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CHKMO 2017 Q3
noobatron3000   7
N 11 minutes ago by Entei
Source: CHKMO
Let ABC be an acute-angled triangle. Let D be a point on the segment BC, I the incentre of ABC. The circumcircle of ABD meets BI at P and the circumcircle of ACD meets CI at Q. If the area of PID and the area of QID are equal, prove that PI*QD=QI*PD.
7 replies
noobatron3000
Dec 31, 2016
Entei
11 minutes ago
Geometry
Jackson0423   1
N 12 minutes ago by ricarlos
Source: Own
In triangle ABC with circumcenter O, if the intersection point of lines BO and AC is N, then BO = 2ON, and BMN = 122 degrees with respect to the midpoint M of AB. Find MNB.
1 reply
Jackson0423
Yesterday at 4:40 PM
ricarlos
12 minutes ago
Not so classic orthocenter problem
m4thbl3nd3r   4
N 29 minutes ago by hanzo.ei
Source: own?
Let $O$ be circumcenter of a non-isosceles triangle $ABC$ and $H$ be a point in the interior of $\triangle ABC$. Let $E,F$ be foots of perpendicular lines from $H$ to $AC,AB$. Suppose that $BCEF$ is cyclic and $M$ is the circumcenter of $BCEF$, $HM\cap AB=K,AO\cap BE=T$. Prove that $KT$ bisects $EF$
4 replies
m4thbl3nd3r
Yesterday at 4:59 PM
hanzo.ei
29 minutes ago
Numbers not power of 5
Kayak   33
N 38 minutes ago by ihategeo_1969
Source: Indian TST D1 P2
Show that there do not exist natural numbers $a_1, a_2, \dots, a_{2018}$ such that the numbers \[ (a_1)^{2018}+a_2, (a_2)^{2018}+a_3, \dots, (a_{2018})^{2018}+a_1 \]are all powers of $5$

Proposed by Tejaswi Navilarekallu
33 replies
1 viewing
Kayak
Jul 17, 2019
ihategeo_1969
38 minutes ago
Chile TST IMO prime geo
vicentev   4
N 39 minutes ago by Retemoeg
Source: TST IMO CHILE 2025
Let \( ABC \) be a triangle with \( AB < AC \). Let \( M \) be the midpoint of \( AC \), and let \( D \) be a point on segment \( AC \) such that \( DB = DC \). Let \( E \) be the point of intersection, different from \( B \), of the circumcircle of triangle \( ABM \) and line \( BD \). Define \( P \) and \( Q \) as the points of intersection of line \( BC \) with \( EM \) and \( AE \), respectively. Prove that \( P \) is the midpoint of \( BQ \).
4 replies
1 viewing
vicentev
Today at 2:35 AM
Retemoeg
39 minutes ago
Cute orthocenter geometry
MarkBcc168   77
N 42 minutes ago by ErTeeEs06
Source: ELMO 2020 P4
Let acute scalene triangle $ABC$ have orthocenter $H$ and altitude $AD$ with $D$ on side $BC$. Let $M$ be the midpoint of side $BC$, and let $D'$ be the reflection of $D$ over $M$. Let $P$ be a point on line $D'H$ such that lines $AP$ and $BC$ are parallel, and let the circumcircles of $\triangle AHP$ and $\triangle BHC$ meet again at $G \neq H$. Prove that $\angle MHG = 90^\circ$.

Proposed by Daniel Hu.
77 replies
MarkBcc168
Jul 28, 2020
ErTeeEs06
42 minutes ago
A weird inequality
Eeightqx   0
an hour ago
For all $a,\,b,\,c>0$, find the maximum $\lambda$ which satisfies
$$\sum_{cyc}a^2(a-2b)(a-\lambda b)\ge 0.$$hint
0 replies
Eeightqx
an hour ago
0 replies
Student's domination
Entei   0
an hour ago
Given $n$ students and their test results on $k$ different subjects, we say that student $A$ dominates student $B$ if and only if $A$ outperforms $B$ on all subjects. Assume that no two of them have the same score on the same subject, find the probability that there exists a pair of domination in class.
0 replies
Entei
an hour ago
0 replies
The Curious Equation for ConoSur
vicentev   3
N an hour ago by AshAuktober
Source: TST IMO-CONO CHILE 2025
Find all triples \( (x, y, z) \) of positive integers that satisfy the equation
\[
x + xy + xyz = 31.
\]
3 replies
vicentev
2 hours ago
AshAuktober
an hour ago
You just need to throw facts
vicentev   3
N an hour ago by MathSaiyan
Source: TST IMO CHILE 2025
Let \( a, b, c, d \) be real numbers such that \( abcd = 1 \), and
\[
a + \frac{1}{a} + b + \frac{1}{b} + c + \frac{1}{c} + d + \frac{1}{d} = 0.
\]Prove that one of the numbers \( ab, ac \) or \( ad \) is equal to \( -1 \).
3 replies
1 viewing
vicentev
an hour ago
MathSaiyan
an hour ago
A number theory problem from the British Math Olympiad
Rainbow1971   5
N an hour ago by Rainbow1971
Source: British Math Olympiad, 2006/2007, round 1, problem 6
I am a little surprised to find that I am (so far) unable to solve this little problem:

