[/quote]
,
be six positive real numbers. Prove that
cards on a table, flipped face down. Madina knows that on each card a single number is written and that the numbers are different integers from 
to 
cards, and she is told a number that is written on one of the chosen cards, but not which specific card it is on. After several moves, Madina must determine the written numbers on as many cards as possible. What is the maximum number of cards Madina can ensure to determine?
cube is given, and a 
grid of square unit cells is drawn on each of its six faces. A beam is a 
rectangular prism. Several beams are placed inside the cube subject to the following conditions:
faces of each beam coincide with unit cells lying on opposite faces of the cube. (Hence, there are 
possible positions for a beam.)
faces of each beam touch either a face of the cube or the interior of the face of another beam.
be nonzero real numbers such that 
.
Show that for any odd integer 
,
prove that the triangle must have a right angle.
denote the set of natural numbers, and let 
,
,
.
whose product is a perfect cube.
and 
be prime numbers. Prove that if 
is a perfect square, then 
is also a perfect square.
points in 
can be partitioned into two sets whose convex hulls intersect.
and 
be the unit circle in the complex plane. Let 
be the map given by 
.
and 
for 
.
such that 
is called the period of 
.
of period 
.
)
and 
are two sequences between 
and they satisfy 
,
,
Y by Billybillybobjoejr., mathmax12, Johnson100, OronSH, guineapig2013, Rounak_iitr, ItsBesi, PikaPika999
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.
.
be the altitude. 
is cyclic and 
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 
.
all lie on 
,
.
,
.
,
are similar. Then 
,
and 
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 
,
.
.
,
,
.
.
by radical axis. Then, 
so 
is cyclic, now by POP:- 
,
is perpendicular bisector of 
.
