|
|
AOPS
Geometry Unbound Problems for Section 11.5
import geometry;
unitsize(2cm);
pair A,B,C,I,K,L,M,R,pS,t;
A=(0,0);
C=(4,0);
B=(-1,3);
triangle tri=triangle(A,B,C);
draw(tri);
circle o=incircle(tri);
draw(o);
I=o.C;
triangle tr=intouch(tri);
draw(tr);
K=tr.A;
L=tr.B;
M=tr.C;
draw(I--K);
draw(I--L);
draw(I--M);
label("$A$",A,S);
label("$B$",B,NW);
label("$C$",C,S);
label("$I$",I,NW);
label("$K$",K,NE);
label("$L$",L,S);
label("$M$",M,W);
line t=parallel(B,line(K,L));
R=intersectionpoint(line(M,K),t);
pS=intersectionpoint(line(M,L),t);
label("$R$",R,pS);
label("$S$",pS,W);
draw(t);
draw(pS--M);
draw(M--R);
Let $I$ be the incenter of triangle $ABC$. Let $K,L$ and $M$ be the points of tangency of the incircle of $ABC$ with $AB,BC$ and $CA$, respectively. The line $t$ passes through $B$ and is parallel to $KL$. The lines $MK$ and $ML$ intersect $t$ at the points $R$ and $S$. Prove that $\angle RIS$ is acute.
Evan Chen, §2.2 IMO 1998/5
First simple proof (grobber)
The problem is equivalent to showing $BI^2 > BR \cdot BS$.
But from
\[ \triangle BRK \sim \triangle MKL \sim \triangle BLS \]
we conclude
\[ BR = t \cdot \frac{MK}{ML},
\qquad BS = t \cdot \frac{ML}{MK} \]
where $t = BK = BL$ is the length of the tangent from $B$.
Hence $BR \cdot BS = t^2$.
Since $BI > t$ is clear, we are done.
Second projective proof
Let $N$ be the midpoint of $\overline{KL}$, and let ray $MN$ meet the incircle again at $P$.
Note that line $\overline{RBS}$ is the polar of $N$.
By Brokard's theorem, lines $MK$ and $PL$ should thus meet the polar of $N$, so we conclude $R = \overline{MK} \cap \overline{PL}$.
Analogously, $S = \overline{ML} \cap \overline{PK}$.
Again by Brokard's theorem, $\triangle NRS$ is self-polar, so $N$ is the orthocenter of $\triangle RIS$.
Since $N$ lies between $I$ and $B$ we are done. |
|
Author: 乌贼
hbghlyj
Post time 2023-3-3 01:17
![GergonneLine[2].jpg](data/attachment/forum/202305/03/123159c2epng5anwzge5dd.jpg "GergonneLine[2].jpg")
