56th International Mathematical Olympiad Shortlisted Problems

Let $S$ be the center of the parallelogram $BPTQ$, and let $B^{\prime }\ne B$ be the point on the ray $BM$ such that $BM=M{B}^{\prime }$ (see Figure 1). It follows that $ABC{B}^{\prime }$ is a parallelogram. Then, $\angle AB{B}^{\prime }=\angle PQM$ and $\angle B{B}^{\prime }A=\angle B^{\prime }BC=\angle MPQ$, and so the triangles $AB{B}^{\prime }$ and $MQP$ are similar. It follows that $AM$ and $MS$ are corresponding medians in these triangles. Hence, $$\angle SMP=\angle B^{\prime }AM=\angle BCA=\angle BTA.$$ Since $\angle ACT=\angle PBT$ and $\angle TAC=\angle TBC=\angle BTP$, the triangles $TCA$ and $PBT$ are similar. Again, as $TM$ and $PS$ are corresponding medians in these triangles, we have $$\angle MTA=\angle TPS=\angle BQP=\angle BMP.$$ Now we deal separately with two cases.

Case 1. $S$ does not lie on $BM$. Since the configuration is symmetric between $A$ and $C$, we may assume that $S$ and $A$ lie on the same side with respect to the line $BM$. Applying the previous results, we get $$\angle BMS=\angle BMP-\angle SMP=\angle MTA-\angle BTA=\angle MTB,$$ and so the triangles $BSM$ and $BMT$ are similar. We now have $B{M}^2=BS\cdot BT=B{T}^2/2$, so $BT=\sqrt{2}BM$.

Case 2. $S$ lies on $BM$. It follows from the previous results that $\angle BCA=\angle MTA=\angle BQP=\angle BMP$ (see Figure 2). Thus, $PQ\parallel AC$ and $PM\parallel AT$. Hence, $BS/BM=BP/BA=BM/BT$, so $B{T}^2=2B{M}^2$ and $BT=\sqrt{2}BM$.

Figure 1 Figure 2

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Alternative solution.

Again, we denote by $\Omega$ the circumcircle of the triangle $ABC$. Choose the points $X$ and $Y$ on the rays $BA$ and $BC$ respectively, so that $\angle MXB=\angle MBC$ and $\angle BYM=\angle ABM$ (see Figure 4). Then the triangles $BMX$ and $YMB$ are similar. Since $\angle XPM=\angle BQM$, the points $P$ and $Q$ correspond to each other in these triangles. So, if $\overrightarrow{BP}=\mu \cdot \overrightarrow{BX}$, then $\overrightarrow{BQ}=(1-\mu )\cdot \overrightarrow{BY}$. Thus $$\overrightarrow{BT}=\overrightarrow{BP}+\overrightarrow{BQ}=\overrightarrow{BY}+\mu \cdot (\overrightarrow{BX}-\overrightarrow{BY})=\overrightarrow{BY}+\mu \cdot \overrightarrow{YX},$$ which means that $T$ lies on the line $XY$. Let $B^{\prime }\ne B$ be the point on the ray $BM$ such that $BM=M{B}^{\prime }$. Then $\angle M{B}^{\prime }A=\angle MBC=\angle MXB$ and $\angle C{B}^{\prime }M=\angle ABM=\angle BYM$. This means that the triangles $BMX$, $BA{B}^{\prime }$, $YMB$, and $B^{\prime }CB$ are all similar; hence $BA\cdot BX=BM\cdot B{B}^{\prime }=BC\cdot BY$. Thus there exists an inversion centered at $B$ which swaps $A$ with $X$, $M$ with $B^{\prime }$, and $C$ with $Y$. This inversion then swaps $\Omega$ with the line $XY$, and hence it preserves $T$. Therefore, we have $B{T}^2=BM\cdot B{B}^{\prime }=2B{M}^2$, and $BT=\sqrt{2}BM$.

Figure 4

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Alternative solution.

We begin with the following lemma.

Lemma. Let $ABCT$ be a cyclic quadrilateral. Let $P$ and $Q$ be points on the sides $BA$ and $BC$ respectively, such that $BPTQ$ is a parallelogram. Then $BP\cdot BA+BQ\cdot BC=B{T}^2$.

Proof. Let the circumcircle of the triangle $QTC$ meet the line $BT$ again at $J$ (see Figure 5). The power of $B$ with respect to this circle yields $$BQ\cdot BC=BJ\cdot BT.$$ We also have $\angle TJQ=180^{\circ }-\angle QCT=\angle TAB$ and $\angle QTJ=\angle ABT$, and so the triangles $TJQ$ and $BAT$ are similar. We now have $TJ/TQ=BA/BT$. Therefore, $$TJ\cdot BT=TQ\cdot BA=BP\cdot BA.$$ Combining the above, we get the desired result. $\square$

Let $X$ and $Y$ be the midpoints of $BA$ and $BC$ respectively (see Figure 6). Applying the lemma to the cyclic quadrilaterals $PBQM$ and $ABCT$, we obtain $$BX\cdot BP+BY\cdot BQ=B{M}^2$$ and $$BP\cdot BA+BQ\cdot BC=B{T}^2.$$ Since $BA=2BX$ and $BC=2BY$, we have $B{T}^2=2B{M}^2$, and so $BT=\sqrt{2}BM$.

Figure 5 Figure 6