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 
be the midpoint 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 
,
.
.
,
,
.
