Let $\mathcal{C}$ be a circle with center $O$, and let $A$ be a point outside the circle. Let the two tangents from the point $A$ to the circle $\mathcal{C}$ meet this circle at the points $S$ and $T$, respectively. Given a point $M$ on the circle $\mathcal{C}$ which is different from the points $S$ and $T$, let the line $MA$ meet the perpendicular from the point $S$ to the line $MO$ at $P$. Prove that the reflection of the point $S$ in the point $P$ lies on the line $MT$.
Problem
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Tags: geometry, circumcircle, geometric transformation, reflection, similar triangles, geometry proposed
e.lopes
11.02.2006 23:57
my solution is with complex numbers (easy and ugly)
Megus
12.02.2006 01:01
Denote by $M'$ midpoint of $ST$, $S'$ intersection of $SP$ with a circle, $N$ intersection between $MT$ and $SP$. We have to prove that $PM'$ is parallel to $MT$. Now denote $\alpha = \angle SMT$, $\beta = \angle SMP$ and $\gamma = \angle PST$. Now we have $\alpha = \angle MS'S = \angle MSS'$ (angles in circle and definition of $S'$), $\beta = \angle TMM'$ (as $MP$ is symmedian in triangle $MST$ - well known property), $\gamma = \angle S'MT$ (same arc). Now let's play with some similar triangles - our thesis is equivalent to $\frac{SP}{PM'}=\frac{SN}{NT}$ and as $SNT$ and $MNS'$ are similar this is equivalent to $\frac{SP}{PM'}=\frac{MN}{NS'}$. So it is enough to prove that triangles $SPM'$ and $MNS'$ are similar or as $\angle S'MN = \angle PSM'$ it is equivalent to $\frac{SP}{SM'}=\frac{MN}{MS'}$. We have: $\frac{MN}{MS'}=\frac{MN}{MS}=\frac{MS}{MT}$ ($MS'=MS$ and similar triangles $MNS$ and $MST$). As $M'$ is midpoint of $ST$ we finally have to prove $\frac{SP}{M'T}=\frac{MS}{MT}$ - but this is true because triangles $MTM'$ and $MPS$ are similar. P.S. I never thought I could handle so many similar triangles
Sailor
12.02.2006 18:04
Let $Y$ be the midpoint of $[ST]$ and let $X$ be the point of intersection of the perpendicular through $S$ to $[MO]$ with $MT$. Then $\triangle{MXS}\sim\triangle{MST}$. Since $\angle{XMA}=\angle{SMY}$ we are done!
Virgil Nicula
12.02.2006 21:58
Quote:
A equivalent enunciation. Given are the triangle $ABC$, the its incircle $w$ and the points: $D\in BC\cap w$, $E\in CA\cap w$, $F\in AB\cap w$; $L\in AD$, $FL\parallel BC$; the reflection $R$ of the point $F$ w.r.t. the point $L$. Prove that $R\in DE$.
Demonstration. Denote: the midpoints $M$, $N$ of the segments $[FE]$, $[FD]$ respectively; the intersection $S\in DE\cap FL$. I will show that $L\in MN$ (what is equivalently with the conclusion of our problem), i.e. $\boxed {FS=2\cdot FL}\ .$
$\blacktriangleright FL\parallel BC,\ BD=p-b,\ AF=p-a,\ \frac{FL}{BD}=\frac{AF}{AB}\Longrightarrow\boxed {FL=\frac{(p-a)(p-b)}{c}}\ .$
$\blacktriangleright$ $FD=2\cdot BD\sin \frac B2$, $m(\widehat {FDS})=$ $m(\widehat {FDE})=$ $90^{\circ}-\frac A2$, $m(\widehat {DSF})=$ $m(\widehat {EDC})=$ $90^{\circ}-\frac C2$, $\frac{FS}{\sin \widehat {FDS}}=$ $\frac{FD}{\sin \widehat {DSF}}$ $\Longrightarrow$ $FS=\frac{\cos \frac A2}{\cos \frac C2}\cdot 2(p-b)\sin \frac B2$ $\Longrightarrow$ $FS=2(p-b)\sqrt{\frac{p(p-a)}{bc}\cdot \frac{ab}{p(p-c)}\cdot \frac{(p-a)(p-c)}{ac}}$ $\Longrightarrow$ $\boxed {FS=2\cdot \frac{(p-a)(p-b)}{c}}$ $\Longrightarrow$ $FS=2\cdot FL\Longrightarrow L\in MN\Longrightarrow \boxed {\ R\in DE\ }\ .$
Remark. See http://www.mathlinks.ro/Forum/viewtopic.php?t=74477
armpist
12.02.2006 22:29
Darij gives out a hint 'classical tangents to the circumcircle'. Do you make a good use of it anywhere? T.Y. M.T.
