tangent circles wanted, 2 altitudes, 2 tangents to circumcircle

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Source: 2017 Maths Beyond Limits Camp - Olympic Challenge - Older Division - Geometry p6

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#1 h Dec 12, 2020, 11:42 PM

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Let $H_B$ and $H_C$ be the foots of the altitudes from the vertexes $B$ and $C$ in $\vartriangle ABC$ respectively. Denote by $\Gamma$ the circumcircle of $\vartriangle ABC$ . Prove that circumcircle of triangle bounded by the lines tangent to $\Gamma$ at the points $B$ and $C$ and the line $H_BH_C$ is tangent to $\Gamma$

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DNCT1

#2 h Dec 16, 2020, 2:56 PM

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$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8cm); 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 = 1.5072424195747889, xmax = 28.578164473583794, ymin = -16.168606762157857, ymax = 7.874042655301023; /* image dimensions */ pen ffqqff = rgb(1.,0.,1.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); /* draw figures */ draw(circle((13.1,-2.34), 7.218226929101078)); draw((9.92,4.14)--(7.748747015588143,-7.184263772424369)); draw((7.748747015588143,-7.184263772424369)--(19.46183280817208,-5.75026147397312)); draw((19.46183280817208,-5.75026147397312)--(9.92,4.14)); draw((8.429275379017609,-3.63493868344627)--(19.46183280817208,-5.75026147397312)); draw((7.748747015588143,-7.184263772424369)--(14.531785056370618,-0.6401885640045156)); draw(circle((17.01901801966904,-3.674928165376932), 11.358364057893708), ffqqff); draw((5.733349397382493,-4.957939396656096)--(14.622707113181457,-14.7776363084878)); draw((5.733349397382493,-4.957939396656096)--(24.938765703358065,4.466940827572655)); draw((24.938765703358065,4.466940827572655)--(14.622707113181457,-14.7776363084878)); draw(circle((10.42528991188011,0.012737376801253678), 4.158078240714236), red); draw((6.267288781698625,-0.012585682071521151)--(7.748747015588143,-7.184263772424369)); draw((6.267288781698625,-0.012585682071521151)--(8.429275379017609,-3.63493868344627)); draw((6.267288781698625,-0.012585682071521151)--(14.531785056370618,-0.6401885640045156)); draw((6.267288781698625,-0.012585682071521151)--(9.92,4.14)); draw((5.733349397382493,-4.957939396656096)--(6.267288781698625,-0.012585682071521151)); draw((6.267288781698625,-0.012585682071521151)--(24.938765703358065,4.466940827572655)); draw(circle((10.201149904109993,-2.9388173603531627), 4.902866015444295), qqwuqq); draw(circle((15.419214807189064,2.9933410583248325), 9.63293026781981), linetype("2 2") + blue); draw((6.267288781698625,-0.012585682071521151)--(3.8870540672573717,-7.000364484613414)); /* dots and labels */ dot((9.92,4.14),linewidth(3.pt) + dotstyle); label("$A$", (10.286174204321211,4.63164973131802), NE * labelscalefactor); dot((7.748747015588143,-7.184263772424369),linewidth(3.pt) + dotstyle); label("$B$", (7.074369892828618,-8.093213064690751), NE * labelscalefactor); dot((19.46183280817208,-5.75026147397312),linewidth(3.pt) + dotstyle); label("$C$", (19.676878238875748,-6.777902727603305), NE * labelscalefactor); dot((14.531785056370618,-0.6401885640045156),linewidth(3.pt) + dotstyle); label("$H_b$", (14.935643302862871,0.22688953269959924), NE * labelscalefactor); dot((8.429275379017609,-3.63493868344627),linewidth(3.pt) + dotstyle); label("$H_c$", (8.848509417272146,-2.648440041398536), NE * labelscalefactor); dot((5.733349397382493,-4.957939396656096),linewidth(3.pt) + dotstyle); label("$Z$", (5.239053143404279,-5.737889902929512), NE * labelscalefactor); dot((14.622707113181457,-14.7776363084878),linewidth(3.pt) + dotstyle); label("$X$", (14.017984928150701,-15.55683451234974), NE * labelscalefactor); dot((24.938765703358065,4.466940827572655),linewidth(3.pt) + dotstyle); label("$Y$", (25.06047403718714,4.662238343808426), NE * labelscalefactor); dot((6.267288781698625,-0.012585682071521151),linewidth(3.pt) + dotstyle); label("$D$", (5.422584818346713,0.10453508273797645), NE * labelscalefactor); dot((10.930579823760214,-4.1145252463974895),linewidth(3.pt) + dotstyle); label("$H$", (11.417952866466221,-4.942585978178964), NE * labelscalefactor); label("$\alpha$", (3.7402111313744015,-5.6155354529678885), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
$\textit{Proof}$
Let's denote the triangle bounded by the lines tangent to $\Gamma$ at the points $B$ and $C$ and the line $H_BH_C$ is $\bigtriangleup XYZ$
$H$ is the orthocenter of $\bigtriangleup ABC\implies H\in(AH_cH_b)$
Let $\{D,A\}=(AH_bH_c)\cap(ABC)$ so we easily get that $D$ is the spiral similarity center which $(AH_bH_c)\mapsto (ABC), H_cH_b\mapsto BC$
Since $\overline{ZX}$ is tagent to $(ABC)$ at $B$ we get that $\angle DBZ=\angle DCB=\angle DH_bH_c=\angle DH_bZ\implies DZBH_b\quad\text{concyclic} (1)$ Similarly we also get that $DYCH_c\quad\text{concyclic}$ .
By $(1)\implies \angle DZB=180^o-\angle DH_bB=\angle DH_cH=\angle DH_cC=180^o-\angle DYX\implies D\in(XYZ)$
Now let $D\alpha$ is tagent to $(ABC)$ at $D$ .
$\angle \alpha DZ= \angle \alpha DB-\angle ZDB=\angle DCB-\angle H_cH_bB=\angle DCB-\angle H_cCB=\angle DCH_c=\angle DYH_c=\angle DYZ\implies D\alpha \quad\text{tagents to}\quad (XYZ)$ So $(XYZ)$ tagents $(ABC)$ at $D$ (as needed). $\blacksquare$

This post has been edited 1 time. Last edited by DNCT1, Dec 19, 2020, 3:26 PM

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