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IMO 2014 Problem 4
ipaper   169
N an hour ago by YaoAOPS
Let $P$ and $Q$ be on segment $BC$ of an acute triangle $ABC$ such that $\angle PAB=\angle BCA$ and $\angle CAQ=\angle ABC$. Let $M$ and $N$ be the points on $AP$ and $AQ$, respectively, such that $P$ is the midpoint of $AM$ and $Q$ is the midpoint of $AN$. Prove that the intersection of $BM$ and $CN$ is on the circumference of triangle $ABC$.

Proposed by Giorgi Arabidze, Georgia.
169 replies
ipaper
Jul 9, 2014
YaoAOPS
an hour ago
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Tangents forms triangle with two times less area
NO_SQUARES   1
N 2 hours ago by Luis González
Source: Kvant 2025 no. 2 M2831
Let $DEF$ be triangle, inscribed in parabola. Tangents in points $D,E,F$ forms triangle $ABC$. Prove that $S_{DEF}=2S_{ABC}$. ($S_T$ is area of triangle $T$).
From F.S.Macaulay's book «Geometrical Conics», suggested by M. Panov
1 reply
NO_SQUARES
Today at 9:08 AM
Luis González
2 hours ago
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FE solution too simple?
Yiyj1   9
N 2 hours ago by jasperE3
Source: 101 Algebra Problems from the AMSP
Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that the equality $$f(f(x)+y) = f(x^2-y)+4f(x)y$$holds for all pairs of real numbers $(x,y)$.

My solution

I feel like my solution is too simple. Is there something I did wrong or something I missed?
9 replies
Yiyj1
Apr 9, 2025
jasperE3
2 hours ago
J
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interesting function equation (fe) in IR
skellyrah   2
N 2 hours ago by jasperE3
Source: mine
find all function F: IR->IR such that $$ xf(f(y)) + yf(f(x)) = f(xf(y)) + f(xy) $$
2 replies
skellyrah
Today at 9:51 AM
jasperE3
2 hours ago
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Complicated FE
XAN4   1
N 2 hours ago by jasperE3
Source: own
Find all solutions for the functional equation $f(xyz)+\sum_{cyc}f(\frac{yz}x)=f(x)\cdot f(y)\cdot f(z)$, in which $f$: $\mathbb R^+\rightarrow\mathbb R^+$
Note: the solution is actually quite obvious - $f(x)=x^n+\frac1{x^n}$, but the proof is important.
Note 2: it is likely that the result can be generalized into a more advanced questions, potentially involving more bash.
1 reply
XAN4
Today at 11:53 AM
jasperE3
2 hours ago
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Find all sequences satisfying two conditions
orl   34
N 2 hours ago by YaoAOPS
Source: IMO Shortlist 2007, C1, AIMO 2008, TST 1, P1
Let $ n > 1$ be an integer. Find all sequences $ a_1, a_2, \ldots a_{n^2 + n}$ satisfying the following conditions:
\[ \text{ (a) } a_i \in \left\{0,1\right\} \text{ for all } 1 \leq i \leq n^2 + n; \]

\[ \text{ (b) } a_{i + 1} + a_{i + 2} + \ldots + a_{i + n} < a_{i + n + 1} + a_{i + n + 2} + \ldots + a_{i + 2n} \text{ for all } 0 \leq i \leq n^2 - n. \]
Author: Dusan Dukic, Serbia
34 replies
orl
Jul 13, 2008
YaoAOPS
2 hours ago
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IMO Shortlist 2011, G4
WakeUp   125
N 2 hours ago by Davdav1232
Source: IMO Shortlist 2011, G4
Let $ABC$ be an acute triangle with circumcircle $\Omega$. Let $B_0$ be the midpoint of $AC$ and let $C_0$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$ and let $G$ be the centroid of the triangle $ABC$. Let $\omega$ be a circle through $B_0$ and $C_0$ that is tangent to the circle $\Omega$ at a point $X\not= A$. Prove that the points $D,G$ and $X$ are collinear.

