Prove that in a triangle $ABC$ $\forall P, P\neq M_{a},M_{b} ,M_{c}:min\{ \frac{|PA|}{|PM_{a}|},\frac{|PB|}{|PM_{b}|},\frac{|PC|}{|PM_{c}} \}\le 2.$

Question

Suppose a triangle $ABC$. Define $M_{a},M_{b} ,M_{c}$ as the midpoints of the sides $a,b,c$ respectively. Then prove that for any point $P$ different from $M_{a},M_{b} ,M_{c}$, the smallest ratio of magnitudes of lines from $P$ to $M_{x}$ and from $P$ to $x$ is $\le 2$.

I have had certain success in prooving it for special cases, such as triangle $ABP$ being isosceles, but I cannot see a way to generalize this properly. I would think this is a pretty important (or at least very interesting result) but I have not seen it anywhere before (or its proof) so I'm intrigued to hear your ideas.

Answer 1

Hint: "3-way Equality" occurs at the centroid $G$.

Hint: Construct Apollonius circles about $ A, M_a$ with the desired $2:1$ ratio.
Show that these 3 circles will cover triangle $ABC$.

From the first hint, these 3 circles go through a common point.

Find the other point of intersection of 2 of these circles.

This other point is the reflection of the centroid in the corresponding side.

Hence show that if $ P \in \Delta AGD_b$ (where $D_b$ needs to be suitably defined, then $ \frac{ PA}{PM_a} \leq 2$.

Answer 2

This follows from the fact that centroid divides median into $2:1$.

WLOG, point $P \in \triangle ACG$ (or in $\triangle BCG$ or in $\triangle ABG$) where $G$ is the centroid of the $\triangle ABC$.

So $AG + CG \geq AP + CP \implies AG \geq AP \, $ or $\, CG \geq CP$ or both.

If $AG \geq AP,$ (you can show the same for $CG \geq CP$)

Using triangular inequality,

$AP + PD \geq AG + GD \implies PD \geq GD \,$ (as $AP \leq AG$)

So, $\frac{AP}{PD} \leq \frac{AG}{GD} = 2$

Equality occurs when point $P$ is the centroid.

Answer 3

It seems to me (I may be wrong) that the two geometric solutions given to this problem so far have not been very clear to beginners so I want to give an algebraic solution.

The triangle $\triangle{ABC}$ can be given by its vertices $A (0,0), B (a, 0), C (b, c)$ with the condition of $c\ne0$. The locus of the points $P(x,y)$ such that $\dfrac{\overline{PA}}{\overline{PM_A}}=2$ is a circle and the equations of the three circles involved, noted $\Gamma_A,\Gamma_B,\Gamma_C$ are $$\Gamma_A:\hspace 0.5cm 3x^2-4(a+b)x+3y^2-4cy+(a+b)^2+c^2=0..........(1)\\\Gamma_B:\hspace 0.5cm 3x^2-2(2b-a)x+3y^2-4cy+b^2+c^2-a^2=0..........(2)\\\Gamma_C:\hspace 0.5cm3x^2-2(2a-b)x+3y^2+2cy+a^2-b^2-c^2=0..........(3)$$ The set union of intersections $X=(\Gamma_A\cap\Gamma_ B)\cup(\Gamma_A\cap\Gamma_ C)\cup(\Gamma_B\cap\Gamma_ C)$ turns out to be a kind of irregular trifolium and we prove algebraically that the only point belonging to the three foliums is the centroid $G$ of the triangle (which is graphically evident) so that the only solution of the system of the three equations above is given by the known formula $G=\left(\dfrac{a+b}{3}, \dfrac{c}{3}\right)$.

This is straightforward because $(1)-(2)$ gives a single possible value $x=\left(\dfrac{a+b}{3}\right)$ and $(1)-(3)$ and $(2)-(3)$ give a linear system in $x$ and $y$ whose solution is $(x,y)=\left(\dfrac{a+b}{3}, \dfrac{c}{3}\right)$.

Now, noting $\overset{\circ}\Gamma_A$ the interior of the circle $\Gamma_A$, by construction of $\Gamma_A$ we have $\dfrac{\overline{PA}}{\overline{PM_A}}\gt2\iff P\in\overset{\circ}\Gamma_A$ and this is so for the other two circles. It follows that for two of $\left\{ \dfrac{|PA|}{|PM_{a}|},\dfrac{|PB|}{|PM_{b}|},\dfrac{|PC|}{|PM_{c}} \right\}$ be greater than $2$ it is necessaire that the point $P$ belongs to the interior of the intersection of two circles.

For example, if $P(x,y)\in\overset{\circ}\Gamma_A\cap\overset{\circ}\Gamma_ B$ then $\dfrac{\overline{PA}}{\overline{PM_A}}\gt2$ and $\dfrac{\overline{PB}}{\overline{PM_B}}\gt2$ but in this case $P$ must be outside the interior of $\Gamma_C$ since $\{G\}=\Gamma_A\cap\Gamma_ B\cap\Gamma_ C$ and $P\ne G$ (because $\dfrac{\overline{GA}}{\overline{GM_A}}=2)$. This means that $\dfrac{\overline{PC}}{\overline{PM_C}}\lt2$.

We are done.