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4 lines concurrent
Zavyk09   4
N an hour ago by pingupignu
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
4 replies
Zavyk09
Yesterday at 11:51 AM
pingupignu
an hour ago
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Two circles and Three line concurrency
mofidy   0
an hour ago
Two circles $W_1$ and $W_2$ with equal radii intersect at P and Q. Points B and C are located on the circles$W_1$ and $W_2$ so that they are inside the circles $W_2$ and $W_1$, respectively. Also, points X and Y distinct from P are located on $W_1$ and $W_2$, respectively, so that:
$$\angle{CPQ} = \angle{CXQ} \text{ and } \angle{BPQ} = \angle{BYQ}.$$The intersection point of the circumcircles of triangles XPC and YPB is called S. Prove that BC, XY and QS are concurrent.
Thanks.
0 replies
mofidy
an hour ago
0 replies
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Tilted Students Thoroughly Splash Tiger part 2
DottedCaculator   17
N 2 hours ago by HoRI_DA_GRe8
Source: ELMO 2024/5
In triangle $ABC$ with $AB<AC$ and $AB+AC=2BC$, let $M$ be the midpoint of $\overline{BC}$. Choose point $P$ on the extension of $\overline{BA}$ past $A$ and point $Q$ on segment $\overline{AC}$ such that $M$ lies on $\overline{PQ}$. Let $X$ be on the opposite side of $\overline{AB}$ from $C$ such that $\overline{AX} \parallel \overline{BC}$ and $AX=AP=AQ$. Let $\overline{BX}$ intersect the circumcircle of $BMQ$ again at $Y \neq B$, and let $\overline{CX}$ intersect the circumcircle of $CMP$ again at $Z \neq C$. Prove that $A$, $Y$, and $Z$ are collinear.

