Let 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 C meet this circle at the points S and T, respectively. Given a point M on the circle 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
Source:
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 α=∠SMT, β=∠SMP and γ=∠PST. Now we have α=∠MS′S=∠MSS′ (angles in circle and definition of S′), β=∠TMM′ (as MP is symmedian in triangle MST - well known property), γ=∠S′MT (same arc). Now let's play with some similar triangles - our thesis is equivalent to SPPM′=SNNT and as SNT and MNS′ are similar this is equivalent to SPPM′=MNNS′. So it is enough to prove that triangles SPM′ and MNS′ are similar or as ∠S′MN=∠PSM′ it is equivalent to SPSM′=MNMS′. We have: MNMS′=MNMS=MSMT (MS′=MS and similar triangles MNS and MST). As M′ is midpoint of ST we finally have to prove SPM′T=MSMT - 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 △MXS∼△MST. Since ∠XMA=∠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∈BC∩w, E∈CA∩w, F∈AB∩w; L∈AD, FL∥BC; the reflection R of the point F w.r.t. the point L. Prove that R∈DE.
Demonstration. Denote: the midpoints M, N of the segments [FE], [FD] respectively; the intersection S∈DE∩FL. I will show that L∈MN (what is equivalently with the conclusion of our problem), i.e. FS=2⋅FL .
▸
\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).