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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
A tangent problem
hn111009   0
8 minutes ago
Source: Own
Let quadrilateral $ABCD$ with $P$ be the intersection of $AC$ and $BD.$ Let $\odot(APD)$ meet again $\odot(BPC)$ at $Q.$ Called $M$ be the midpoint of $BD.$ Assume that $\angle{DPQ}=\angle{CPM}.$ Prove that $AB$ is the tangent of $\odot(APD)$ and $BC$ is the tangent of $\odot(AQB).$
0 replies
hn111009
8 minutes ago
0 replies
Inspired by Bet667
sqing   4
N 10 minutes ago by ytChen
Source: Own
Let $ a,b $ be a real numbers such that $a^3+kab+b^3\ge a^4+b^4.$Prove that
$$1-\sqrt{k+1} \leq  a+b\leq 1+\sqrt{k+1} $$Where $ k\geq 0. $
4 replies
sqing
Thursday at 1:03 PM
ytChen
10 minutes ago
3-var inequality
sqing   4
N 10 minutes ago by sqing
Source: Own
Let $ a,b>0 $ and $\frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \leq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{1}{a^3+3}+ \frac{1}{b^3+ 3}\leq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
4 replies
sqing
May 7, 2025
sqing
10 minutes ago
Inspired by Kosovo 2010
sqing   2
N 12 minutes ago by sqing
Source: Own
Let $ a,b>0  , a+b\leq k $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq\left(1+\frac{4}{k(k+2)}\right)^2$$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq\left(1+\frac{2}{k+2}\right)^2$$Let $ a,b>0  , a+b\leq 2 $. Prove that
$$\left(1+\frac{1}{a(b+1)}\right)\left(1+\frac{1}{b(a+1)}\right)\geq \frac{9}{4} $$$$\left(1+\frac {a}{b(a+1)}\right)\left(1+\frac {b}{a(b+1)}\right) \geq \frac{9}{4} $$
2 replies
sqing
Yesterday at 3:56 AM
sqing
12 minutes ago
Triangle on a tetrahedron
vanstraelen   2
N Yesterday at 7:51 PM by ReticulatedPython

Given a regular tetrahedron $(A,BCD)$ with edges $l$.
Construct at the apex $A$ three perpendiculars to the three lateral faces.
Take a point on each perpendicular at a distance $l$ from the apex such that these three points lie above the apex.
Calculate the lenghts of the sides of the triangle.
2 replies
vanstraelen
Yesterday at 2:43 PM
ReticulatedPython
Yesterday at 7:51 PM
shadow of a cylinder, shadow of a cone
vanstraelen   2
N Yesterday at 6:33 PM by vanstraelen

a) Given is a right cylinder of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the botom base?

b) Given is a right cone of height $2R$ and radius $R$.
The sun shines on this solid at an angle of $45^{\circ}$.
What is the area of the shadow that this solid casts on the plane of the base?
2 replies
vanstraelen
Yesterday at 3:08 PM
vanstraelen
Yesterday at 6:33 PM
2023 Official Mock NAIME #15 f(f(f(x))) = f(f(x))
parmenides51   3
N Yesterday at 5:13 PM by jasperE3
How many non-bijective functions $f$ exist that satisfy $f(f(f(x))) = f(f(x))$ for all real $x$ and the domain of f is strictly within the set of $\{1,2,3,5,6,7,9\}$, the range being $\{1,2,4,6,7,8,9\}$?

Even though this is an AIME problem, a proof is mandatory for full credit. Constants must be ignored as we dont want an infinite number of solutions.
3 replies
parmenides51
Dec 4, 2023
jasperE3
Yesterday at 5:13 PM
Geometry
AlexCenteno2007   3
N Yesterday at 4:18 PM by AlexCenteno2007
Let ABC be an acute triangle and let D, E and F be the feet of the altitudes from A, B and C respectively. The straight line EF and the circumcircle of ABC intersect at P such that F is between E and P, the straight lines BP and DF intersect at Q. Show that if ED = EP then CQ and DP are parallel.
3 replies
AlexCenteno2007
Apr 28, 2025
AlexCenteno2007
Yesterday at 4:18 PM
Cube Sphere
vanstraelen   4
N Yesterday at 2:37 PM by pieMax2713