Tiger Zhang
17 replies
DottedCaculator
Jun 21, 2024
HoRI_DA_GRe8
2 hours ago
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radii relationship
steveshaff   0
2 hours ago
Two externally tangent circles with radii a and b are each internally tangent to a semicircle and its diameter. The two points of tangency on the semicircle and the two points of tangency on its diameter lie on a circle of radius r. Prove that r^2 = 3ab.
0 replies
steveshaff
2 hours ago
0 replies
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NT Function with divisibility
oVlad   3
N 2 hours ago by sangsidhya
Source: Romanian District Olympiad 2023 9.4
Determine all strictly increasing functions $f:\mathbb{N}_0\to\mathbb{N}_0$ which satisfy \[f(x)\cdot f(y)\mid (1+2x)\cdot f(y)+(1+2y)\cdot f(x)\]for all non-negative integers $x{}$ and $y{}$.
3 replies
oVlad
Mar 11, 2023
sangsidhya
2 hours ago
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Minimum with natural numbers
giangtruong13   1
N 2 hours ago by Ianis
Let $x,y,z,t$ be natural numbers such that: $x^2-y^2+t^2=21$ and $x^2+3y^2+4z^2=101$. Find the min: $$M=x^2+y^2+2z^2+t^2$$
1 reply
giangtruong13
3 hours ago
Ianis
2 hours ago
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angle wanted, right ABC, AM=CB , CN=MB
parmenides51   3
N 2 hours ago by Mathzeus1024
Source: 2022 European Math Tournament - Senior First + Grand League - Math Battle 1.3
In a right-angled triangle $ABC$, points $M$ and $N$ are taken on the legs $AB$ and $BC$, respectively, so that $AM=CB$ and $CN=MB$. Find the acute angle between line segments $AN$ and $CM$.
3 replies
parmenides51
Dec 19, 2022
Mathzeus1024
2 hours ago
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polonomials
Ducksohappi   0
2 hours ago
Let $P(x)$ be the real polonomial such that all roots are real and distinct. Prove that there is a rational number $r\ne 0 $ that all roots of $Q(x)=$ $P(x+r)-P(x)$ are real numbers
0 replies
Ducksohappi
2 hours ago
0 replies
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IMO ShortList 2002, algebra problem 4
orl   62
N 2 hours ago by Ihatecombin
Source: IMO ShortList 2002, algebra problem 4
Find all functions $f$ from the reals to the reals such that \[ \left(f(x)+f(z)\right)\left(f(y)+f(t)\right)=f(xy-zt)+f(xt+yz) \] for all real $x,y,z,t$.
62 replies
orl
Sep 28, 2004
Ihatecombin
2 hours ago
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inequality ( 4 var
SunnyEvan   11
N 2 hours ago by SunnyEvan
Let $ a,b,c,d \in R $ , such that $ a+b+c+d=4 . $ Prove that :
$$ a^4+b^4+c^4+d^4+3 \geq \frac{7}{4}(a^3+b^3+c^3+d^3) $$$$ a^4+b^4+c^4+d^4+ \frac{76}{25} \geq \frac{44}{25}(a^3+b^3+c^3+d^3) $$
11 replies
SunnyEvan
Apr 4, 2025
SunnyEvan
2 hours ago
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Simple parallelogram geo
aleksam   11
N Feb 9, 2022 by EulersTurban
Source: IOM 2017 #1
Let $ABCD$ be a parallelogram in which angle at $B$ is obtuse and $AD>AB$. Points $K$ and $L$ on $AC$ such that $\angle ADL=\angle KBA$(the points $A, K, C, L$ are all different, with $K$ between $A$ and $L$). The line $BK$ intersects the circumcircle $\omega$ of $ABC$ at points $B$ and $E$, and the line $EL$ intersects $\omega$ at points $E$ and $F$. Prove that $BF||AC$.
11 replies
aleksam
Sep 5, 2017
EulersTurban
Feb 9, 2022
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Simple parallelogram geo
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Reveal topic tags
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Angle Chase
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Source: IOM 2017 #1
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aleksam
101 posts
aleksam
#1 h  Sep 5, 2017, 1:39 PM • 4 Y
Y by tenplusten, dooo203, Adventure10, Mango247
Let $ABCD$ be a parallelogram in which angle at $B$ is obtuse and $AD>AB$. Points $K$ and $L$ on $AC$ such that $\angle ADL=\angle KBA$(the points $A, K, C, L$ are all different, with $K$ between $A$ and $L$). The line $BK$ intersects the circumcircle $\omega$ of $ABC$ at points $B$ and $E$, and the line $EL$ intersects $\omega$ at points $E$ and $F$. Prove that $BF||AC$.
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rmtf1111
698 posts
rmtf1111
#3 h  Sep 5, 2017, 3:23 PM • 2 Y
Y by Adventure10, Mango247
Note that it is enough to show that $\triangle{ABC} \sim \triangle{CFA}$. Let $L'$ be the isogonal conjugate of $L$ wrt $\triangle{DAC}$. Performing a reflection over the midpoint of segment $AC$ we see that $L'$ is the isotomic conjugate of $K$ wrt $\triangle{ABC}$, thus by Steiner's Theorem we have:
$$\left(\frac{AB}{BC}\right)^2=\left(\frac{DC}{AC}\right)^2=\frac{CL\cdot CL'}{AL \cdot AL'}=\frac{CL\cdot AK}{AL \cdot CK} = (L,K;C,A) \stackrel{E}{=} (F,B;C,A)=\frac{FC\cdot AB}{FA \cdot BC}\implies \frac{FC}{FA}=\frac{AB}{BC} \implies \triangle{ABC} \sim \triangle{CFA} \ \ \ \blacksquare $$
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tenplusten
1000 posts
tenplusten
#4 h  Sep 5, 2017, 5:26 PM • 1 Y
Y by Adventure10
Just intersect $CE $ and $AD $.And see two cyclic quadrilaterals.
Is it really simple?I think it is harder than IMO 2017 P4.Quite hard for p1. (It took me one hour and half of hour)
This post has been edited 1 time. Last edited by tenplusten, Sep 5, 2017, 5:58 PM
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DVDthe1st
341 posts
DVDthe1st
#5 h  Sep 6, 2017, 2:38 AM • 1 Y
Y by Adventure10
Just note that $F$ and $D$ are symmetric about $AC$?
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tenplusten
1000 posts
tenplusten
#6 h  Sep 6, 2017, 5:03 AM • 2 Y
Y by Adventure10, Mango247
DVDthe1st wrote:
Just note that $F$ and $D$ are symmetric about $AC$?

Yeah.Official solution takes the reflection of D wrt AC. Then shows that it is just F.
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madmathlover
145 posts
madmathlover
#7 h  Sep 6, 2017, 2:12 PM • 2 Y
Y by Adventure10, Mango247
Let $F'$ be the reflection of $D$ across $AC$.
$AC$ passes through the midpoints of $DB$ and $DF’$. So, $AC \parallel BF'$.

Also, $AB = CD = CF'$. So, $ABF'C$ is an isosceles trapezoid i.e. $ABF'C$ is cyclic.

