Lots of Tangent Circles

by MP8148, Jun 19, 2020, 10:38 PM

Source: own? Won’t be surprised if it has appeared somewhere else though. Nonetheless it’s a pretty cool problem and I encourage you to try.

Let $ABC$ be a triangle with circumcircle $\Gamma$ . Suppose that $M$ is the midpoint of $\overline{BC}$ , $E$ is the foot of the $B$ -altitude, and $F$ is the foot of the $C$ -altitude. Denote $\omega_1$ , $\omega_2$ to be the two circles centered at $M$ tangent to $\Gamma$ . Show that $(AEF)$ is tangent to both $\omega_1$ and $\omega_2$ .

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Dang that's cool

by Frestho, Jun 19, 2020, 10:44 PM

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I’ll post my solution if no one solves in three days (from now) :ninja:

This post has been edited 1 time. Last edited by MP8148, Jun 21, 2020, 5:06 PM

by MP8148, Jun 21, 2020, 5:05 PM

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wait is this right

This post has been edited 1 time. Last edited by Frestho, Jun 27, 2020, 4:35 AM
Reason: shrinked the diagram size so it's not squished

by Frestho, Jun 21, 2020, 7:17 PM

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^yep

(fyi I was not aware of your solution and had a more complicated one which proves the below statement)

Extension: show that the tangency points lie on the internal and external bisectors of $\angle BAC$ .

This post has been edited 1 time. Last edited by MP8148, Jun 21, 2020, 7:31 PM

by MP8148, Jun 21, 2020, 7:28 PM

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Let the tangency point on segment $MN$ be $X$ and the other tangency point be $Y$ . Then we have $\triangle ANX \sim \triangle DMX$ , which proves collinearity.

Next, we just need to prove $Y$ , $A$ , $G$ are collinear. This also follows from similar triangles/homothety: $\triangle YAN \sim \triangle YGM$ by SAS similarity.

This post has been edited 1 time. Last edited by Frestho, Mar 3, 2021, 7:58 PM

by Frestho, Jun 21, 2020, 7:36 PM

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Nice

One more:

Show that $H$ , $G$ , the reflection $S$ of $D$ over $M$ , and $K = (AEF) \cap \Gamma$ are concyclic.

This post has been edited 1 time. Last edited by MP8148, Jun 21, 2020, 7:43 PM

by MP8148, Jun 21, 2020, 7:40 PM

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We have $\measuredangle KHS = \measuredangle KHX = \measuredangle KAX = \measuredangle KAD = \measuredangle KGD = \measuredangle KGS$ , as desired.

by Frestho, Jun 21, 2020, 7:51 PM

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^wow I somehow missed that :huh:

my solution uses the fact that $K$ , $M$ , $H$ collinear

by MP8148, Jun 21, 2020, 7:58 PM

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Last one?

Show that $\overline{KG}$ , $\overline{HS}$ , $\overline{BC}$ concur at a point $R$ , and that $R$ , $K$ , $X$ , $M$ , $D$ are concyclic.

by MP8148, Jun 21, 2020, 8:04 PM

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Wow what a rich configuration!

Concurrency:
We use the radical axis theorem. We know that (K, G, H, S) and (B, C, G, K) are concyclic, so we only need to prove B, H, S, C are concyclic.

We have $\measuredangle BHC = 180 - (90 - \measuredangle ABC) - (90 - \measuredangle BCA) = \measuredangle ABC + \measuredangle BCA = \measuredangle BDC = \measuredangle BSC$ .