Given the cube $\left(\begin{array}{ll} EFGH \\ ABCD \end{array}\right)$ with edge $6$ cm.
Find the volume of the sphere passing through $A,B,C,D$ and tangent to the plane $(EFGH)$.
4 replies
vanstraelen
Yesterday at 1:10 PM
pieMax2713
Yesterday at 2:37 PM
Square number
linkxink0603   3
N Yesterday at 2:36 PM by Zok_G8D
Find m is positive interger such that m^4+3^m is square number
3 replies
linkxink0603
Yesterday at 11:20 AM
Zok_G8D
Yesterday at 2:36 PM
Combinatorics
AlexCenteno2007   0
Yesterday at 2:05 PM
Adrian and Bertrand take turns as follows: Adrian starts with a pile of ($n\geq 3$) stones. On their turn, each player must divide a pile. The player who can make all piles have at most 2 stones wins. Depending on n, determine which player has a winning strategy.
0 replies
AlexCenteno2007
Yesterday at 2:05 PM
0 replies
How many pairs
Ecrin_eren   6
N Yesterday at 12:57 PM by Ecrin_eren


Let n be a natural number and p be a prime number. How many different pairs (n, p) satisfy the equation:

p + 2^p + 3 = n^2 ?



6 replies
Ecrin_eren
May 2, 2025
Ecrin_eren
Yesterday at 12:57 PM
parallelogram in a tetrahedron
vanstraelen   1
N Yesterday at 12:19 PM by vanstraelen
Given a tetrahedron $ABCD$ and a plane $\mu$, parallel with the edges $AC$ and $BD$.
$AB \cap \mu=P$.
a) Prove: the intersection of the tetrahedron with the plane is a parallelogram.
b) If $\left|AC\right|=14,\left|BD\right|=7$ and $\frac{\left|PA\right|}{\left|PB\right|}=\frac{3}{4}$,
calculates the lenghts of the sides of this parallelogram.
1 reply
vanstraelen
May 5, 2025
vanstraelen
Yesterday at 12:19 PM
Find max
tranlenhanhbnd   0
Yesterday at 11:50 AM
Let $x,y,z>0$ and $x^2+y^2+z^2=1$. Find max
$D=\dfrac{x}{\sqrt{2 y z+1}}+\dfrac{y}{\sqrt{2 z x+1}}+\dfrac{z}{\sqrt{2 x y+1}}$.
0 replies
tranlenhanhbnd
Yesterday at 11:50 AM
0 replies
Trapezium with two right-angles: prove < AKB = 90° and more
Leonardo   5
N Jun 14, 2023 by UI_MathZ_25
Source: Mexico 2002
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.

[Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .]
5 replies
Leonardo
May 8, 2004
UI_MathZ_25
Jun 14, 2023
Trapezium with two right-angles: prove < AKB = 90° and more
G H J
G H BBookmark kLocked kLocked NReply
Source: Mexico 2002
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Leonardo
128 posts
#1 • 2 Y
Y by Adventure10, Mango247
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.

[Moderator edit: The proposed solution can be found at http://erdos.fciencias.unam.mx/mexproblem3.pdf .]
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darij grinberg
6555 posts
#2 • 2 Y
Y by Adventure10, Mango247
Proving < AKB = 90° is very easy: Since < MAD = 90° and < MKD = 90°, the points A and K lie on the circle with diameter MD, and thus < AKM = < ADM. In the right-angled triangle DAM, we have < ADM = 90° - < DMA. Thus, < AKM = 90° - < DMA. Similarly, < BKM = 90° - < CMB. Hence,

< AKB = < AKM + < BKM = (90° - < DMA) + (90° - < CMB)
= 180° - < DMA - < CMB = < CMD = 90°.

Now we are going to show $\frac{KP}{PA}+\frac{KQ}{QB}=1$. In fact, let U be the orthogonal projection of the point K on the line AB, and let the line AC meet the line KU at T. Also let the line AK meet the line BC at Q.

