Problem. Let K and M be points on the sides AB and BC of a triangle ABC, and let $N=AM\cap CK$. Assume that the quadrilaterals AKMC and KBMN are cyclic, and their circumradii are equal. Prove that < ABC = 45°. Solution. By the extended law of sines, the circumradius R of a triangle ABC can be found by the formula $R=\frac{a}{2\sin A}$, where a = BC is a side and A = < CAB is the corresponding angle of the triangle ABC. Applying this extended law of sines to the triangles KAM and KBM, we find that the circumradius of triangle KAM equals $\frac{KM}{2\sin\measuredangle KAM}$, and that the circumradius of triangle KBM equals $\frac{KM}{2\sin\measuredangle KBM}$. Now, the circumradius of triangle KAM is, of course, the circumradius of the cyclic quadrilateral AKMC, and the circumradius of triangle KBM is the circumradius of the cyclic quadrilateral KBMN. Since the circumradii of the quadrilaterals AKMC and KBMN are equal, we thus can say that the circumradii of the triangles KAM and KBM are equal. Thus, $\frac{KM}{2\sin\measuredangle KAM}=\frac{KM}{2\sin\measuredangle KBM}$. Thus, sin < KAM = sin < KBM. Hence, either < KAM = < KBM, or < KAM + < KBM = 180°. Now, < KAM + < KBM = 180° is impossible, since it would mean < BAM + < ABM = 180°, but the sum of two angles of a non-degenerate triangle is never 180°. Thus, we must have < KAM = < KBM. In other words, < BAM = < ABC. Similarly, < BCK = < ABC. Now, < KNM = < ANC = 180° - < NAC - < NCA (sum of angles in triangle ANC) = 180° - (< CAB - < BAM) - (< ACB - < BCK) = 180° - (< CAB - < ABC) - (< ACB - < ABC) = (180° - < CAB - < ACB) + 2 < ABC. By the sum of angles in triangle ABC, we have 180° - < CAB - < ACB = < ABC, and thus we get < KNM = < ABC + 2 < ABC = 3 < ABC. On the other hand, since the quadrilateral KBMN is cyclic, < KNM = 180° - < KBM, so that < KNM = 180° - < ABC. Thus, 3 < ABC = 180° - < ABC. This becomes 4 < ABC = 180°, what is equivalent to < ABC = 45°. And the problem is solved. darij

Correct answer and maginifcent solution!!!

Rushil wrote: On the sides $AB$ and $BC$ of triangle $ABC$, points $K$ and $M$ are chosen such that the quadrilaterals $AKMC$ and $KBMN$ are cyclic , where $ N = AM \cap CK $ . If these quads have the same ctrcumradii, find $ \angle ABC$ By the informations we have $\angle CKA = \angle AMC = 180^{\circ} - \angle BMA = \angle BKC \iff \angle CKA = \angle CKB = 90^{\circ}$. Hence $N$ is the orthocenter of $ABC$. Denote $I$ and $J$ are respectively the midpoints of $AC$, $BN$. We also have the quadrilateral $IMJK$ is cyclic, hence triangle $IMJ$ is right at $M$, with $MI = MJ$ we have $MIJ$ is isocoles right triangle. Hence $\angle MjK = 90^{\circ}$, so $\angle ABC = 45^{\circ}$.