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Problem 2106

National Olympiad second round; IMO P1/P4

Geometry Difficulty 7.6 Prove it European Girls' Mathematical Olympiad (EGMO) competition problems · European Girls' Mathematical Olympiad (EGMO)

Let $ABCD$ be a convex quadrilateral with $\angle DAB = \angle BCD = 90^{\circ}$ and $\angle ABC > \angle CDA$ . Let $Q$ and $R$ be points on the segments $BC$ and $CD$ , respectively, such that the line $QR$ intersects lines $AB$ and $AD$ at points $P$ and $S$ , respectively. It is given that $PQ = RS$ . Let the midpoint of $BD$ be $M$ and the midpoint of $QR$ be $N$ . Prove that $M, N, A$ and $C$ lie on a circle.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solutions — 3

Solution 1

Solution:

Note that $N$ is also the midpoint of $PS$ . From right-angled triangles $PAS$ and $CQR$ we obtain $\angle ANP = 2 \angle ASP$ , $\angle CNQ = 2 \angle CRQ$ , hence
$\angle ANC = \angle ANP + \angle CNQ = 2(\angle ASP + \angle CRQ) = 2(\angle RSD + \angle DRS) = 2 \angle ADC.$
Similarly, using right-angled triangles $BAD$ and $BCD$ , we obtain $\angle AMC = 2 \angle ADC$ .
Thus $\angle AMC = \angle ANC$ , and the required statement follows.

Solution 2

Solution:

In this proof we show that we have $\angle NCM = \angle NAM$ instead. From right-angled triangles $BCD$ and $QCR$ we get $\angle DRS = \angle CRQ = \angle RCN$ and $\angle BDC = \angle DCM$ . Hence $\angle NCM = \angle DCM - \angle RCN$ . From right-angled triangle $APS$ we get $\angle PSA = \angle SAN$ . From right-angled triangle $BAD$ we have $\angle MAD = \angle BDA$ . Moreover, $\angle BDA = \angle DRS + \angle RSD - \angle RDB$ .
Therefore $\angle NAM = \angle NAS - \angle MAD = \angle CDB - \angle DRS = \angle NCM$ , and the required statement follows.

Solution 3

Solution:

As $N$ is also the midpoint of $PS$ , we can shrink triangle $APS$ to a triangle $A_0 QR$ (where $P$ is sent to $Q$ and $S$ is sent to $R$ ). Then $A_0, Q, R$ and $C$ lie on a circle with center $N$ . According to the shrinking, the line $A_0R$ is parallel to the line $AD$ . Therefore $\angle CNA = \angle CNA_0 = 2 \angle CRA_0 = 2 \angle CDA = \angle CMA$ . The required statement follows.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.