Proposed by Ismail Isaev and Mikhail Isaev, Russia
125 replies
WakeUp
Jul 13, 2012
Davdav1232
2 hours ago
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Z[x], P(\sqrt[3]5+\sqrt[3]25)=5+\sqrt[3]5
jasperE3   5
N 3 hours ago by Assassino9931
Source: VJIMC 2013 2.3
Prove that there is no polynomial $P$ with integer coefficients such that $P\left(\sqrt[3]5+\sqrt[3]{25}\right)=5+\sqrt[3]5$.
5 replies
jasperE3
May 31, 2021
Assassino9931
3 hours ago
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IMO problem 1
iandrei   77
N 3 hours ago by YaoAOPS
Source: IMO ShortList 2003, combinatorics problem 1
Let $A$ be a $101$-element subset of the set $S=\{1,2,\ldots,1000000\}$. Prove that there exist numbers $t_1$, $t_2, \ldots, t_{100}$ in $S$ such that the sets \[ A_j=\{x+t_j\mid x\in A\},\qquad j=1,2,\ldots,100 \] are pairwise disjoint.
77 replies
iandrei
Jul 14, 2003
YaoAOPS
3 hours ago
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Divisibility on 101 integers
BR1F1SZ   4
N 3 hours ago by BR1F1SZ
Source: Argentina Cono Sur TST 2024 P2
There are $101$ positive integers $a_1, a_2, \ldots, a_{101}$ such that for every index $i$, with $1 \leqslant i \leqslant 101$, $a_i+1$ is a multiple of $a_{i+1}$. Determine the greatest possible value of the largest of the $101$ numbers.
4 replies
BR1F1SZ
Aug 9, 2024
BR1F1SZ
3 hours ago
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2^x+3^x = yx^2
truongphatt2668   2
N 3 hours ago by CM1910
Prove that the following equation has infinite integer solutions:
$$2^x+3^x = yx^2$$
2 replies
truongphatt2668
Yesterday at 3:38 PM
CM1910
3 hours ago
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Prove perpendicular
shobber   29
N 3 hours ago by zuat.e
Source: APMO 2000
Let $ABC$ be a triangle. Let $M$ and $N$ be the points in which the median and the angle bisector, respectively, at $A$ meet the side $BC$. Let $Q$ and $P$ be the points in which the perpendicular at $N$ to $NA$ meets $MA$ and $BA$, respectively. And $O$ the point in which the perpendicular at $P$ to $BA$ meets $AN$ produced.

Prove that $QO$ is perpendicular to $BC$.
29 replies
shobber
Apr 1, 2006
zuat.e
3 hours ago
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Two Orthocenters and an Invariant Point
Mathdreams   2
N Apr 6, 2025 by hukilau17
Source: 2025 Nepal Mock TST Day 1 Problem 3
Let $\triangle{ABC}$ be a triangle, and let $P$ be an arbitrary point on line $AO$, where $O$ is the circumcenter of $\triangle{ABC}$. Define $H_1$ and $H_2$ as the orthocenters of triangles $\triangle{APB}$ and $\triangle{APC}$. Prove that $H_1H_2$ passes through a fixed point which is independent of the choice of $P$.

(Kritesh Dhakal, Nepal)
2 replies
Mathdreams
Apr 6, 2025
hukilau17
Apr 6, 2025
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Two Orthocenters and an Invariant Point
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Reveal topic tags
geometry
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G H BBookmark kLocked kLocked NReply
Source: 2025 Nepal Mock TST Day 1 Problem 3
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Mathdreams
1466 posts
Mathdreams
#1 h  Apr 6, 2025, 1:30 PM • 3 Y
Y by RANDOM__USER, PikaPika999, khan.academy
Let $\triangle{ABC}$ be a triangle, and let $P$ be an arbitrary point on line $AO$, where $O$ is the circumcenter of $\triangle{ABC}$. Define $H_1$ and $H_2$ as the orthocenters of triangles $\triangle{APB}$ and $\triangle{APC}$. Prove that $H_1H_2$ passes through a fixed point which is independent of the choice of $P$.

(Kritesh Dhakal, Nepal)
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RANDOM__USER
7 posts
RANDOM__USER
#2 h  Apr 6, 2025, 2:14 PM • 1 Y
Y by PikaPika999
Let us linearly move \(P\) on the line \(AO\), then \(H_1\) and \(H_2\) move linearly on the perpendicular lines from \(B\) and \(C\) onto \(AO\) respectively. Additionally the lines on which \(H_1\) and \(H_2\) move are parallel to each other, because the perpendicular lines from \(B\) and \(C\) onto \(AO\) are parallel.

