Using only a compass and a ruler, reconstruct triangle $ABC$ given the following three points: point $M$ the intersection of its medians, point $I$ is the center of its inscribed circle and the point $Q_a$ is touch point of the inscribed circle to side $BC$.
Problem
Source: 2016 XIX All-Ukrainian Tournament of Young Mathematicians named after M. Y. Yadrenko, Qualifying p14
Tags: geometry, incenter, construction, Centroid, Ukrainian TYM
Mathematicsislovely
06.05.2021 15:50
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 3.5, xmax = 16., ymin = -5., ymax = 6.; /* image dimensions */ /* draw figures */ draw((6.54,5.22)--(4.76,-3.58), linewidth(0.4)); draw((4.76,-3.58)--(14.6,-3.7), linewidth(0.4)); draw((14.6,-3.7)--(6.54,5.22), linewidth(0.4)); draw((6.54,5.22)--(9.68,-3.64), linewidth(0.4)); draw(circle((8.192512705889316,-0.8070348862393066), 2.814615736356606), linewidth(0.4)); draw((6.54,5.22)--(11.201809324152952,-3.6585586502945473), linewidth(0.4)); draw((8.226834735931586,2.0073715772268397)--(8.158190675847045,-3.6214413497054525), linewidth(0.4)); draw((8.192512705889316,-0.8070348862393066)--(9.514974588221365,-0.4459302275213894), linewidth(0.4)); /* dots and labels */ dot((6.54,5.22),linewidth(2.pt) + dotstyle); label("$A$", (6.608659465737516,5.283504297328684), NE * labelscalefactor); dot((4.76,-3.58),linewidth(2.pt) + dotstyle); label("$B$", (4.378694308943092,-3.8221867595818795), NE * labelscalefactor); dot((14.6,-3.7),linewidth(2.pt) + dotstyle); label("$C$", (14.650048970541654,-4.109379241896312), NE * labelscalefactor); dot((8.192512705889316,-0.8070348862393066),linewidth(2.pt) + dotstyle); label("$I$", (7.909472473867597,-0.9671556118678066), NE * labelscalefactor); dot((9.68,-3.64),linewidth(2.pt) + dotstyle); label("$D$", (9.548158990602895,-4.126272917326573), NE * labelscalefactor); dot((8.633333333333335,-0.6866666666666666),linewidth(2.pt) + dotstyle); label("$M$", (8.703475219089855,-0.61238842783233), NE * labelscalefactor); dot((8.158190675847045,-3.6214413497054525),linewidth(2.pt) + dotstyle); label("$Q_a$", (8.247345982472813,-4.143166592756835), NE * labelscalefactor); dot((11.201809324152952,-3.6585586502945473),linewidth(2.pt) + dotstyle); label("$J$", (11.119270805617148,-4.109379241896312), NE * labelscalefactor); dot((9.514974588221365,-0.4459302275213894),linewidth(2.pt) + dotstyle); label("$N$", (9.581946341463416,-0.3758769718086791), NE * labelscalefactor); dot((8.226834735931586,2.0073715772268397),linewidth(2.pt) + dotstyle); label("$Q'_a$", (8.298027008763595,2.073705965579135), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Construct $ABC$ using the following steps: Let $N$ be the point on $IM$ such that $M$ devides $NI$ in $2:1$ ratio internally.Then $N$ is the Nagel point of $ABC$.Draw this point. Construct the incircle $\omega$ using the compass at center $I$ and with radius $IQ_a$. Draw $Q'_a$ , reflection of $Q_a$ across $I$.Then draw the line parpendicular to $IQ_a$ at $Q_a$.This line will essentially the line $BC$.Call this line $\ell$.Let $NQ'_a$ cut $\ell$ at $J$. Next draw the midpoint $D$ of $JQ_a$.$D$ will be midpoint of $BC$ as well. Extend $DM$ and $JN$.They will cut at $A$. Next draw the 2 tangents from $A$ to $\omega$.They will cut $\ell$ at $B,C$.