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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

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0 replies
jlacosta
May 1, 2025
0 replies
Need help with barycentric
Sadigly   0
13 minutes ago
Hi,is there a good handout/book that explains barycentric,other than EGMO?
0 replies
Sadigly
13 minutes ago
0 replies
Combinatorics
P162008   3
N 36 minutes ago by P162008
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
3 replies
P162008
4 hours ago
P162008
36 minutes ago
Find min and max
lgx57   0
an hour ago
Source: Own
$x_1,x_2, \cdots ,x_n\ge 0$,$\displaystyle\sum_{i=1}^n x_i=m$. $k_1,k_2,\cdots,k_n >0$. Find min and max of
$$\sum_{i=1}^n(k_i\prod_{j=1}^i x_j)$$
0 replies
lgx57
an hour ago
0 replies
Find min
lgx57   0
an hour ago
Source: Own
$a,b>0$, $a^4+ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
$a,b>0$, $a^4-ab+b^4=60$. Find min of
$$4a^2-ab+4b^2$$
0 replies
lgx57
an hour ago
0 replies
No more topics!
triangleABC BC=1/2(AB+AC)
RaleD   11
N Aug 19, 2022 by franzliszt
Source: Bosnia and Herzegovina 2011
In triangle $ABC$ it holds $|BC|= \frac{1}{2}(|AB|+|AC|)$. Let $M$ and $N$ be midpoints of $AB$ and $AC$, and let $I$ be the incenter of $ABC$. Prove that $A, M, I, N$ are concyclic.
11 replies
RaleD
May 16, 2011
franzliszt
Aug 19, 2022
triangleABC BC=1/2(AB+AC)
G H J
Source: Bosnia and Herzegovina 2011
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RaleD
118 posts
#1 • 2 Y
Y by Adventure10, Mango247
In triangle $ABC$ it holds $|BC|= \frac{1}{2}(|AB|+|AC|)$. Let $M$ and $N$ be midpoints of $AB$ and $AC$, and let $I$ be the incenter of $ABC$. Prove that $A, M, I, N$ are concyclic.
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Number1
355 posts
#2 • 3 Y
Y by Durjoy1729, Adventure10, Mango247
Let $S$ be circumcenter of triangle $ABC$ and let angle bisector at $A$ cut circumcircle of triangle $ABC$ at $A'$.
Since $A,M,N$ and $S$ are concyclic with $\angle MSA = 90^{\circ}$ it is enought to prove that $I$ bisect $AA'$.
But this could be easly proved by Ptolomy:

$AA'\cdot BC = AB\cdot A'C+AC\cdot A'B$

Since $A'I = A'B =A'C$ this follows. :)
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hatchguy
555 posts
#3 • 1 Y
Y by Adventure10
From $BC= \frac{AB+AC}{2}$ we have that $BC = BM+CN$. Let $E$ be the point in $BC$ such that $BM= CE$ and $CN = CE$.

Since $BI$ and $CI$ are angle bisectors of angles $\angle ABC$ and $\angle ACB$, respectively, and triangles $BME$ and $CNE$ are isosceles we have that $BI$ and $CI$ are the perpendicular bisectors of $ME$ and $NE$, respectively. Hence $I$ is the circumcenter of triangle $MEN$.

Therefore, since $I$ lies on the perpendicular bisector of $MN$ and on the angle bisector of $\angle MAN$ we have that $AMIN$ is cyclic.
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jgnr
1343 posts
#4 • 2 Y
Y by Adventure10, Mango247
Let AI meets BC at D and the circumcircle of ABC at K. We have AB/BD = c/(ac/(b+c)) = 2. Since ABD is similar to AKC, then AK=2KC. This also implies AK=2KI. The homothety centered at A of ratio 1/2 brings K,B,C to I,M,N, and also the circumcircle of ABC to the circle passing through A,M,I,N, so they are concyclic, as desired.
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master_Hjom
174 posts
#5 • 2 Y
Y by Adventure10, Mango247
Btw, I had this problem on my geometry test..
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arshakus
769 posts
#6 • 2 Y
Y by Adventure10, Mango247
master_Hjom wrote:
Btw, I had this problem on my geometry test..
We know that $AIXZ$ is cyclic, where $X,Z$ are the points of perpendiculars from $I$ to $AB$ and $AC$ respectively. So we need to prove is $\angle XIM=\angle ZIN$. From the condition $2BC=AB+AC$ we get that $XM=ZN$. Thus $\triangle INZ= \triangle IMX$. So we get that $\angle XIM=\angle ZIN$, from which follows that $AMIN$ is cyclic.
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Babai
488 posts
#7 • 2 Y
Y by Adventure10, Mango247
The problem is equivalent to prove that $IM=IN$. i.e.$IM^2=IN^2$
Use barycentric coordinate and if we compare it remains to prove that $S_B(s-2b)=S_C(s-2c)$ where $S_B=\frac{1}{2}(a^2+c^2-b^2)$ and like wise $S_C$.
If we calculate $\frac{S_B}{S_C}=\frac{5c-3b}{5b-3c}$ which is nothing but $\frac{s-2c}{s-2b}$.
[While calculating we use the fact $a=\frac{b+c}{2}$]
Hence proved.
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Gomes17
132 posts
#8 • 2 Y
Y by Projeto_Ramanujan, Adventure10
Open the gates
This post has been edited 1 time. Last edited by Gomes17, Aug 27, 2017, 11:54 PM
Reason: typo
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vsathiam
201 posts
#9 • 2 Y
Y by Adventure10, Mango247
Let line AI intersect (ABC) at A and P. It follows that BP = CP = IP = r by the incenter excenter lemma.

