## Not too unkind

For students upto class 5 (age upto 12)
SN.Pushpita
Posts: 11
Joined: Thu Aug 10, 2017 1:44 pm

### Not too unkind

. Note that $EF\| ST$ implies that $M$ lies on the median from $A$ of $\triangle {AST}$. As the reflection of $M$ in $ST$ lies on the$\odot{AST}$, we conclude that $AK$ is a symmedian in $\triangle{AST}$, and hence in $\triangle{AEF}$as well. Since $K$ lies on the perpendicular bisector of $EF$, we see that $KE$,$KF$ are tangents to $\odot{AEF}$. Let $BE$ meet$\odot{AEF}$at $P$. Applying Pascal’s Theorem on the cyclic hexagon $\odot{AEEPFF}$we observe that $AE\bigcap PF$,$T$, and $B$, are collinear. This shows that $P$ lies on the line $CF$. Therefore, as $A,E,F$, and the intersection of $BE$ and $CF$ lie on a circle, $\odot{AEF}$ passes through a special point on the $A$ median.
We will find out some properties of that point.
Lemma: Let $X_A$ be the orthogonal projection of $H$ on $A$ median. Then all circles satisfying the above mentioned property pass through this point.
Proof: It's kind of easy using pop, miquel quadrangle Theorem and are left to the readers.
Let $B_1$ and $C_1$ be the feet of altitudes from $B,C$ to sides $AC$,$AB$ respectively. Evidently, $X_A$ lies on$\odot {AB_1C_1}$. Thus, $X_A$ is the center of a spiral similarity sending $B_1C_1$ to $EF$.
Therefore, we have
$\frac{B_1E }{C_1F}=\frac{{X_A}{B_1}}{{X_A}C_1}=\frac{B_1A}{C_1A}=\frac{AB}{AC}$
We are almost done here as from the given condition, it's easy to derive that $AC=\sqrt{67}$. So we get
$C_1F= \frac{2\sqrt{67}}{7}$
So $m+n+P=76$

SN.Pushpita
Posts: 11
Joined: Thu Aug 10, 2017 1:44 pm

### Re: Not too unkind

Solution to a modified G6. Mathmash round 15 P8

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: Not too unkind

SN.Pushpita wrote:
Thu May 17, 2018 2:24 pm
. Note that $EF\| ST$ implies that $M$ lies on the median from $A$ of $\triangle {AST}$. As the reflection of $M$ in $ST$ lies on the$\odot{AST}$, we conclude that $AK$ is a symmedian in $\triangle{AST}$, and hence in $\triangle{AEF}$as well. Since $K$ lies on the perpendicular bisector of $EF$, we see that $KE$,$KF$ are tangents to $\odot{AEF}$. Let $BE$ meet$\odot{AEF}$at $P$. Applying Pascal’s Theorem on the cyclic hexagon $\odot{AEEPFF}$we observe that $AE\bigcap PF$,$T$, and $B$, are collinear. This shows that $P$ lies on the line $CF$. Therefore, as $A,E,F$, and the intersection of $BE$ and $CF$ lie on a circle, $\odot{AEF}$ passes through a special point on the $A$ median.
We will find out some properties of that point.
Lemma: Let $X_A$ be the orthogonal projection of $H$ on $A$ median. Then all circles satisfying the above mentioned property pass through this point.
Proof: It's kind of easy using pop, miquel quadrangle Theorem and are left to the readers.
Let $B_1$ and $C_1$ be the feet of altitudes from $B,C$ to sides $AC$,$AB$ respectively. Evidently, $X_A$ lies on$\odot {AB_1C_1}$. Thus, $X_A$ is the center of a spiral similarity sending $B_1C_1$ to $EF$.
Therefore, we have
$\frac{B_1E }{C_1F}=\frac{{X_A}{B_1}}{{X_A}C_1}=\frac{B_1A}{C_1A}=\frac{AB}{AC}$
We are almost done here as from the given condition, it's easy to derive that $AC=\sqrt{67}$. So we get
$C_1F= \frac{2\sqrt{67}}{7}$
So $m+n+P=76$
This post is perfect at geometry section.