Claim: If \(D\) is the base of the altitude from \(A\) onto \(BC\), then \(D \in H_1H_2\).
Proof: Because of the properties stated above, we only need to verify this statement for two positions of \(P\) to prove it holds for a general position of \(P\). When \(P\) is the antipode of \(A\) in \((ABC)\) the statement is more than trivial, and when \(P \in BC\), \(H_1\) and \(H_2\) lie on \(AD\).

Thus \(H_1H_2\) always passes through \(D\). \(\blacksquare\)
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hukilau17
283 posts
hukilau17
#3 h  Apr 6, 2025, 7:05 PM • 1 Y
Y by PikaPika999
Complex bash with $\triangle ABC$ inscribed in the unit circle, so that
$$|a|=|b|=|c|=1$$$$o=0$$$$p = \lambda a$$for some real number $\lambda$. Then the circumcenter of $\triangle APB$ has coordinate
$$o_1 = \frac{ab(p\overline{p}-1)}{p-a-b+ab\overline{p}} = \frac{ab(1+\lambda)}{a+b}$$and so its orthocenter has coordinate
$$h_1 = a + b + p - 2o_1 = \frac{a^2+b^2+\lambda a^2-\lambda ab}{a+b}$$Similarly,
$$h_2 = \frac{a^2+c^2+\lambda a^2-\lambda ac}{a+c}$$To guess the fixed point, we pick some convenient values of $\lambda$. When $\lambda = -1$, we have $h_1 = b$ and $h_2 = c$. When $\lambda = 0$, we have $h_1 = \frac{a^2+b^2}{a+b}$ and $h_2 = \frac{a^2+c^2}{a+c}$. Thus, we let $Q$ be the intersection of these two lines. Since $Q$ is on line $BC$, we have
$$\overline{q} = \frac{b+c-q}{bc}$$Since $Q$ lies on the line through $\frac{a^2+b^2}{a+b}$ and $\frac{a^2+c^2}{a+c}$, we have
$$\frac{\frac{a^2+b^2}{a+b} - q}{\frac{a^2+b^2}{a+b} - \frac{a^2+c^2}{a+c}} \in \mathbb{R}$$We set this equal to its conjugate, and use the fact that $\overline{q} = \frac{b+c-q}{bc}$, to get
$$\frac{(a+c)(a^2+b^2-aq-bq)}{(b-c)(-a^2+ab+ac+bc)} = \frac{(a+c)(b^2c-a^2b-ab^2-abc+a^2q+abq)}{a(c-b)(a^2+ab+ac-bc)}$$$$a(a^2+b^2-aq-bq)(a^2+ab+ac-bc) = (-a^2+ab+ac+bc)(a^2b+ab^2+abc-b^2c-a^2q-abq)$$$$q = \frac{a(a^2+b^2)(a^2+ab+ac-bc) - b(a^2+ab+ac-bc)(-a^2+ab+ac+bc)}{a(a+b)(a^2+ab+ac-bc) - a(a+b)(-a^2+ab+ac+bc)}$$We simplify this to
$$q = \frac{(a^2+ab+ac-bc)(a^3+a^2b-abc-b^2c)}{a(a+b)(2a^2-2bc)} = \frac{a^2+ab+ac-bc}{2a}$$So $Q$ is the foot of the altitude from $A$ onto line $BC$. It remains to show that $Q$ always lies on line $H_1H_2$ as $\lambda$ varies. Indeed, we find
$$h_1-q = \frac{2a(a^2+b^2+\lambda a^2-\lambda ab) - (a+b)(a^2+ab+ac-bc)}{2a(a+b)} = \frac{(a-b)(2\lambda a^2+a^2-ab-ac-bc)}{2a(a+b)}$$and similarly
$$h_2-q = \frac{(a-c)(2\lambda a^2+a^2-ab-ac-bc)}{2a(a+c)}$$So the factor involving $\lambda$ cancels out, and
$$\frac{h_1-q}{h_2-q} = \frac{(a-b)(a+c)}{(a-c)(a+b)}$$which is real. $\blacksquare$
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