Concyclicity:
idk gonna keep trying

by Frestho, Jun 21, 2020, 8:13 PM

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wait oops is should be $\measuredangle BHC = (90 - \measuredangle ABC) + (90 - \measuredangle BCA) = -\measuredangle ABC - \measuredangle BCA = -\measuredangle BDC = \measuredangle BSC$

by Frestho, Jun 21, 2020, 8:16 PM

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For aesthetic purposes:

$[asy] size(10cm); defaultpen(fontsize(8pt)); defaultpen(linewidth(0.3)); dotfactor *= 1.5; markscalefactor = 0.02; pair A = dir(130), B = dir(205), C = dir(335), M = dir(270), N = dir(90), D = (B+C)/2, R = 2D-M, H = orthocenter(A,B,C), E = foot(B,C,A), F = foot(C,A,B), S = extension(R,H,B,C), K = S+dir(S--N)*abs(S-B)*abs(S-C)/abs(S-N), P = extension(A,M,R,S), G = extension(A,M,B,C), Q = extension(A,H,S,M), Y = extension(A,N,D,P); draw(A--B--C--A); draw(unitcircle); draw(circumcircle(A,E,F), purple); draw(circumcircle(K,D,M), orange); draw(circle(D,abs(D-M))^^circle(D,abs(D-N)), heavyred); draw(circumcircle(K,H,R)^^circumcircle(S,K,H), heavyblue); draw(Q--A^^M--N^^B--S--R^^B--E^^C--F); draw(N--S--M--A^^K--D--Y--N, heavygreen); draw(circumcircle(C,P,E)^^circumcircle(B,P,F), magenta); draw(arc(circumcenter(C,P,F),circumradius(C,P,F),40,140)^^arc(circumcenter(B,P,E),circumradius(B,P,E),-125,10), lightblue); dot("$A$", A, dir(100)); dot("$B$", B, dir(230)); dot("$C$", C, dir(335)); dot("$D$", M, dir(310)); dot("$G$", N, dir(60)); dot("$M$", D, dir(45)); dot("$N$", (A+H)/2, dir(20)); dot("$S$", R, dir(315)); dot("$H$", H, dir(270)); dot("$E$", E, dir(30)); dot("$F$", F, dir(220)); dot("$R$", S, dir(180)); dot("$K$", K, dir(170)); dot("$X$", P, dir(330)); dot("$J$", G, dir(240)); dot("$Q$", Q, dir(270)); dot("$Y$", Y, dir(135)); [/asy]$

(The additional facts on the diagram are quite easy to prove I believe)

