Mein gott what have you done to my diagram
by agbdmrbirdyface, Nov 6, 2016, 12:49 AM
I inverted it.
I don't know if it's great to write solutions with motivation involved, but that's what I do:
Solution:![[asy]
size(250);
pair A, B, C, F, H, K, Q, L, M, O, N, T;
A = origin;
B = (17/14) * dir(310);
C = dir(250);
F = foot(A, B, C);
H = orthocenter(A, B, C);
path circumcircle = circumcircle(A, B, C);
O = circumcenter(A, B, C);
M = midpoint(B--C);
T = midpoint(H--A);
path ninepoint = circumcircle(F, M, T);
path circ1 = circle(T, distance(A, T));
pair[] intersection1 = intersectionpoints(circumcircle, circ1);
Q = intersection1[1];
path circ2 = circle(midpoint(H--Q), distance(H, Q)/2);
pair[] intersection2 = intersectionpoints(circumcircle, circ2);
K = intersection2[1];
N = circumcenter(Q, K, H);
path circQKH = circumcircle(Q, K, H);
path circFKM = circumcircle(K, F, M);
L = rotate(180, midpoint(H--O)) * N;
draw(A--B--C--cycle);
draw(circumcircle,red);
draw(ninepoint, red);
draw(A--F);
draw(A--Q--M);
draw(Q--K--L);
draw(circQKH,blue);
draw(circFKM,blue);
draw(M--N--T--L--cycle, orange);
draw(circumcircle(A, Q, L), dashed);
draw(M--L, dashed);
dot(H);
dot(F);
dot(M);
dot(O);
dot(T);
dot(Q);
dot(K);
dot(N);
dot(L);
label("$A$", A, dir(90));
label("$B$", B, SE);
label("$C$", C, SW);
label("$F$", F, S);
label("$H$", H, dir(245)* 1.8);
label("$O$", O, NE);
label("$M$", M, SE);
label("$Q$", Q, dir(130)*0.05);
label("$K$", K, W);
label("$N$", N, dir(190));
label("$T$", T, NW);
label("$L$", L, dir(30));
[/asy]](//latex.artofproblemsolving.com/a/8/7/a87739d383bcf40ac51b7c4ba2556488440b39ea.png)
Since a lot of the diagram is centered around
, if we would like to invert, we should do it around the orthocenter. Although a lot of the circles pass through
, the other points don't go to very nice places, and it mangles the tangency we want to preserve.
Specifically, we should invert in a way that should overlay the diagram onto itself, and note that the negative inversion at
that takes the red circles to each other (nine-point circle and circumcircle) will take a lot of the points to themselves. This inversion swaps
and
,
and
, and
and
. (
is the point where
.)
This will convert our desired tangency into line
and the circumcircle of
, which can be handled with synthetic geo. Noting
, we see that it now suffices to show
, because then this will imply the tangency.
If we let
be the midpoint of
, we know it lies on the nine-point circle; and similarly for the midpoint of
,
. It's well known
is a diameter of the nine-point circle, so then
must be a rectangle, with a little extra effort from angle chasing.
then must lie on
, because
,
, and
concur at
; therefore the line
is the perpendicular bisector of
, completing the proof.
IMO 2015/3 wrote:
Let
be an acute triangle with
. Let
be its cirumcircle,
its orthocenter, and
the foot of the altitude from
. Let
be the midpoint of
. Let
be the point on
such that
and let
be the point on
such that
. Assume that the points
,
,
,
and
are all different and lie on
in this order. Prove that the circumcircles of triangles
and
are tangent to each other.






















I don't know if it's great to write solutions with motivation involved, but that's what I do:
Solution:
![[asy]
size(250);
pair A, B, C, F, H, K, Q, L, M, O, N, T;
A = origin;
B = (17/14) * dir(310);
C = dir(250);
F = foot(A, B, C);
H = orthocenter(A, B, C);
path circumcircle = circumcircle(A, B, C);
O = circumcenter(A, B, C);
M = midpoint(B--C);
T = midpoint(H--A);
path ninepoint = circumcircle(F, M, T);
path circ1 = circle(T, distance(A, T));
pair[] intersection1 = intersectionpoints(circumcircle, circ1);
Q = intersection1[1];
path circ2 = circle(midpoint(H--Q), distance(H, Q)/2);
pair[] intersection2 = intersectionpoints(circumcircle, circ2);
K = intersection2[1];
N = circumcenter(Q, K, H);
path circQKH = circumcircle(Q, K, H);
path circFKM = circumcircle(K, F, M);
L = rotate(180, midpoint(H--O)) * N;
draw(A--B--C--cycle);
draw(circumcircle,red);
draw(ninepoint, red);
draw(A--F);
draw(A--Q--M);
draw(Q--K--L);
draw(circQKH,blue);
draw(circFKM,blue);
draw(M--N--T--L--cycle, orange);
draw(circumcircle(A, Q, L), dashed);
draw(M--L, dashed);
dot(H);
dot(F);
dot(M);
dot(O);
dot(T);
dot(Q);
dot(K);
dot(N);
dot(L);
label("$A$", A, dir(90));
label("$B$", B, SE);
label("$C$", C, SW);
label("$F$", F, S);
label("$H$", H, dir(245)* 1.8);
label("$O$", O, NE);
label("$M$", M, SE);
label("$Q$", Q, dir(130)*0.05);
label("$K$", K, W);
label("$N$", N, dir(190));
label("$T$", T, NW);
label("$L$", L, dir(30));
[/asy]](http://latex.artofproblemsolving.com/a/8/7/a87739d383bcf40ac51b7c4ba2556488440b39ea.png)
Since a lot of the diagram is centered around


Specifically, we should invert in a way that should overlay the diagram onto itself, and note that the negative inversion at









This will convert our desired tangency into line




If we let