Since < AKB = 90°, the point K lies on the circle with diameter AB. The center of this circle is the midpoint M of the segment AB. Hence, MK = MA = MB. Now, since MK = MB, the triangle KMB is isosceles, so that < MKB = < MBK. Thus, < CKB = 90° - < MKB = 90° - < MBK = < CBK, and it follows that the triangle KCB is isosceles, so that CK = CB. Similarly, DK = DA.

The lines AD, BC and KU are parallel to each other, since all of them are perpendicular to the line AB. Thus, from AD || BC, we have by Thales CQ : DA = CK : DK. But DK = DA; thus, CQ = CK. Together with CK = CB, this yields CQ = CB, and thus the point C is the midpoint of the segment BQ.

Since KU || BC, Thales yields TK : TU = CQ : CB. Since CQ = CB, we thus have TK = TU, and hence the point T is the midpoint of the segment KU.

Now, we have defined the point T as the point of intersection of the lines AC and KU and proved that this point T is the midpoint of the segment KU. Instead, we could have defined the point T as the midpoint of the segment KU and would be able to conclude that this point T lies on the line AC. Similarly, the same point T lies on the line BD. Altogether, the point T lies on the three lines AC, BD and KU.

Now, the points P and Q join the scene. Applying the van Aubel theorem to triangle AKB, whose cevians AQ, BP and KU concur at the point T, we get $\frac{KT}{TU} = \frac{KP}{PA}+\frac{KQ}{QB}$. Since the point T is the midpoint of the segment KU, we have $\frac{KT}{TU}=1$; thus, $\frac{KP}{PA}+\frac{KQ}{QB}=1$, and we are done.

Darij
This post has been edited 1 time. Last edited by darij grinberg, Jan 28, 2005, 5:18 PM
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Huynh Anh Hao
31 posts
#3 • 2 Y
Y by Adventure10, Mango247
Leonardo wrote:
Let $ABCD$ be a quadrilateral with $\measuredangle DAB=\measuredangle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\measuredangle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.
We can easily prove $\angle AKB = 90^{\circ}$
We call $I$ is the intersection of $AC$ and $BD$.
=> $\frac{IB}{IA}= \frac{AB}{CD}= \frac{BK}{CK}$
=> $IK$ // $CD$ // $AB$
=> $\frac{KP}{PA}= \frac{IK}{AD}= \frac{KC}{DC}$
$\frac{KQ}{QB}= \frac{IK}{CB}= \frac{KD}{DC}$
=> The proof is completed.
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Virgil Nicula
7054 posts
#4 • 1 Y
Y by Adventure10
Prove easily that $KA\perp KB\ .$ Denote $B'\in BC\cap AK$ and $A'\in AD\cap BK\ .$ Then $CB=CB'$ and $DA=DA'\ .$

Apply the Menelaus' theorem to the transversal $\overline{AQC}$ for the triangle $BB'K$, where $CB'=CB$ :

$\frac{AK}{AB'}\cdot \frac{CB'}{CB}\cdot\frac{QB}{QK}=1$ $\Longrightarrow$ $\frac{QK}{QB}=\frac{AK}{AB'}$ $\Longrightarrow$ $\boxed{\ \frac{QK}{QB}=\frac{AA'}{AA'+BB'}\ }\ \ (1)\ .$

Apply the Menelaus' theorem to the transversal $\overline{BPD}$ for the triangle $AA'K$, where $DA'=DA$ :

$\frac{BK}{BA'}\cdot\frac{DA'}{DA}\cdot\frac{PA}{PK}=1$ $\Longrightarrow$ $\frac{PK}{PA}=\frac{BK}{BA'}$ $\Longrightarrow$ $\boxed{\ \frac{PK}{PA}=\frac{BB'}{AA'+BB'}\ }\ \ (2)\ .$

Therefore, from the sum of the relations $(1)$ and $(2)$ we find the required relation $\boxed{\ \frac{PK}{PA}+\frac{QK}{QB}=1\ }\ \ (*)\ .$