by MP8148, Jun 21, 2020, 8:36 PM

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$[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.2; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.748882177063523, xmax = 9.191274205313425, ymin = -3.771978931579647, ymax = 6.72123666494164; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-0.76,1.88), 4.173104360065778), linewidth(0.7) + wrwrwr); draw((-2.754064896607228,5.545856678611264)--(-4.04,-0.7), linewidth(0.7) + wrwrwr); draw((-4.04,-0.7)--(2.5640107474082594,-0.6430443022496428), linewidth(0.7) + wrwrwr); draw((2.5640107474082594,-0.6430443022496428)--(-2.754064896607228,5.545856678611264), linewidth(0.7) + wrwrwr); draw((xmin, -115.949957743549*xmin-86.24196788509722)--(xmax, -115.949957743549*xmax-86.24196788509722), linewidth(0.7) + wrwrwr); /* line */ draw(circle((-0.7379946262958703,-0.6715221511248214), 1.621487319020532), linewidth(0.7) + wrwrwr); draw(circle((-0.7379946262958703,-0.6715221511248214), 6.724721401111023), linewidth(0.7) + wrwrwr); draw(circle((-2.7320595229030986,2.994334527486442), 2.551617041045246), linewidth(0.7) + wrwrwr); draw((-2.754064896607228,5.545856678611264)--(-2.71005414919897,0.4428123763616201), linewidth(0.7) + wrwrwr); draw(circle((-2.4537070072350224,3.4869403973641884), 3.0549024968803984), linewidth(0.7) + wrwrwr); draw((xmin, 1.0539949148164767*xmin + 6.891917765901611)--(xmax, 1.0539949148164767*xmax + 6.891917765901611), linewidth(0.7) + wrwrwr); /* line */ draw((xmin, 0.25897492057519744*xmin + 1.1446484344049077)--(xmax, 0.25897492057519744*xmax + 1.1446484344049077), linewidth(0.7) + wrwrwr); /* line */ draw((xmin, 0.008624410215066559*xmin-0.665157382731131)--(xmax, 0.008624410215066559*xmax-0.665157382731131), linewidth(0.7) + wrwrwr); /* line */ draw(circle((-3.9765492812196994,-1.5102265856621562), 3.345394054732774), linewidth(0.7) + linetype("4 4") + wrwrwr); /* dots and labels */ dot((-0.76,1.88),dotstyle); label("$O$", (-0.6844368801777315,2.0930721934282164), NE * labelscalefactor); dot((-4.04,-0.7),dotstyle); label("$B$", (-3.9619654500350583,-0.49445036172231166), NE * labelscalefactor); dot((-2.754064896607228,5.545856678611264),dotstyle); label("$A$", (-2.668204172459798,5.758729146558131), NE * labelscalefactor); dot((2.5640107474082594,-0.6430443022496428),dotstyle); label("$C$", (2.6577797535583585,-0.42976229784354847), NE * labelscalefactor); dot((-0.7379946262958703,-0.6715221511248214),linewidth(4pt) + dotstyle); label("$M$", (-0.6413115042585562,-0.49445036172231166), NE * labelscalefactor); dot((-3.7601962452589217,0.6590220421226699),linewidth(4pt) + dotstyle); label("$F$", (-3.681650506560419,0.82087360381254), NE * labelscalefactor); dot((-0.2129085969141928,2.5885905980904096),linewidth(4pt) + dotstyle); label("$E$", (-0.12380699322845194,2.761515520175436), NE * labelscalefactor); dot((-0.7240107745510422,-2.292949170029703),linewidth(4pt) + dotstyle); label("$D$", (-0.6413115042585562,-2.1116519586913918), NE * labelscalefactor); dot((-0.7959892254489576,6.052949170029703),linewidth(4pt) + dotstyle); label("$G$", (-0.7059995681373192,6.233108281669061), NW * labelscalefactor); dot((-2.71005414919897,0.4428123763616201),linewidth(4pt) + dotstyle); label("$H$", (-2.6250787965406226,0.6052467242166627), NE * labelscalefactor); dot((-2.732059522903099,2.994334527486442),linewidth(4pt) + dotstyle); label("$N$", (-2.64664148450021,3.171206591407603), NE * labelscalefactor); dot((-1.5128017396990068,0.7528707240203364),linewidth(4pt) + dotstyle); label("$X$", (-1.3744428948845373,0.8855616676913033), NW * labelscalefactor); dot((-0.7519784780406985,0.9499048677800599),linewidth(4pt) + dotstyle); label("$S$", (-0.6628741922181439,1.1227512352467683), NE * labelscalefactor); dot((-4.929076597897483,1.6966960969767648),linewidth(4pt) + dotstyle); label("$K$", (-4.846035656378153,1.877445313832339), NE * labelscalefactor); dot((-7.229087787888357,-0.7275040012946093),linewidth(4pt) + dotstyle); label("$R$", (-7.153243268054035,-0.5591384256010749), NE * labelscalefactor); dot((-3.9513173061071907,5.235798330952546),linewidth(4pt) + dotstyle); label("$Y$", (-3.875714698196708,5.413726139204727), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

dang i'm still stumped rip

by Frestho, Jun 21, 2020, 8:40 PM

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^using the fact that $K$ , $M$ , $H$ collinear solves

by MP8148, Jun 21, 2020, 8:42 PM

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wait how, im dumb

by Frestho, Jun 22, 2020, 7:47 PM

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So first $DMXR$ cyclic from $\measuredangle RXM = \measuredangle SXM = \measuredangle MSX = \measuredangle RDM.$ Then using the collinear fact $\measuredangle XKM = \measuredangle XKH = \measuredangle XAH = 90^\circ - \measuredangle AHX = \measuredangle HRM = \measuredangle XRM,$ so $MXKR$ is also cyclic and done.

by MP8148, Jun 22, 2020, 7:55 PM

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(The collinear fact follows from $KFHE$ harmonic and $M$ being the intersection of tangents to $(AEF)$ and $E$ and $F$ )

edit: oops what was I doing it literally just follows from the reflection of $H$ over $M$ is the $A$ -antipode

This post has been edited 1 time. Last edited by MP8148, Jul 4, 2020, 11:28 PM

by MP8148, Jun 22, 2020, 7:57 PM

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wow

nice problem!