Now, $\angle CF'L = \angle CDL = \angle CBK = \angle CBE = \angle CF'E$. As $L$ aand $E$ lies on the same side of $F'$, so, $F', L, E$ are collinear.

So, $F' = F$ and we got the desired result
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anantmudgal09
1979 posts
anantmudgal09
#8 h  Sep 6, 2017, 5:27 PM • 2 Y
Y by Adventure10, Mango247
Let $EV \parallel BC$ with $V$ on $\omega$. Reflect $L$ across mid-point of $AC$ to get $J$; then $BK, BJ$ are isogonal in angle $ABC$. Particularly, we see $B, J, V$ are collinear. Note that lines $BV$ and $EL$ are symmetric about the perpendicular bisector of $AC$. Consequently, $BFVE$ is an isosceles trapezoid, so $BF \parallel EV \parallel AC$.

Remark. Motivated by "removing the floating $D$ from the picture". Also too easy for #1 in my opinion.
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Ferid.---.
1008 posts
Ferid.---.
#9 h  Sep 10, 2017, 12:56 PM • 2 Y
Y by Adventure10, Mango247
Let $AD\cap CE=S.$ and $\angle ABK=x,\angle CBK=y,\angle CAD=z,\angle CAB=t.$
We have $\angle ADL=\angle ABE=\angle ACS=x,$ then $DSCL$ is cyclic.Then $AL\cdot AC=AD\cdot AS,(1)$ since PoP.
Also we can find easily ( with angle )
$\triangle LDC\sim \triangle CBE,(2)$ and $\triangle ABE\sim \triangle LDA.(3)$
Then $$AL\cdot AC=^{(1)} AD\cdot AS=BC\cdot AS=^{(2)} \frac{LD\cdot CE}{LC} \cdot AS=^{(3)} \frac{CE}{LC} \cdot AS\cdot \frac{AB\cdot LA}{AE}$$$\implies \frac{AC}{CB}\cdot \frac{AE}{AS}\cdot \frac{LC}{CE}=1\implies Sin(z+x)\cdot Sin(t)=Sin(x+t)\cdot Sin(z)\implies Cos(z+x-t)=Cos(x+t-z)\implies z=t.$
This means $t=\angle FAC=\angle FBC=z=\angle CAD=\angle ACB\to BF\parallel AC.$
This post has been edited 1 time. Last edited by Ferid.---., Sep 10, 2017, 12:56 PM
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IndoMathXdZ
691 posts
IndoMathXdZ
#10 h  Jul 26, 2019, 3:11 PM • 2 Y
Y by Adventure10, Mango247
Here's my synthetic solution by angle chasing:
Let $BK$ intersects $\omega$ at $E$. Notice that it suffices to prove that $\measuredangle AEF = \measuredangle BEC$.

Claim 01. $\triangle ALD \sim \triangle EAB$.
Proof. Notice that
\[ \measuredangle DAL \equiv \measuredangle DAC = \measuredangle BCA = \measuredangle BEA \]Notice that
\[ \measuredangle LDA = \measuredangle ABK \equiv \measuredangle ABE \]
Claim 02. $\triangle ALE \sim \triangle DLC$
Proof. Notice that by Claim 01, we have
\[ \frac{AL}{AE} = \frac{DL}{AB} = \frac{DL}{DC} \]Using this and $\measuredangle EAL \equiv \measuredangle EAC = \measuredangle EBC \equiv \measuredangle KBC = \measuredangle CDL$, we have what we want.

Now, to finish, notice that
\[ \measuredangle AEF \equiv \measuredangle AEL = \measuredangle DCL \equiv \measuredangle DCA = \measuredangle BAC = \measuredangle BEC \]
This post has been edited 2 times. Last edited by IndoMathXdZ, Jul 26, 2019, 4:04 PM
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Ali3085
214 posts
Ali3085
#11 h  Jul 18, 2020, 5:13 PM
Y by
a nice problem but easy :blush:
let $F'$ be the reflection of $D$ on $AC$
claim(1):$BE'||AC$
proof:
since $F'C=CD=AB$ and $\angle ACF'=\angle ACD=\angle CAB$ so $ABF'C$ is an isoscele trapezoid
$\blacksquare$
claim(2):$F'=F$
proof:
$\angle AF'C = \angle ADC =\angle ABC$ so $F' \in (ABC)$
$\angle AF'L=\angle ADL=\angle ABK=\angle ABE$ so $F',L,E$ are collinear
$\blacksquare$
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HKIS200543
380 posts
HKIS200543
#12 h  Aug 17, 2020, 11:35 AM
Y by
Let $F'$ denote the point on $\omega$ such that $BF' \parallel AC$. A solution purely with complex numbers is not difficult. Here, we just use it to prove the basic fact that $F'$ is the reflection of $D$ over $AC$.

