#10) Orthocenter and Circumcenter on the Incircle.
by ayan.nmath, Feb 13, 2018, 11:46 AM
Problem. If the orthocenter 
and the circumcenter 
lie on the incircle of 
prove that this triangle is unique and one of it's angles is 
(Lonesan)
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(14cm); 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 = -11.56, xmax = 11.56, ymin = -8.04, ymax = 8.04; /* image dimensions */pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */draw(circle((-1.47,0.54), 6.618194617869741), linewidth(0.8) + linetype("4 4") + rvwvcq); draw((3.3824279616279336,5.040493625949423)--(-7.16,-2.84), linewidth(2) + wrwrwr); draw((-7.16,-2.84)--(4.307449273562116,-2.688247185612742), linewidth(2) + wrwrwr); draw((3.3824279616279336,5.040493625949423)--(4.307449273562116,-2.688247185612742), linewidth(2) + wrwrwr); draw(circle((1.2264297598997855,0.009779353729876573), 2.7385589808979787), linewidth(1.2) + dtsfsf); draw((xmin, 2.333357360032656*xmin-2.8519195530953714)--(xmax, 2.333357360032656*xmax-2.8519195530953714), linewidth(1.2) + linetype("4 4") + sexdts); /* line */ /* dots and labels */dot((-1.47,0.54),dotstyle); label("$O$", (-1.38,0.74), NE * labelscalefactor); dot((-7.16,-2.84),dotstyle); label("$B$", (-7.08,-2.64), NE * labelscalefactor); dot((4.307449273562116,-2.688247185612742),dotstyle); label("$A$", (4.38,-2.48), NE * labelscalefactor); dot((3.3824279616279336,5.040493625949423),dotstyle); label("$C$", (3.46,5.24), NE * labelscalefactor); dot((1.2264297598997855,0.009779353729876573),linewidth(4pt) + dotstyle); label("$I$", (1.3,0.16), NE * labelscalefactor); dot((3.469877235190048,-1.5677535596633183),linewidth(4pt) + dotstyle); label("$H$", (3.54,-1.4), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]](//latex.artofproblemsolving.com/8/9/1/891b4de8fa68ae5581ff80ecd7f93e77598400c9.png)
and the circumcenter 
lie on the incircle of 
prove that this triangle is unique and one of it's angles is 
(Lonesan)
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(14cm); 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 = -11.56, xmax = 11.56, ymin = -8.04, ymax = 8.04; /* image dimensions */pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */draw(circle((-1.47,0.54), 6.618194617869741), linewidth(0.8) + linetype("4 4") + rvwvcq); draw((3.3824279616279336,5.040493625949423)--(-7.16,-2.84), linewidth(2) + wrwrwr); draw((-7.16,-2.84)--(4.307449273562116,-2.688247185612742), linewidth(2) + wrwrwr); draw((3.3824279616279336,5.040493625949423)--(4.307449273562116,-2.688247185612742), linewidth(2) + wrwrwr); draw(circle((1.2264297598997855,0.009779353729876573), 2.7385589808979787), linewidth(1.2) + dtsfsf); draw((xmin, 2.333357360032656*xmin-2.8519195530953714)--(xmax, 2.333357360032656*xmax-2.8519195530953714), linewidth(1.2) + linetype("4 4") + sexdts); /* line */ /* dots and labels */dot((-1.47,0.54),dotstyle); label("$O$", (-1.38,0.74), NE * labelscalefactor); dot((-7.16,-2.84),dotstyle); label("$B$", (-7.08,-2.64), NE * labelscalefactor); dot((4.307449273562116,-2.688247185612742),dotstyle); label("$A$", (4.38,-2.48), NE * labelscalefactor); dot((3.3824279616279336,5.040493625949423),dotstyle); label("$C$", (3.46,5.24), NE * labelscalefactor); dot((1.2264297598997855,0.009779353729876573),linewidth(4pt) + dotstyle); label("$I$", (1.3,0.16), NE * labelscalefactor); dot((3.469877235190048,-1.5677535596633183),linewidth(4pt) + dotstyle); label("$H$", (3.54,-1.4), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]](http://latex.artofproblemsolving.com/8/9/1/891b4de8fa68ae5581ff80ecd7f93e77598400c9.png)
.
Let 
be the reflection of 
Note that 
since 
Also, 
which means 
Therefore, 
and 
)
be the relfection of 
be the second intersection of the altitude with 
be the second intersection of 
Since in this case 
because 
by assumption of this case. Which is absurd because 
intersects 
Hence this case is not possible.
so they have a side of equal length. Place the equal sides as chords of the circle such that they coincide, let the chord be 
![\[\angle AIB=\angle 90^{\circ}+\frac{60^{\circ}}{2}=120^{\circ}=\angle AOB\]](http://latex.artofproblemsolving.com/0/b/0/0b0cf8b2055883b2c2c664182c0d684296f9c456.png)
Hence 
lies on circle 
So as we move 
along the arc 
of 
Drop a perpendicular from 
,
(![\[R^2-2Rr=r^2\implies r=\mbox{ fixed quantity in terms of R} (1)\]](http://latex.artofproblemsolving.com/4/b/8/4b8a075dab2bf8bb03160a83d0c6207e19158584.png)
Since we fixed the circumradius of both the triangles, hence 
of 
twice(it is more easier if we drop a perp from This post has been edited 4 times. Last edited by pro_4_ever, Mar 12, 2018, 9:58 AM
March 2018
February 2018