\triangle DXP$ we get $\angle XDP=\angle XPD$, so $\angle BAY=\angle AYD-60^{\circ}$ and $\angle DAP=90^{\circ}-\angle AYD \implies \angle XAY=60^{\circ}$. As $\triangle AXY$ is equilateral and $XY$ is the midline in $\triangle PAQ$, the required follows.