Consider two circles, tangent at $T$, both inscribed in a rectangle of height $2$ and width $4$. A point $E$ moves counterclockwise around the circle on the left, and a point $D$ moves clockwise around the circle on the right. $E$ and $D$ start moving at the same time; $E$ starts at $T$, and $D$ starts at $A$, where $A$ is the point where the circle on the right intersects the top side of the rectangle. Both points move with the same speed. Find the locus of the midpoints of the segments joining $E$ and $D$.
Problem
Source: 2011 Lusophon Mathematical Olympiad - Problem 5
Tags: geometry, rectangle, symmetry, geometric transformation, parallelogram, reflection, rotation
Luis González
18.09.2011 06:35
Let $(O_1,1) \cong (O_2,1)$ denote the given circles touching the top side of the rectangle at $A,B.$ Their common internal tangent (symmetry axis) cuts $\overline{AB}$ at $P.$ Since $D,E$ move with equal angular velocities, then $\angle AO_1D=-\angle TO_2E$ $\Longrightarrow$ $\mathcal{L}: E \mapsto D$ is the composition of the axial symmetries about $PT,$ $PO_1,$ $AO_1,$ in that order. Thus, $\mathcal{L}$ is equivalent to the composition of an axial symmetry about certain axis $\ell$ and translation parallel to $\ell$ $\Longrightarrow$ midpoint $M$ of $\overline{DE}$ lies on $\ell.$ Now, if $Q$ denotes the antipode of $B$ WRT $(O_2),$ we have $\mathcal{L}: T \mapsto A,$ $Q \mapsto T$ $\Longrightarrow$ $\ell \equiv AT.$ Note that $M$ is always an ordinary point (not lying at infinity), so the locus of $M$ will be actually a segment contained in the line $AT,$ precisely, the diameter of the circle $(T,1)$ intersected by the line $AT.$
yetti
19.09.2011 16:44
Circle $(T,1)$ touches top rectangle side at $P$. $PTO_1A$ is a square. Translate $D \in (O_1,1)$ on the right by $\overrightarrow{O_1T}$ to $D' \in (T,1)$ and $E \in (O_2,1)$ on the left by $\overrightarrow{O_2T}$ to $E' \in (T,1)$. This takes $A, T$ to $P, O_1$. $DD'EE'$ is parallelogram $\Longrightarrow$ segments $DE, D'E'$ have common midpoint $M$. $\measuredangle D'TP = \measuredangle DO_1A = -\measuredangle EO_2T = -\measuredangle E'TO_1$ $\Longrightarrow$ $D'E' \parallel PO_1$ $\Longrightarrow$ $M$ is on diameter of $(T,1)$ perpendicular to $PO_1$ (or parallel to $AT$).
Number1
19.09.2011 18:25
We can get $D$ form $E$ with compositon reflection about $O_1O_2$, rotation about $O_2$ with rotation angle $90$ and translation $\overrightarrow{O_1O_2}$. Then use http://en.wikipedia.org/wiki/Hjelmslev's_theorem
nunoarala
19.09.2011 23:06
Both solutions seem to be correct. Besides, I'd like to know how to post those problems from Lusophon Mathematical Olympiad in the Contest section, since this contest isn't on the list. Could someone tell me how to do it?
Thelink_20
16.06.2024 17:38
(Define points as in the attachement)
Proof: See that $\angle BTA'=\angle ATA'=90^{\circ}\Rightarrow$ $A$, $T$, $B$ are colinear. Because of the symmetry of the condition, $TE=AD$ and $\angle BTE=\angle DAT\Rightarrow \Delta BTE\equiv\Delta BAT.$ It suffices to show that $MD=ME$, but by the Ratio Lemma: $\frac{MD}{ME}=\frac{TE}{TD}\cdot\frac{\sin\angle ETM}{\sin\angle DTM}$ But $\frac{\sin\angle ETM}{\sin\angle DTM}=\frac{\sin\angle ETB}{\sin\angle TBE}=\frac{TD}{TE}$, by Law of Sines on $\Delta BTE$, thus $\frac{MD}{ME}=1$, as desired.
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