of the plane, which are announced to Bob. Next, Bob marks infinitely many points in the plane, designating each a city. He may not place two cities within distance at most one unit of each other, and no three cities he places may be collinear. Finally, roads are constructed between the cities as follows: for each pair 
of cities, they are connected with a road along the line segment 
if and only if the following condition holds:
and 
such[/center]
is directly similar to either 
or 
.
is directly similar to 
if there exists a sequence of rotations, translations, and dilations sending 
to 
,
to 
,
to 
.
and 
be positive integers with 
.
be a polynomial of degree 
such that the polynomial 
divides 
is zero. Prove that 
and 
be a 
grid of unit squares.
or 
rectangle. A subset 
of grid squares in 
edges of the grid squares.
Y by Billybillybobjoejr., mathmax12, Johnson100, OronSH, guineapig2013, Rounak_iitr, ItsBesi
In an acute triangle 
, let 
be the midpoint of 
. Let 
be the foot of the perpendicular from 
to 
. Suppose that the circumcircle of triangle 
intersects line 
at two distinct points 
and 
. Let 
be the midpoint of 
. Prove that 
.
Proposed by Holden Mui
, let 
be the midpoint of 
. Let 
be the foot of the perpendicular from 
to 
. Suppose that the circumcircle of triangle 
intersects line 
at two distinct points 
and 
. Let 
be the midpoint of 
. Prove that 
.Proposed by Holden Mui
This post has been edited 2 times. Last edited by GoodMorning, Mar 23, 2023, 10:31 PM
be the foot of the 
.
to 
be the actual foot; we will prove that 
is cyclic. This is just angle chasing:
Thus 
.
is the 
,
.
by PoP, and then trivial by midline.
,
and 
.
gg
![[asy]
size(10cm);
pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair D = foot(A, B, C);
pair M = midpoint(B--C);
pair P = foot(C, A, M);
pair Q = 2*M-D;
filldraw(A--B--C--cycle, palecyan, blue);
draw(circumcircle(A, B, P), orange+dashed);
draw(A--P--C, deepgreen);
draw(A--Q, red+dashed);
pair N = midpoint(A--Q);
draw(B--N--C, red+dotted);
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M$", M, dir(45));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(315));
dot("$N$", N, dir(45));
[/asy]](http://latex.artofproblemsolving.com/5/9/9/5998b96c7d2e2c9dad2e6182ea721b52e82fe8e4.png)
(oppositely oriented).
from the given concyclicity of 
.![\[ \frac{BM}{BP} = \frac{AM}{AQ} \implies
\frac{2BM}{BP} = \frac{AM}{AQ/2} \implies
\frac{BC}{BP} = \frac{AM}{AN} \]](http://latex.artofproblemsolving.com/9/0/3/9036afb1ab1889e5fe1e3513c3120ee6cbc5102d.png)
implying the similarity (since 
)
![\[ \measuredangle \left( BC, MN \right) = -\measuredangle \left( PC, AM \right) = 90^\circ \]](http://latex.artofproblemsolving.com/5/c/6/5c6a7b674dc37264b8261b22bd71f752e5e40b77.png)
as desired.
,
,
.
is 
,
is 
.
.
.
,
,
.
,
,
and hence 
and get 
cyclic by PoP and then 
and finish by homothety centered at Q which sends 
to 
so 
which means we’re done
.
be the intersection of 
,
finishes.
,
is a cyclic quadrilateral. By power of a point, this means 
.
are concyclic, and therefore by power of a point 
.
,
,
to get 
.
and 
.
.
.
.
and 
as desired. 
Then clearly 
therefore 
Hence by the Midpoint Theorem it follows that 
Thus 
is the perpendicular bisector of 
.
be the altitude. 
implies 
.
is cyclic.Apply power of point twice on 
.
and 
.
solving our problem
be the reflection of 
.
,
are concyclic. Observe that 
by MPT, so 
.
.
,
.
are similar right triangles.
and 
,![\[BM\cdot QM=CM\cdot DM\iff DM=QM,\]](http://latex.artofproblemsolving.com/9/2/8/928063d6a6f181fe42f9164dc871a0ea3138c836.png)
done. 
![\[ AM\cdot MP = DM\cdot MC.\]](http://latex.artofproblemsolving.com/6/9/e/69ea80ac8d732d75d3032a5334f50fcd2ab6c504.png)
We similarly find ![\[ AM\cdot MP = BM\cdot MQ, \]](http://latex.artofproblemsolving.com/8/2/e/82e9d7a8a3c418b761e5829f344b646d37f71cca.png)
so ![\[ DM\cdot MC = BM\cdot MQ \implies DM=MQ. \]](http://latex.artofproblemsolving.com/1/d/3/1d30221d147c533d16cf7add35d2453ee6fe945f.png)
![\[ \triangle{ADQ}\sim \triangle{NMQ}, \]](http://latex.artofproblemsolving.com/2/f/1/2f13ce18a45f88f8abe5ab3407c7b35465c8e89b.png)
so 
.
![\[MD \cdot \frac{BC}2 = MA \cdot MP = MQ \cdot \frac{BC}2 \implies \boxed{MD = MQ}.\]](http://latex.artofproblemsolving.com/3/3/b/33b005d8cb1288953bfe561f1cfa6c89a93d375e.png)
In particular, 
.
is cyclic. Applying PoP on M with respect to the circumcenter to 
.
,
.
,
.
,
are similar. Then 
,
as desired.
.
.
,![\[AP^2 + PC^2 = AC^2 \iff (1 - \lambda)^2(p^2 + q^2) + (1 - \lambda x)^2 + \lambda^2q^2 = (p - 1)^2 + q^2.\]](http://latex.artofproblemsolving.com/7/3/7/7372f7abed64b6edf582e7a5d881bd7b1afd04fc.png)
This quadratic in 
has a solution 
and leading coefficient 
and constant term 
,
.
,
be the coordinates of the circumcenter of 
and ![\[y - \frac q2\left( 1 + \frac p{p^2 + q^2} \right) = -\frac pq\left( x - \frac p2\left( 1 + \frac p{p^2 + q^2} \right) \right)\]](http://latex.artofproblemsolving.com/8/0/5/805c78f99d4705f793d0c8a1ff30ed7fe4cc1645.png)
and![\[y - \frac q2 = -\frac{p + 1}q\left( x - \frac{p - 1}2 \right).\]](http://latex.artofproblemsolving.com/3/4/f/34f6664230941dc85c53e2edd16294f630ec433e.png)
![\[(x,y) = \left( -\frac{p + 1}2,\frac{p(p + 1)}{2q} + \frac q2 \right)\]](http://latex.artofproblemsolving.com/f/3/c/f3c52e08169e04292b0c93af420a406225296681.png)
is a solution. So the 
-
,
,
,
,
is 
and 
be the foot of the perpendicular from 
is cyclic. By power of point on 
,
But 
,
.
.
,
,
.