Applying a homothety about A with scale factor 2 maps M to B and N to C. If AMIN were cyclic, the homothety would also map I to P. So it is sufficient to prove that AI = IP, or AP = 2r.

Let AB = c, AC = b, and BC = a. The problem's condition tells us that 2a = b+c. Applying Ptolemy's on ABPC gives:

$$AB \cdot PC + BP \cdot AC = AP \cdot BC \rightarrow cr+br=a \cdot AN \rightarrow AN = 2r$$as desired.
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jayme
9792 posts
#10 • 1 Y
Y by Adventure10
Dear Mathlinkers,
I being the midpoint of AA'...we are done by homothety...

Sincerely
Jean-Louis
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Taco12
1757 posts
#11 • 2 Y
Y by RedFireTruck, megarnie
MAN IS MANI
We use barycentric coordinates, with $ABC$ as the reference triangle. We have $M = \left(\frac{1}{2}, \frac{1}{2}, 0\right), N = \left(\frac{1}{2}, 0, \frac{1}{2}\right), I = (a:b:c)$. Consider the circumcircle of $AMN$. We must show that $I$ also lies on this circle. Plugging in each of the points $A, M, N$ into the circle formula, we get $u=0, w=\frac{b^2}{2}, v=\frac{c^2}{2}$. The equation of $(AMN)$ is $$-a^2yz-b^2zx-c^2xy+\left(\frac{c^2y+b^2z}{2}\right)(x+y+z)=0.$$Now, plugging in $(a:b:c)$, we have
\begin{align*}
-a^2bc-ab^2c-abc^2+\left(\frac{bc^2+b^2c}{2}\right)(x+y+z)=0 \\
abc(a+b+c)=\frac{(bc^2+b^2c)(a+b+c)}{2} \\
\frac{bc^2+b^2c}{2}=abc \\
a=\frac{1}{2}(b+c), \\
\end{align*}which is clear from the problem statement, so we are done. $\blacksquare$
This post has been edited 1 time. Last edited by Taco12, Jul 19, 2022, 8:17 PM
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franzliszt
23531 posts
#12 • 2 Y
Y by Mango247, Mango247
Literally same as @above, but posting for storage.

Clearly $M,N$, the midpoints of $AB,AC,$ in addition to the circumcenter $O$ and $A$ are concyclic with diameter $OA.$ Thus, it suffices to show that $I$ lies on $(AMN)$.


The general equation of a circle is $$-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$$for some constants $u,v,w$. To find the constants $u,v,w$ which determine $\Omega$, we can plug in each of $A,M,N$ in the general form and solve for $u,v,w$. Note that the equation of a circle is homogenous so we can use homogenized coordinates here.
  • Plugging in $A=(1,0,0)$ gives $u=0$.
  • Plugging in $M=(1:0:1)$ gives $u+w=\frac{b^2}2$.
  • Plugging in $N=(1:1:0)$ gives $u+v=\frac{c^2}2$.

Plugging in these constants in the general equation, we find that $\Omega$ has equation $$-a^2yz-b^2zx-c^2xy+(x+y+z)\left(\frac{c^2}2\cdot y+\frac{b^2}2\cdot z\right)=0.$$Hence, using Barycentric Power of a Point, we can compute $$\text{Pow}_{\Omega}(I)=-(a+b+c)(abc)+(a+b+c)\left(\frac{c^2}2\cdot b+\frac{b^2}2\cdot c\right)$$since $I=(a:b:c)$. But this is $0$ since $a=\frac{b+c}2$. Done.
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