Kicking the IMO P1 while it's down
by agbdmrbirdyface, Oct 9, 2016, 10:16 PM
Here's the really easy barycentric solution to this, and a better asymptote diagram:
http://www.artofproblemsolving.com/community/c282525h1314108
Solution:![[asy]
size(250);
pair A, B, C, J, K, L, M, F, G, S, T, I, O, P;
A = (0,0);
B = dir(250);
C = 1.5*dir(320);
draw(A--B--C--A, blue);
O = circumcenter(A, B, C);
I = incenter(A, B, C);
P = extension(O, midpoint(B--C), A, I);
J = -I + 2*P;
path Aexcircle = CP(J, foot(J, B, C));
draw(Aexcircle, red);
M = IP(Aexcircle, B--C);
K = extension(A, B, J, foot(J, A, B));
L = extension(A, C, J, foot(J, A, C));
draw(A--K, blue);
draw(A--L, blue);
F = extension(B, J, M, L);
draw(J--F--L, orange);
G = extension(C, J, M, K);
draw(K--G--J, orange);
S = extension(A, F, B, C);
draw(A--S, magenta);
draw(S--B, blue);
T = extension(A, G, B, C);
draw(A--T, magenta);
draw(T--C, blue);
dot(J);
dot(M);
dot(K);
dot(L);
dot(F);
dot(G);
dot(S);
dot(T);
label("$A$", A, N);
label("$B$", B, dir(215)*1.5);
label("$C$", C, dir(40)*1.5);
label("$J$", J, S);
label("$M$", M, N);
label("$K$", K, NW);
label("$L$", L, NE);
label("$F$", F, N);
label("$G$", G, N);
label("$S$", S, N);
label("$T$", T, N);
[/asy]](//latex.artofproblemsolving.com/1/e/3/1e3323cadbcfe9b15f9eb08e6744ee8b9dd04d6f.png)
Suppose we're not good at synth/proj geo (like me) and we're like what how do we approach this oh my lowd this looks rlly hard
Okay, so we can bary this. It looks like everything revolves around cevians and the point
, the A-excenter, which has really really nice coordinates in bary. So why not lol.
Let's list out the points needed to be calculated, and the order we should do them in. We know
, and can easily derive but it's well-known. We can then easily get 
, and then smash these together to get 
, and then we can smash 
together to get 
. If everything goes well, we can easily see 
is the midpoint here; otherwise, we can apply distance formula easily.
Let's begin!
We use
as our reference triangle, so 
by areas. We conveniently know things about that makes them easy to compute, because they're tangency points of the excircle to the lines 
.
After these observations, we should have
. Wow that was easy.
On to computing
! We can see that line 
is easy to compute because it's a "cevian"; the hard thing is to compute 
. This line is given by the determinant 
. Expanding gives us 
. We note has to be of the form 
for some constant 
, and we can plug this into to get the coordinates we need. 
. And after algebra we get 
. Cool now we have 
.
A similar thing happens to
and we get 
. Now since are both on line 
we get 
. Hmm it's not immediately obvious that these points are equidistant from , but one look at the displacement vectors 
gives us 
and now it's clear that is the midpoint of 
, so we're done.
Tidbit
http://www.artofproblemsolving.com/community/c282525h1314108
Solution:
![[asy]
size(250);
pair A, B, C, J, K, L, M, F, G, S, T, I, O, P;
A = (0,0);
B = dir(250);
C = 1.5*dir(320);
draw(A--B--C--A, blue);
O = circumcenter(A, B, C);
I = incenter(A, B, C);
P = extension(O, midpoint(B--C), A, I);
J = -I + 2*P;
path Aexcircle = CP(J, foot(J, B, C));
draw(Aexcircle, red);
M = IP(Aexcircle, B--C);
K = extension(A, B, J, foot(J, A, B));
L = extension(A, C, J, foot(J, A, C));
draw(A--K, blue);
draw(A--L, blue);
F = extension(B, J, M, L);
draw(J--F--L, orange);
G = extension(C, J, M, K);
draw(K--G--J, orange);
S = extension(A, F, B, C);
draw(A--S, magenta);
draw(S--B, blue);
T = extension(A, G, B, C);
draw(A--T, magenta);
draw(T--C, blue);
dot(J);
dot(M);
dot(K);
dot(L);
dot(F);
dot(G);
dot(S);
dot(T);
label("$A$", A, N);
label("$B$", B, dir(215)*1.5);
label("$C$", C, dir(40)*1.5);
label("$J$", J, S);
label("$M$", M, N);
label("$K$", K, NW);
label("$L$", L, NE);
label("$F$", F, N);
label("$G$", G, N);
label("$S$", S, N);
label("$T$", T, N);
[/asy]](http://latex.artofproblemsolving.com/1/e/3/1e3323cadbcfe9b15f9eb08e6744ee8b9dd04d6f.png)
Suppose we're not good at synth/proj geo (like me) and we're like what how do we approach this oh my lowd this looks rlly hard
Okay, so we can bary this. It looks like everything revolves around cevians and the point
, the A-excenter, which has really really nice coordinates in bary. So why not lol.Let's list out the points needed to be calculated, and the order we should do them in. We know
, and can easily derive but it's well-known. We can then easily get 
, and then smash these together to get 
, and then we can smash 
together to get 
. If everything goes well, we can easily see 
is the midpoint here; otherwise, we can apply distance formula easily.Let's begin!
We use
as our reference triangle, so 
by areas. We conveniently know things about that makes them easy to compute, because they're tangency points of the excircle to the lines 
.After these observations, we should have
. Wow that was easy.On to computing
! We can see that line 
is easy to compute because it's a "cevian"; the hard thing is to compute 
. This line is given by the determinant 
. Expanding gives us 
. We note has to be of the form 
for some constant 
, and we can plug this into to get the coordinates we need. 
. And after algebra we get 
. Cool now we have 
.A similar thing happens to
and we get 
. Now since are both on line 
we get 
. Hmm it's not immediately obvious that these points are equidistant from , but one look at the displacement vectors 
gives us 
and now it's clear that is the midpoint of 
, so we're done.Tidbit
I should do 104 now because I'm actual trash at NT and I'm not doing anything to improve. Also, I'm trying to get onto the TJ Varsity Math Team and I should practice Duke and PuMAC problems oops.
May 2017
April 2017