Let $A,B$ be two distinct points on a given circle $O$ and let $P$ be the midpoint of the line segment AB. Let $O_1$ be the circle tangent to the line $AB$ at $P$ and tangent to the circle $O$. Let $l$ be the tangent line, different from the line $AB$, to $O_1$ passing through $A$. Let $C$ be the intersection point, different from $A$, of $l$ and $O$. Let $Q$ be the midpoint of the line segment $BC$ and $O_2$ be the circle tangent to the line $BC$ at $Q$ and tangent to the line segment $AC$. Prove that the circle $O_2$ is tangent to the circle $O$.
Problem
Source: APMO 2006, Problem 4
Tags: trigonometry, geometry, circumcircle, geometry unsolved
25.03.2006 03:58
angle chasing and sine law...
25.03.2006 18:13
Trigonometry finishes off this question very quickly. Let the center of $O$ be $X$ and let the radii of $O$ and $O_1$ be $R$ and $R_1$ respectively. Let $XP$ intersect $O$ at $D$. We see that $O_1$ is tangential to $O$ iff $2R = 2R_1 + PD = 2AP \tan \frac{\alpha}{2} + AP \tan \frac{\gamma}{2} = R(2 \sin \gamma + \tan \frac{\alpha}{2} + \sin \gamma \tan \frac{\gamma}{2})$. Using the substitution $\tan \frac{\alpha}{2} = t, \tan \frac{\gamma}{2} = u$, this reduces to $tu = \frac{1}{2}$. Now apply the result to $O_2$.
25.03.2006 18:34
Since the circle tangent to $AB$ (at its midpoint) is also tangent to $AC$( uppose at point $Y$) and to $O$, from the Casey's theorem we get: \[ CY=\frac{a-b}{2}\Longleftrightarrow a+c=3b (1) \] (where $a,b,c$ are the sides of $\triangle{ABC}$) Now if $O_2$ is tangent to $AC$ at $T$ and to $BC$ at $Q$ then, using (1) and the fact that $CT=CQ=\frac{a}{2}$, we obtain: \[ CT\cdot{c}=b\cdot{BQ}+a\cdot{AT} (2) \] However, the only circle which is tangent to $CA,BC$ and the circumcircle of $\triangle{ABC}$ should also satisfy $(2)$ (from the Casey's theorem), therefore $O_2$ is tangent to $O$.
09.04.2006 23:43
This is another intresting property involving a triangle with $a+c=3b$
11.04.2006 12:36
Yeah,that triangle has many many good properties.
12.04.2006 01:32
what a boring solution with an inversion on point a with the AB*AC it is obviously
12.04.2006 21:56
ooiler wrote: what a boring solution with an inversion on point a with the AB*AC it is obviously Congratulations for a magnificent solution!
17.04.2006 08:04
ooiler wrote: what a boring solution with an inversion on point a with the AB*AC it is obviously How? I don't understand?
07.03.2014 06:22
ooiler wrote: what a boring solution with an inversion on point a with the AB*AC it is obviously
17.02.2015 19:47
Sailor wrote: Since the circle tangent to $AB$ (at its midpoint) is also tangent to $AC$( uppose at point $Y$) and to $O$, from the Casey's theorem we get: \[ CY=\frac{a-b}{2}\Longleftrightarrow a+c=3b (1) \] (where $a,b,c$ are the sides of $\triangle{ABC}$) Now if $O_2$ is tangent to $AC$ at $T$ and to $BC$ at $Q$ then, using (1) and the fact that $CT=CQ=\frac{a}{2}$, we obtain: \[ CT\cdot{c}=b\cdot{BQ}+a\cdot{AT} (2) \] However, the only circle which is tangent to $CA,BC$ and the circumcircle of $\triangle{ABC}$ should also satisfy $(2)$ (from the Casey's theorem), therefore $O_2$ is tangent to $O$. Can you just state "by Casey's Theorem..." Is it well-known enough to use without proof? For example, in APMO #3, Wolstenholme's Theorem need proof! Can someone tell me which theorems you need proof for?
07.04.2017 02:22
Wait you actually don't need anything complicated (it's all straightforward bashing): *My actual proof's a bit long so I'll just write my thought processes: 1) First you rephrase the problem statement: Triangle ABC is drawn such that the A mixtilinear circle touches AB at the midpoint, P. Prove that the C mixtilinear circle touches BC at the midpoint Q. 2) Then you find a series of equivalent statements. The path I went along used the fact that the tangency pts of mixtilinear circles on the sides of the triangle are bisected by the incenter. Anyways, it suffices to prove that the incenter I bisects C-mixtilinear tangency points $R_2$ and Q, which reduces to $R_2, I, Q$ need to be collinear. 3) After some simplifications you arrive at a+c=3b. This statement is independent of everything I just did and is essentially a much easier problem than the one we started with. 4) Specifically, the new problem states that "the A mixtilinear circle is tangent to AB at the midpoint P. Prove that a+c = 3b". Here, you note that AIP is a right triangle and you length bash. (using [ABC]=rs, and other incircle stuff) This works out in the end and you're done.
17.02.2020 13:10
It's very easy using bary first note that the problem equivalet to "$\triangle ABC$ is drawn such that the A-mixtilinear circle touches $AB$ at the midpoint, Prove that the C-mixtilinear circle touches $BC$ at the midpoint Let the A-mixtilinear touches $(ABC)$ at $M$ and $E$ the ex-touch point on $BC$ then easitly get $M(-a:\frac{b^2}{s-b}:\frac{c^2}{s-c})$ (using that $AM$ and $AE$ are isogonal but $MP$ is the perpindacular bisetor of $AB$ thus $M$ is the midpoint of arc $ACB$ then $M(-a:b:\frac{c^2}{a-b})$ $b=s-b \implies a+c=3b$ which is symmetric in $A,C$ so this is equivalent to "the C-mixtilinear circle touches $BC$ at the midpoint "
09.07.2021 22:43
It is obvious that $O_1$ is $A$-mixtilinear circle. The problem is equivalent to proving that if the $A$-mixtilinear circle touches $AB$ at its midpoint, then $C$-mixtilinear circle touches BC at its midpoint. Let $F$, $G$ be touchpoints of $A$-mixtilinear and $C$-mixtilinear with $AC$ and $Q$ touchpoint of $C$-mixtilinear and $BC$. Since $GQ$ and $FP$ intersect each other at middle (the incenter-well-known), we get that FGPQ is parallelogram, thus PQ|| AC. $P$ is midpoint => $Q$ is midpoint.