Virgil Nicula
12.02.2006 22:55
What do you wish, Armpist ? I solved this problem and nothing something. There is nothing to be done with you: you are incorrigible. Please Armpist, say me something in connection with the my solution and you don't comment besides ...
sprmnt21
13.02.2006 15:47
if t is the tangent through M to c (then t // SP), let's call U and V the intersections of ST with Am and t. It is well known that {S,T; U,V} is a harmonic quadruple then M{S,T; U,V} is a harmonic pencil; then {S,S'; P, oo} is a harmonic quadruple i.e. SP=S'P.
Virgil Nicula
13.02.2006 16:02
I like the your solution, Sprmnt21 ! And I will find a pure syntetical solution for this nice problem ... because the "likeable" Armpist don't let me !
Sailor
13.02.2006 20:22
Virgil Nicula wrote:
Quote:
A equivalent enunciation. Given are the triangle $ABC$, the its incircle $w$ and the points: $D\in BC\cap w$, $E\in CA\cap w$, $F\in AB\cap w$; $L\in AD$, $FL\parallel BC$; the reflection $R$ of the point $F$ w.r.t. the point $L$. Prove that $R\in DE$.
Demonstration. Denote: the midpoints $M$, $N$ of the segments $[FE]$, $[FD]$ respectively; the intersection $S\in DE\cap FL$. I will show that $L\in MN$ (what is equivalently with the conclusion of our problem), i.e. $\boxed {FS=2\cdot FL}\ .$
$\blacktriangleright FL\parallel BC,\ BD=p-b,\ AF=p-a,\ \frac{FL}{BD}=\frac{AF}{AB}\Longrightarrow\boxed {FL=\frac{(p-a)(p-b)}{c}}\ .$
$\blacktriangleright$ $FD=2\cdot BD\sin \frac B2$, $m(\widehat {FDS})=$ $m(\widehat {FDE})=$ $90^{\circ}-\frac A2$, $m(\widehat {DSF})=$ $m(\widehat {EDC})=$ $90^{\circ}-\frac C2$, $\frac{FS}{\sin \widehat {FDS}}=$ $\frac{FD}{\sin \widehat {DSF}}$ $\Longrightarrow$ $FS=\frac{\cos \frac A2}{\cos \frac C2}\cdot 2(p-b)\sin \frac B2$ $\Longrightarrow$ $FS=2(p-b)\sqrt{\frac{p(p-a)}{bc}\cdot \frac{ab}{p(p-c)}\cdot \frac{(p-a)(p-c)}{ac}}$ $\Longrightarrow$ $\boxed {FS=2\cdot \frac{(p-a)(p-b)}{c}}$ $\Longrightarrow$ $FS=2\cdot FL\Longrightarrow L\in MN\Longrightarrow \boxed {\ R\in DE\ }\ .$
Remark. See http://www.mathlinks.ro/Forum/viewtopic.php?t=74477
It was a typo, it is corrected now!
armpist
13.02.2006 21:12
here is my solution(?) : MA is a symmedian . SP is an antiparallel to ST, it is perp. to MO. Symmedians divides antiparallels in half, so S' is on MT. T.Y. M.T.
Virgil Nicula
13.02.2006 21:35
Nicely, Armpist ! I renounce to solve in a different way (pure syntetically) this problem. Why are you so much "unbearable" man ? It is a gainsaying (see my bottom signature).
armpist
13.02.2006 22:08
Hi there Virgil! I remember your previous nik - Ph-An , i.e. FUN. This is my final answer to your rhetorical question. T.Y. M.T.
Virgil Nicula
13.02.2006 22:36
The first nik was Levi. It is O.K. You are the best ! So long ! I will see and read you soon !
armpist
14.02.2006 01:04
Virgil Nicula wrote: The first nik was Levi. It is O.K. You are the best ! So long ! I will see and read you soon ! Dear Virgil, Adulation will take you nowhere. T.Y. M.T.
Virgil Nicula
14.02.2006 12:22
Generally, I don't like the compliments. Quote: I am not in the habit of kissing the hand of a woman, only of the my mother or of the my queen ! (Oscar Wilde).