Remark. Denote the intersections $R\in MK\cap AC$ and $S\in MK\cap BD\ .$

Apply the Menelaus' theorem to the transversal $\overline{ARQ}$ for the triangle $MBK$ : $\frac{AM}{AB}\cdot\frac{QB}{QK}\cdot\frac{RK}{RM}=1$ $\Longrightarrow$ $\boxed{\ \frac{RK}{RM}=2\cdot\frac{QK}{QB}\ }\ \ (3)\ .$

Apply the Menelaus' theorem to the transversal $\overline{BSP}$ for the triangle $MAK$ : $\frac{BM}{BA}\cdot\frac{PA}{PK}\cdot\frac{SK}{SM}=1$ $\Longrightarrow$ $\boxed{\ \frac{SK}{SM}=2\cdot\frac{PK}{PA}\ }\ \ (4)\ .$

Therefore, from the sum of the relations $(3)$ , $(4)$ and using the relation $(*)$ obtain a new and interesting relation $\boxed{\ \frac{SK}{SM}+\frac{RK}{RM}=2\ }\ \ (**)\ .$
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parmenides51
30651 posts
#5
Y by
Let $ABCD$ be a quadrilateral with $\angle DAB=\angle ABC=90^{\circ}$. Denote by $M$ the midpoint of the side $AB$, and assume that $\angle CMD=90^{\circ}$. Let $K$ be the foot of the perpendicular from the point $M$ to the line $CD$. The line $AK$ meets $BD$ at $P$, and the line $BK$ meets $AC$ at $Q$. Show that $\angle{AKB}=90^{\circ}$ and $\frac{KP}{PA}+\frac{KQ}{QB}=1$.
This post has been edited 1 time. Last edited by parmenides51, Jan 8, 2023, 12:00 AM
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UI_MathZ_25
116 posts
#6
Y by
Since $AMKD$ and $MBCK$ are cyclic \[\angle AKM = \angle ADM = 90^{\circ} - \angle AMD = \angle BMC = \angle BKC = 90^{\circ} - \angle BKM \square \]By Pappus' theorem on $A-M-B$ and $D-K-C,$ \[E = AK \cap MD, F = AC \cap BD, G = MC \cap BK \]are collinear. Hence, by angle chasing $DM$ and $MC$ are perpendicular bisector of $AK$ and $BK$, respectively. Thereby $EG \parallel AB$.
We say that the parallel to $AB$ that passes through $K$ cut to $BD$ and $AC$ at $J$ and $L$, respectively. Therefore \[\frac{KP}{PA} + \frac{KQ}{QB} = \frac{KJ}{AB} + \frac{KL}{AB} = \frac{JL}{AB} = 1\]where the last equation is given because \[\frac{JL}{AB} = \frac{FL}{AF} = \frac{EK}{AE} = 1 \blacksquare\]
[asy]
import graph; size(13cm); 
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 = -10.366231833421102, xmax = 11.055501321739651, ymin = -4.584365118522112, ymax = 26.38971231092906;  /* image dimensions */
pen zzttff = rgb(0.6,0.2,1); pen zzttqq = rgb(0.6,0.2,0); pen ccqqqq = rgb(0.8,0,0); 

