points on two circles

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Source: 7-th Kyiv math festival, 2008

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rogue

#1 h May 11, 2008, 3:14 PM • 2 Y

Y by Adventure10, Mango247

Let $K,L,M$ and $N$ be the midpoints of sides $AB,$ $BC,$ $CD$ and $AD$ of the convex quadrangle $ABCD.$ Is it possible that points $A,B,L,M,D$ lie on the same circle and points $K,B,C,D,N$ lie on the same circle?

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yetti

#2 h May 14, 2008, 9:13 AM • 2 Y

Y by Adventure10, Mango247

$(O_a)$ is circumcircle of the $\triangle ABD,$ $(O_c)$ is circumcircle of the $\triangle BCD.$ $LM \parallel BD$ is midline of the $\triangle BCD,$ if $L, M \in (O_a),$ then $BMLD$ is isosceles trapezoid and the $\triangle BCD$ is isosceles. Similarly, $NK \parallel BD$ is midline of the $\triangle ABD,$ if $N, K \in (O_b),$ then $BDNK$ is isosceles trapezoid and $\triangle ABD$ is isosceles. Therefore, $ABCD$ is at least a kite, let $E$ be its diagonal intersection. Assuming $ABLMD, CDNKB$ are both cyclic, $LM = \frac{BD}{2} = NK$ implies $\angle DAB = \angle BCD$ (see http://www.mathlinks.ro/viewtopic.php?t=204352) and $ABCD$ is a rhombus. Power of $A$ to $(O_c)$ is

$AK \cdot AB = \frac {AB^2}{2} = (AE + EO_c)^2 - R_c^2$

Substituting $AB = \frac {BD}{2 \sin \frac {A}{2}},$ $AE = \frac {BD}{\tan \frac {A}{2}},$ $EO_c = R_c \cos A,$ $R_c = \frac {BD}{2 \sin A}$ and reducing by $\frac {BD^2}{4 \sin^2 A},$ we get the following equation for the angle $\angle DAB \equiv \angle A$ :

$2 \cos^2 \frac {A}{2} + 1 = \left(2\cos^2 \frac {A}{2} + \cos A\right)^2$

$\cos A + 2 = (2 \cos A + 1)^2$

with roots $\cos A = \frac {_1}{^4}$ and $- 1,$ only the 1st one acceptable.

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