On the sides of the triangle $ABC$, three similar isosceles triangles $ABP \ (AP = PB)$, $AQC \ (AQ = QC)$, and $BRC \ (BR = RC)$ are constructed. The first two are constructed externally to the triangle $ABC$, but the third is placed in the same half-plane determined by the line $BC$ as the triangle $ABC$. Prove that $APRQ$ is a parallelogram.
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Tags: trigonometry, geometry, parallelogram, Triangle, IMO Shortlist
AK1024
10.09.2010 02:19
[mod. edit : Here is alternate link for the above image.]
A classic solution using spiral similarity.
The spiral similarity with centre $C$ that sends $A$ to $B$ must send $Q$ to $R$ since $\triangle AQC$ is similar to $\triangle BRC$. Hence a spiral similarity with centre $C$ that sends $Q$ to $A$ must send $R$ to $B$. This has the consequence that $\triangle QRC$ is similar to $\triangle ABC$.
By a symmetric argument, $\triangle BPR$ is similar to triangle to $\triangle ABC$, implying $\triangle BPR$ is similar to $\triangle QRC$. Moreover they are in the similar in the ratio $\frac{BR}{RC}=1$, i.e. they are congruent. Hence $QR=PB=PA$ and $PR=QC=QA$. It follows that $APRQ$ is a parallelogram.
sayantanchakraborty
05.06.2014 14:33
From the similar isisceles triangles we get $\frac{BP}{AB}=\frac{CQ}{AC}=\frac{BR}{BC}=k$.We also have $\angle{PBR}=B$ Using the cosine rule in $\triangle{BRP}$ we get ${PR^2=BR^2+BR^2-2BP*BRcosB=k^2(AB^2+BC^2-2AB*BCcosB}=(ACk)^2=QC^2=QA^2 \Rightarrow PR=AQ$ Similarly $AP=RQ$ and the result follows.
jayme
05.06.2014 16:12
Dear Mathlinkers, you can se on synthetic proof on http://perso.orange.fr/jl.ayme vol. 16 Deux triangles semblables..., p. 14-17. Sincerely Jean-Louis
Tsikaloudakis
05.06.2014 18:47
Solution (math class A΄ high school) \[\begin{gathered} \left. \begin{gathered} \frac{{PB}}{{AB}} = \frac{{RB}}{{CB}} \hfill \\ P\hat BR = A\hat BC \hfill \\ \end{gathered} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} tr.PBR \approx tr.ACB\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} {{\hat P}_2} = {{\hat A}_1} \hfill \\ \bullet \mathop {}\limits^{\mathop {}\limits^{} } similarly\mathop {}\limits^{} \mathop {}\limits^{} {{\hat Q}_2} = {{\hat A}_1}\mathop {}\limits^{} \mathop {}\limits^{} So\mathop {}\limits^{} :\mathop {}\limits^{} {{\hat P}_1} = {{\hat Q}_1}\mathop {}\limits_{\mathop {}\nolimits_{} } \hfill \\ \left. \begin{gathered} {{\hat P}_1} + {{\hat P}_2} = {{\hat A}_1} + 2\varphi \mathop {}\limits_{} \hfill \\ {A_1} + 2\varphi + {{\hat R}_1} + {{\hat R}_2} + \theta = {360^o} \hfill \\ \end{gathered} \right\}\mathop {}\limits_{} \Rightarrow \mathop {}\limits^{} \theta = {{\hat A}_1} + 2\varphi \mathop {}\limits_{\mathop {}\limits_{\mathop {}\limits_{\mathop {}\limits_{} } } } \hfill \\ \left. \begin{gathered} {{\hat P}_1} = {{\hat Q}_1} \hfill \\ \theta = {{\hat A}_1} + 2\varphi \hfill \\ \end{gathered} \right\}\mathop {}\limits^{} \Rightarrow \mathop {}\limits^{} APRQ\mathop {}\limits^{} \# \hfill \\ \end{gathered} \]
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HamstPan38825
13.12.2022 03:27
Let $a, b, c$ lie on the unit circle without loss of generality. Then,$$p = \frac{a+b}2 + (a+b) \cdot k\frac{|a-b|}{|a+b|}$$and similarly for $q$ for some fixed real constant $k$ by the similarity condition. So, stripping away terms, it suffices to show that$$\sum_{\mathrm{cyc}} \frac{(a+b)|a-b|}{|a+b|} = 0$$after canceling the $k$ term. Notice that each term represents a vector perpendicular to one of the sides with the same length as that side. The angles formed by the three vectors are congruent (directed) to each other, so they must form a closed triangle, as needed.