Note that $d = a + c - b$ and $f' = \frac{ac}{b}$. The reflection of $f'$ over $AC$ is given by
\[ a + c -ac \overline{f} = a + c - b = d. \]
Then the angle chase is immediate.
\[ \angle AF'L = \angle ADL - \angle KBA = \angle EBA = \angle EFA \]so $F' = F$.
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EulersTurban
386 posts
EulersTurban
#13 h  Feb 9, 2022, 3:30 PM
Y by
[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -30.522181898533855, xmax = 25.099726037859206, ymin = -5.046513103909196, ymax = 28.21045012698899; /* image dimensions */ /* draw figures */ draw((-11,1)--(-3,12), linewidth(0.4) + red); draw((-3,12)--(6,12), linewidth(0.4) + red); draw((6,12)--(-2,1), linewidth(0.4) + red); draw((-2,1)--(-11,1), linewidth(0.4) + red); draw((-11,1)--(6,12), linewidth(0.4) + red); draw((-6.639215718736752,3.8216839466997485)--(-2,1), linewidth(0.4) + red); draw((-3,12)--(-3.5077820305878022,5.847905744913775), linewidth(0.4) + red); draw(circle((-6.5,12.681818181818182), 12.51858123083674), linewidth(0.4) + linetype("4 4") + blue); draw((-18.942629407869116,11.30491969191832)--(-6.639215718736752,3.8216839466997485), linewidth(0.4) + red); draw((-18.942629407869116,11.30491969191832)--(2.312195121951221,3.790243902439023), linewidth(0.4) + red); draw((-2,1)--(2.312195121951221,3.790243902439023), linewidth(0.4) + red); draw((2.312195121951221,3.790243902439023)--(6,12), linewidth(0.4) + red); draw((-18.942629407869116,11.30491969191832)--(-11,1), linewidth(0.4) + red); draw((-11,1)--(2.312195121951221,3.790243902439023), linewidth(0.4) + red); /* dots and labels */ dot((-11,1),dotstyle); label("$A$", (-10.843935095971819,1.379777826078773), NE * labelscalefactor); dot((-2,1),dotstyle); label("$B$", (-1.8398664483050566,1.379777826078773), NE * labelscalefactor); dot((6,12),dotstyle); label("$C$", (6.147613803657394,12.380716536736143), NE * labelscalefactor); dot((-3,12),linewidth(4pt) + dotstyle); label("$D$", (-2.8564548440093684,12.308103079900121), NE * labelscalefactor); dot((-6.639215718736752,3.8216839466997485),dotstyle); label("$K$", (-6.487127685810482,4.175395914265629), NE * labelscalefactor); dot((-3.5077820305878022,5.847905744913775),linewidth(4pt) + dotstyle); label("$L$", (-3.3647490418615242,6.13595924883823), NE * labelscalefactor); dot((-18.942629407869116,11.30491969191832),linewidth(4pt) + dotstyle); label("$E$", (-18.795108619516256,11.581968511539898), NE * labelscalefactor); dot((2.312195121951221,3.790243902439023),linewidth(4pt) + dotstyle); label("$F$", (2.4443275050202575,4.0664757290115965), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Too easy geo :D

Rephrase the problem in terms of $F$.
Restated problem wrote:
Let $F$ be a point on circle $\omega$, such that $BF \parallel AC$, on line $AC$ choose $L$, let $E$ be the second intersection of $FL$ with $\omega$.Then let $K$ be the intersection of $AC$ with $BE$. Prove that $\angle KBA = \angle ADL$

Notice that $\angle DAL = \angle DAC = \angle ACB = \angle AEB$, thus we need to show that $\angle EAB = \angle ALD$, but this holds since :
$$\angle EAB = 180 - \angle EFB = 180 - \angle LFB = 180 - \angle FLC = 180 - \angle CLD = \angle ALD$$thus we have that $EAB \sim ALD$, this implies that $\angle KBA = \angle ADL$, hence we are done :D
This post has been edited 1 time. Last edited by EulersTurban, Feb 9, 2022, 3:31 PM
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