b*k 发帖数: 27 | 1 Let AH1, BH2, CH3 be the altitudes of an acute-angled triangle ABC.
The incircle of the triangle ABC touches the sides BC, CA, AB at T1, T2, T3
respectively. Let the lines L1, L2, L3 be the reflections of the lines
H2H3, H3H1, H1H2 in the lines T2T3, T3T1, T1T2 repectively.
Prove that L1, L2, L3 determine a triangle whose vertices lin on the
incircle of the triangle ABC. | l**i 发帖数: 5 | 2 it's not hard to prove that L1//BC etc,
the next thing is to prove the distance
from the center of the incircle to L1
is rcosA,where r is the radius of the
incircle,etc,and we are done.
To prove this,using trigonometrical
relations to find |H3T3| and suppose L1
intersects AB at P3,calculate |T3P3|
we can get the propotion of |AP3| to |AB|
the remaning thing is just checking it
using r and |AH1|.
I believe there should be simpler plane
geometrical method to do this.
【在 b*k 的大作中提到】![](/moin_static193/solenoid/img/up.png) : Let AH1, BH2, CH3 be the altitudes of an acute-angled triangle ABC. : The incircle of the triangle ABC touches the sides BC, CA, AB at T1, T2, T3 : respectively. Let the lines L1, L2, L3 be the reflections of the lines : H2H3, H3H1, H1H2 in the lines T2T3, T3T1, T1T2 repectively. : Prove that L1, L2, L3 determine a triangle whose vertices lin on the : incircle of the triangle ABC.
| u**x 发帖数: 45 | 3 Here is a geometric proof
1) easy to show that L1//BC, L2//AC, L3//AB. omitted
2) Let O be the incenter of ABC, let the incenter of triangle AH2H3 be X.
O, and X are on the bisector of angle A. AO cross T1T2 perpendicularly at
Y.
Since AH2H3 ~= ABC and AH2=AB*cos(A), ----(*)
thus AX=AO*cos(A)
Then XY=AY-AX=AO*cos(A/2)^2-AO*cos(A)=AO*sin(A/2)^2
And since YO=AO*sin(A/2)^2
X is the reflection of O in line T2T3.
So the distance of O to L1 is equal to the distance of X to H2H3, i.e.,
the in-radius of
【在 l**i 的大作中提到】![](/moin_static193/solenoid/img/up.png) : it's not hard to prove that L1//BC etc, : the next thing is to prove the distance : from the center of the incircle to L1 : is rcosA,where r is the radius of the : incircle,etc,and we are done. : To prove this,using trigonometrical : relations to find |H3T3| and suppose L1 : intersects AB at P3,calculate |T3P3| : we can get the propotion of |AP3| to |AB| : the remaning thing is just checking it
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