[quote]Let $n$ be an integer. Show that, if $2 + 2 \sqrt{1+12n^2}$ is an integer, then it is a perfect square.[/quote]

I set $k := \sqrt{1+12n^2}$. If $2 + 2 \sqrt{1+12n^2}$ is an integer, then $k (=\sqrt{1+12n^2})$ is at least rational, so that $1 + 12n^2$ must be a perfect square then. Using Conway's topograph method, I have found out that the smallest non-negative pairs $(n, k)$ for which this happens are $(0,1), (2,7), (28,97)$ and $(390, 1351)$, and that, for every such pair $(n,k)$, the "next" such pair can be calculated as
$$
\begin{bmatrix}
7 & 2 \\
24 & 7 
\end{bmatrix} \begin{bmatrix}
n \\
k 
\end{bmatrix}
.$$The eigenvalues of that matrix are irrational, however, so that any calculation which uses powers of that matrix is a little cumbersome. There must be an easier way, but I cannot find it. Can you?

Thank you.




5 replies
Rainbow1971
Yesterday at 8:39 PM
Rainbow1971
an hour ago
Finding maximum sum of consecutive ten numbers in circle.
Goutham   3
N an hour ago by FarrukhKhayitboyev
Each of $999$ numbers placed in a circular way is either $1$ or $-1$. (Both values appear). Consider the total sum of the products of every $10$ consecutive numbers.
$(a)$ Find the minimal possible value of this sum.
$(b)$ Find the maximal possible value of this sum.
3 replies
Goutham
Feb 8, 2011
FarrukhKhayitboyev
an hour ago
construction of line knowing newton line direction
PROF65   7
N Jan 7, 2023 by ancamagelqueme
Source: own?
Let $ABC$ be a triangle and $\ell ,d \   $ be two lines .
Construct a line $DE$ s.t. $D\in AB,E\in AC ,DE\parallel d $ and the newton line of $ DBCE$ is parallel to $\ell$.
RH HAS
Best regards.
7 replies
PROF65
Jan 4, 2023
ancamagelqueme
Jan 7, 2023
construction of line knowing newton line direction
G H J
G H BBookmark kLocked kLocked NReply
Source: own?
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PROF65
2016 posts
#1 • 1 Y
Y by GeoKing
Let $ABC$ be a triangle and $\ell ,d \   $ be two lines .
Construct a line $DE$ s.t. $D\in AB,E\in AC ,DE\parallel d $ and the newton line of $ DBCE$ is parallel to $\ell$.
RH HAS
Best regards.
This post has been edited 2 times. Last edited by PROF65, Jan 5, 2023, 1:20 AM
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youthdoo
1312 posts
#2
Y by
What is a construction you mean?
Compasses and straightedge?
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PROF65
2016 posts
#3
Y by
youthdoo wrote:
What is a construction you mean?
Compasses and straightedge?

exactly
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PROF65
2016 posts
#4
Y by
any idea
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youthdoo
1312 posts
#5
Y by
Do we just have to show that it is possible?
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GeoKing
515 posts
#6 • 1 Y
Y by PROF65
Sol:- Construct $H$ the orthocenter of $ABC$. Construct line $\ell'$ through $H$ perpendicular to $\ell$. Construct the line through $A$ perpendicular to $d$ and let it intersect $\ell'$ at $H'$. Let the line though $H'$ perpendicular to $AC,AB$ meet $AB,AC$ respectively at $ D,E$ . These are the required points.
Note that the construction works because steiner line and newton line are perpendicular and using this we constructed $H'$ the orthocenter of $ADE$ and using that we finally constructed $D,E$
This post has been edited 1 time. Last edited by GeoKing, Jan 6, 2023, 4:34 AM
Reason: latex
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PROF65
2016 posts
#7 • 1 Y
Y by GeoKing
another way
the isogonal of $M$ ,the miquel point of $ BCED$ is the infinity point of the newton line so we can construct $ M$ then we draw a line parallel to $d$ through $C$; it hits $(ABC)$ at another point say $G$ let the line $MG$ cuts $AB $ at point that is exactly $D$ . Finally we draw a parallel to $d$ to construct $E$.
RH HAS
Best regards.
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ancamagelqueme
104 posts
#8
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Quote:
Given three lines $a, b, c$ and two vectors $\vec{u}, \vec{v}$, construct a line $d$, with direction $u$, such that Newton line of the quadrilateral formed by the four lines $a, b, c, d$ has direction $\vec{v}$.

The three lines $a, b, c$ form a triangle $ABC$. Let ${\mathcal P}$ be the parabola inscribed in its medial triangle and whose axis has the direction of the vector $\vec{u}$. Let $d$ be the conjugate diameter of the direction of the vector $\vec{v}$. ${\mathcal P}$ and $d$ intersects at a point $T_{uv}$. The tangent at $T_{uv}$ to ${\mathcal P}$ is Newton's line of the quadrilateral $abcd$, which has the direction of the vector $\vec{v}$.

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