draw((-4.935382086114243,14.666485582905072)--(-4.471987430283072,14.6832425523788)--(-4.4887443997568,15.146637208209972)--(-4.9521390555879705,15.129880238736243)--cycle, linewidth(2) + ccqqqq); 
draw((8.964466288580857,15.633123269262516)--(8.981223258054586,15.169728613431346)--(9.444617913885757,15.186485582905073)--(9.427860944412028,15.649880238736245)--cycle, linewidth(2) + ccqqqq); 
draw((0.5048812728795858,8.885791671830379)--(0.05681575859019139,9.005176335767642)--(-0.06256890534707155,8.557110821478247)--(0.385496608942323,8.437726157540984)--cycle, linewidth(2) + ccqqqq); 
draw((1.8753492906749505,15.100742375453619)--(2.1644871539575745,14.73823072171654)--(2.526998807694653,15.027368584999165)--(2.237860944412029,15.389880238736243)--cycle, linewidth(2) + ccqqqq); 
 /* draw figures */
draw((-4.9521390555879705,15.129880238736243)--(9.427860944412028,15.649880238736245), linewidth(2) + zzttff); 
draw((2.237860944412029,15.389880238736243)--(9.779170059339375,5.934832022091594), linewidth(2)); 
draw((-4.7597159241415214,9.808640565274834)--(9.779170059339375,5.934832022091594), linewidth(2)); 
draw((-4.9521390555879705,15.129880238736243)--(-4.7597159241415214,9.808640565274834), linewidth(2) + zzttqq); 
draw((9.427860944412028,15.649880238736245)--(9.779170059339375,5.934832022091594), linewidth(2) + zzttqq); 
draw((-4.7597159241415214,9.808640565274834)--(2.237860944412029,15.389880238736243), linewidth(2)); 
draw((2.237860944412029,15.389880238736243)--(0.385496608942323,8.437726157540984), linewidth(2)); 
draw((-4.7597159241415214,9.808640565274834)--(9.427860944412028,15.649880238736245), linewidth(2)); 
draw((-4.9521390555879705,15.129880238736243)--(9.779170059339375,5.934832022091594), linewidth(2)); 
draw((-4.9521390555879705,15.129880238736243)--(0.385496608942323,8.437726157540984), linewidth(2)); 
draw((0.385496608942323,8.437726157540984)--(9.427860944412028,15.649880238736245), linewidth(2)); 
draw((-2.2833212233228233,11.783803198138612)--(4.906678776677175,12.043803198138612), linewidth(2) + linetype("4 4") + zzttff); 
draw((-8.876118698789208,8.113855372426736)--(-4.7597159241415214,9.808640565274834), linewidth(2)); 
draw((-8.876118698789208,8.113855372426736)--(5.483684752146957,8.616005447031823), linewidth(2) + zzttff); ; 
 /* dots and labels */
dot((-4.9521390555879705,15.129880238736243),dotstyle); 
label("$A$", (-4.857786164951195,15.350962062606659), NE * labelscalefactor); 
dot((9.427860944412028,15.649880238736245),dotstyle); 
label("$B$", (9.525377524942455,15.875575935794258), NE * labelscalefactor); 
dot((2.237860944412029,15.389880238736243),linewidth(4pt) + dotstyle); 
label("$M$", (2.33379567999563,15.569551176434825), NE * labelscalefactor); 
dot((9.779170059339375,5.934832022091594),dotstyle); 
label("$C$", (9.875120107067527,6.148360370440855), NE * labelscalefactor); 
dot((-4.7597159241415214,9.808640565274834),linewidth(4pt) + dotstyle); 
label("$D$", (-4.682914873888658,9.973669862433766), NE * labelscalefactor); 
dot((0.385496608942323,8.437726157540984),linewidth(4pt) + dotstyle); 
label("$K$", (0.7380951490499821,7.394318319261403), NE * labelscalefactor); 
dot((-1.7095574770328914,11.064436864858926),linewidth(4pt) + dotstyle); 
label("$P$", (-1.6226672802942643,11.24148672263713), NE * labelscalefactor); 
dot((2.749073801386117,10.322905991297157),linewidth(4pt) + dotstyle); 
label("$Q$", (2.8365506418004234,10.498283735621365), NE * labelscalefactor); 
dot((-2.2833212233228233,11.783803198138612),linewidth(4pt) + dotstyle); 
label("$E$", (-2.1909989762475086,11.96283079827008), NE * labelscalefactor); 
dot((0.2611707105781351,11.875815423842962),linewidth(4pt) + dotstyle); 
label("$F$", (0.3446347441592743,12.050266443801346), NE * labelscalefactor); 
dot((4.906678776677175,12.043803198138612),linewidth(4pt) + dotstyle); 
label("$G$", (5.000582868699316,12.22513773486388), NE * labelscalefactor); 
dot((-8.876118698789208,8.113855372426736),dotstyle); 
label("$J$", (-8.79239021385827,8.334251508722518), NE * labelscalefactor); 
dot((5.483684752146957,8.616005447031823),linewidth(4pt) + dotstyle); 
label("$L$", (5.568914564652561,8.793288647761669), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
 [/asy]
Z K Y
N Quick Reply
G
H
=
a