by Frestho, Jun 22, 2020, 7:58 PM

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Yay

In case you are wondering how I came up with all these, I just messed around in Geogebra and hunted for non-trivial concyclic/collinear points.

by MP8148, Jun 22, 2020, 8:03 PM

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lol nice

by Frestho, Jun 22, 2020, 8:04 PM

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rip jun-dec post
by mathisawesome2169, Jan 15, 2023, 5:22 AM
rip may post
by MP8148, Jun 1, 2022, 5:27 AM
The question is, how will it end?
and the smart answer is "I won't be alive to see it end so I won't care and I will just get this midterm done"
by RedFlame2112, Feb 15, 2022, 11:57 PM
well the universe will end anyways
by kevinmathz, Feb 8, 2022, 2:44 PM
@below this is why you shouldn't waste your time thinking about life, because it doesn't give happy results
by MP8148, Jan 28, 2022, 3:51 AM
why is life so short
by MP8148, Jan 26, 2022, 11:25 PM
https://www.urbandictionary.com/define.php?term=anti-simp

lol what

didn't know this is an actual word
by MP8148, Nov 28, 2021, 4:01 PM
you should antisimp for college board tbh
by mathisawesome2169, Nov 28, 2021, 2:31 PM
good idea
by v4913, Nov 27, 2021, 10:48 PM
tired of simping for college board
by MP8148, Nov 2, 2021, 8:11 PM
waitttttttt why default avatar (
by MrOreoJuice, Nov 2, 2021, 2:04 PM
is it just me or did the total number of blogposts mysteriously decrease by 1? I'm pretty sure it went from 178 to 177
by MP8148, Oct 29, 2021, 12:18 PM
woops i should have mentioned, it was the problem with which the bary cs pdf was attached (linked in the recent geo marathon) and also thanks for the asy guide link
by MrOreoJuice, Sep 27, 2021, 8:03 AM
idk which diagram you're referring to, but thx. it takes practice to get better at asy (like everything else lol), but i think this is pretty good
by MP8148, Sep 26, 2021, 10:11 PM
omgg that diagram is so hawt
you should make an asy tutorial lol
by MrOreoJuice, Sep 26, 2021, 6:12 PM
*shouts not shorts
by MP8148, Dec 4, 2019, 5:42 AM
Wow asymptote works in shorts that’s epic :O
by MP8148, Dec 4, 2019, 5:42 AM
$[asy] unitsize(1cm); pair A = dir(110), B = dir(210), C = dir(330); draw(A--B--C--A); [/asy]$
by MP8148, Dec 4, 2019, 5:40 AM
Pen advantages: Easier to write with (IMO, esp. if you have a decent pen that costs more than 25 cents), nicer to look at.
Pencil advantages: Erasable.

I tried erasable pens but they just feel like darker pencils.
by Frestho, Aug 31, 2019, 6:35 PM
I use pens for everything, except at real [math] contests where I use pencils. Weird right
by MP8148, Aug 22, 2019, 5:58 AM
Lol when I mock AMCs and AIMEs I use my pen.
by Frestho, Aug 22, 2019, 4:43 AM
Sure.
by MP8148, Aug 18, 2019, 5:30 PM
Contrib pls?
by kevinmathz, Aug 6, 2019, 10:07 AM
$\text{[math]}$
by Frestho, Jun 5, 2019, 1:34 AM
Apparently it does.
by MP8148, May 29, 2019, 2:50 AM
Just curious: does LaTeX work in shoutboxes? $\dfrac{2}{3}$
by MP8148, May 29, 2019, 2:47 AM
Nice blog name
by Frestho, Mar 11, 2019, 5:11 PM
I'll definitely make much more. 1 per month is the lower bound .
by MP8148, Dec 19, 2018, 5:08 AM
Only one per month?! You should make it more
by hashtagmath, Dec 18, 2018, 5:22 AM

107 shouts

Prove AB, AC, AD are concurrent

Lots of Tangent Circles

by MP8148, Jun 19, 2020, 10:38 PM

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