Let $ ABC$ be a triangle and let $ \mathcal{M}_{a}$, $ \mathcal{M}_{b}$, $ \mathcal{M}_{c}$ be the circles having as diameters the medians $ m_{a}$, $ m_{b}$, $ m_{c}$ of triangle $ ABC$, respectively. If two of these three circles are tangent to the incircle of $ ABC$, prove that the third is tangent as well.
Problem
Source: Romanian TST 5 2008, Problem 2
Tags: geometry, geometry proposed
plane geometry
15.06.2008 23:18
Could somebody solve it?
kaka__2004
17.06.2008 17:21
no one ? :
silouan
18.06.2008 10:52
My idea is the following and I will write it as a hint :
Let $ AA_1,BB_1,CC_1$ be the altitudes and let $ w$ be the incircle and $ K,L,N$ the midpoints of BC,AC,AB
Let that the $ M_b$ and $ M_c$ are tangent to incircle and take ta GPT theorem for the 4-tuples of circles $ (B,w,L,B_1)$
and for the 4-tuple $ (C,w,N,C_1)$ for the circles $ M_b$ and $ M_c$ respectively . Then we have only to prove that
the converse of the GPT for the 4-tuple of circles $ (A_1,K,w,A)$ which will mean that there is a circle which passes through the points $ A_1,\ K,\ A$ (which is $ M_a$) which is tangent to the incircle
Hope I am true ...Could be done like this Cosmin ?
yetti
19.06.2008 11:16
My idea is that the triangle is equilateral. Sorry for being so prosaic. If $ M_a$ is midpoint of $ BC$ and $ H_a$ foot of the A-altitude, the 9-point circle $ (N)$ and the circle $ \mathcal M_a$ define a pencil with 2 different base points $ H_a, M_a$ in a non-isosceles $ \triangle ABC.$ The incircle tangency point $ D_a$ with $ BC$ is in between $ H_a, M_a.$ $ (N)$ is tangent to $ (I).$ If $ \angle A > 90^\circ,$ $ \mathcal M_a$ is smaller than $ (N)$ and intersects $ (I).$ If $ \angle A < 90^\circ,$ $ \mathcal M_a$ is larger than $ (N)$ and does not touch $ (I).$ The only way $ \mathcal M_a$ can touch $ (I)$ is when the angle $ \angle A$ is right, in which case $ \mathcal M_a \equiv (N)$ are identical, or if the 2 base points $ H_a \equiv M_a$ are identical and the $ \triangle ABC$ is isosceles with $ AB = AC.$ Therefore, the $ \triangle ABC$ must be equilateral for 2 of the 3 circles $ \mathcal M_a, \mathcal M_b, \mathcal M_c$ to be tangent to $ (I),$ and then the 3rd one is also tangent to $ (I).$
Altheman
19.06.2008 21:55
Woo who's ready for some algebra... Let $ M$ be the midpoint of $ BC$ and $ M'$ be the midpoint of $ AM$. Since $ \mathcal{M}$ is larger than the incircle, the tangency is equivalent to \[ \frac{AM}{2}-r=M'I\]
Note that $ 4AM^2+a^2=2b^2+2c^2$, $ IM^2=r^2+\left(\frac{b-c}{2}\right)^2$, and $ AI^2=r^2+(s-a)^2$. By the median formula on $ IMA$, \[ 4M'I^2+AM^2+2AI^2+2IM^2\] \[ (AM-2r)^2+AM^2=2(r^2+(s-a)^2)+2\left(r^2+\left(\frac{b-c}{2}\right)^2\right)\]
\[ \frac{2b^2+2c^2-a^2}{4}-2rAM=(s-a)^2+\left(\frac{b-c}{2}\right)^2\]
\[ 2rAM=a(s-a)\]
For ease of computation, I'll do the substitution, $ a=y+z$, etc. Square the relation above, and we get \[ \frac{4xyz}{x+y+z}\cdot\frac{2(x+z)^2+2(x+y)^2-(y+z)^2}{4}=(y+z)^2x^2\]\[ \frac{(x)(y-z)^2(x^2 + x y + x z - y z)}{x+y+z}=0\]
\[ \iff (b-c)^2(b^2+c^2-a^2)=0\] It is easy to see that the only way for this to happen for two vertices is for the triangle to be equilateral. (Otherwise the triangle degenerates)