Consider a trapezoid $ABCD$ with $BC||AD$ and $BC<AD$. Let the lines $AB$ and $CD$ meet at $X$. Let $\omega_1$ be the incircle of the triangle $XBC$, and let $\omega_2$ be the excircle of the triangle $XAD$ which is tangent to the segment $AD$ . Denote by $a$ and $d$ the lines tangent to $\omega_1$ , distinct from $AB$ and $CD$, and passing through $A$ and $D$, respectively. Denote by $b$ and $c$ the lines tangent to $\omega_2$ , distinct from $AB$ and $CD$, passing through $B$ and $C$ respectively. Assume that the lines $a,b,c$ and $d$ are distinct. Prove that they form a parallelogram.
Problem
Source: 2015 Azerbaijan IMO TST
Tags: geometry, trapezoid
rkm0959
29.05.2015 15:43
sunken rock wrote: Wrong! Suppose $\omega_2$ was excircle of $\triangle XBC$, then $c\equiv d$. Best regards, sunken rock Note that in the problem statement, the problem assumes that the lines a,b,c,d are all distinct. Also, w2 is the excircle of triangle XAD.
tranquanghuy7198
30.05.2015 19:36
My solution: Lemma. Given a trapezium $ABCD$ ($AB\parallel{CD}$). If we have: $|AB-CD| = |AD-BC|$, $ABCD$ will be a parallelogram. Proof. $K\in{CD}$ such that $ABCK$ is a parallelogram Now: $|AB- CD| = |CK-CD| = DK \geq |AD-AK| = |AD-BC|$ The equality holds $\Longleftrightarrow D \equiv K$ and $ABCD$ will be a parallelogram. Back to our main problem. $a\cap{BC} = F, c\cap{AD} = E$ We have: $AU = AJ, CH = CV, IH = GJ$ $\Rightarrow AG+AU = AG+AJ = GJ = IH = CI+CH = CI+CV$ $\Rightarrow AK+(AE+EU) = CL+(CF+FV)$ $\Rightarrow (AF+FK)+(AE+EU) = (CE+EL)+(CF+FV)$ $\Rightarrow AF+AE = CE+CF$ ($\because FK = FV, EU = EL$) $\Rightarrow AF-CE = CF-AE$ But $AECF$ is a trapezium ($AE\parallel{CF}$) $\Rightarrow AECF$ is a parallelogram (lemma) $\Rightarrow a\parallel{c}$ Analogously, we have $b\parallel{d}$ and the conclusion follows. Q.E.D
Attachments:
TelvCohl
26.07.2015 14:58
Let $ I_1, I_2 $ be the center of $ \omega_1, \omega_2 $, respectively. Let $ V $ be the reflection of $ C $ in $ \overline{I_1I_2} $. Since $ \measuredangle I_1VX $ $ = \measuredangle XCI_1 $ $ =\measuredangle I_1CB $ $ =\measuredangle I_1I_2A $, so $ I_1, I_2, A, V $ are concyclic $ \Longrightarrow \measuredangle I_1AI_2=\measuredangle I_1VI_2=\measuredangle I_2CI_1 $ $ (\star) $. Since $ AD $ is the image of line $ a $ under $ \mathbf{R}(AI_1) \circ \mathbf{R} (AI_2) $ and $ BC $ is the image of $ c $ under $ \mathbf{R}(CI_2) \circ \mathbf{R} (CI_1) $ ($\mathbf{R}(\ell) :$ reflection in the line $ \ell $) , so from $ AD \parallel BC $ and $ (\star) $ we get $ a \parallel c $. Similarly, we can prove $ b \parallel d \Longrightarrow a, b, c, d $ form the parallelogram.
Attachments:
