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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
7 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
7 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
Let's Invert Some
Shweta_16   8
N Apr 1, 2025 by ihategeo_1969
Source: STEMS 2020 Math Category B/P4 Subjective
In triangle $\triangle{ABC}$ with incenter $I$, the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$, respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$. The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$. Prove that $FQ$ is parallel to $BI$.

Proposed by Anant Mudgal
8 replies
Shweta_16
Jan 26, 2020
ihategeo_1969
Apr 1, 2025
Let's Invert Some
G H J
Source: STEMS 2020 Math Category B/P4 Subjective
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Shweta_16
15 posts
#1 • 4 Y
Y by Smita, mijail, Adventure10, Funcshun840
In triangle $\triangle{ABC}$ with incenter $I$, the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$, respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$. The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$. Prove that $FQ$ is parallel to $BI$.

Proposed by Anant Mudgal
This post has been edited 2 times. Last edited by Shweta_16, Jan 26, 2020, 1:34 PM
Reason: let's chase some angels
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GeoMetrix
924 posts
#2 • 5 Y
Y by mueller.25, amar_04, AlastorMoody, sameer_chahar12, Adventure10
[asy]
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dot((1.88,-6.82),dotstyle); 
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[/asy]
IMO the title doesnt suit the prob.
Proof: Let $I_A$ be the $A$-excentre. Its well known(ISL 2002 G7) that $K=I_AD \cap \omega$. Now we'll use a phantom approach . Let $\ell$ be the line through $F$ parallel to $BI$. and let $\ell \cap BC=Q'$ We just need to show that $Q' \in \odot(KEG)$. Firstly define $\ell \cap AI=L$. Notice that $\angle LQ'C=\angle IBC=\angle IKC=\angle LKC \implies LQ'I_AC$ cyclic. Now notice that $\Delta FAL \cong \Delta EAL \implies \angle AFL=\angle AEL$. But notice that $\angle AEL=\angle AFL=\angle ABI=\angle LQ'C\implies LEQ'C$ cyclic. This combined with the fact that $LQ'I_AC$ is cyclic $\implies LQ'CI_AE$ is cyclic. So now we just need to show that $K\in \odot(LQ'CI_AE)$ But notice that $\angle I_AEQ'=\angle I_ACQ'=90-\frac{\angle C}{2}$ and $\angle Q'I_AE=\angle Q'CE=\angle C \implies I_AQ=I_AE$. FInally $\angle EKD =\angle CDE=\angle I_ACD=\angle I_AEQ'=\angle I_AQ'E\implies K \in \odot(I_AQ'E)$ as desired. $\blacksquare$
This post has been edited 9 times. Last edited by GeoMetrix, Jan 27, 2020, 6:13 AM
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anantmudgal09
1980 posts
#3 • 8 Y
Y by Pluto1708, biomathematics, GammaBetaAlpha, amar_04, Sumitrajput0271, DPS, Adventure10, Funcshun840
My problem.

My solution was to apply $\sqrt{DE \cdot DF}$ inversion in contact triangle $\triangle DEF$. It is quite simple from here :)
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TheDarkPrince
3042 posts
#4 • 2 Y
Y by amar_04, Adventure10
Shweta_16 wrote:
In triangle $\triangle{ABC}$ with incenter $I$, the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$, respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$. The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$. Prove that $FQ$ is parallel to $BI$.

Proposed by Anant Mudgal

Quick sketch:

Let $I_a$ be the A-excenter. We know that $I_a,D,K$ are collinear. Angle chase to show that $I_a$ lies on $\odot(KEC)$. This will gives us that $\angle QKD = \angle DIC$ and also we have $\angle BKD = DKC$.

Now we'll fix $K,D$ and move $C$ linearly. Therefore $B$ and $Q$ move linearly, so just work when $C = D$ and $C$ is point of infinity to get that $BQ = BD = BF$, so we are done.
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Wizard_32
1566 posts
#9 • 4 Y
Y by GeoMetrix, amar_04, Adventure10, Mango247
Here's my solution, which is much more of a "complete the configuration" type, while ignoring $A.$ Nice problem btw.
Shweta_16 wrote:
In triangle $\triangle{ABC}$ with incenter $I$, the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$, respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$. The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$. Prove that $FQ$ is parallel to $BI$.

Proposed by Anant Mudgal
Clearly, it suffices to show $BF=BQ.$ Let $\omega$ be tangent to $BC$ at $D.$ Since $BF=BD,$ hence it suffices to show that $BQ=BD.$ Call the circle through $B,C$ tangent to $\omega$ as $\gamma.$ We now rephrase the problem without $A:$
Rephrased Problem wrote:
Let $\gamma$ be a circle through two points $B,C,$ and let $\omega$ be a circle tangent to $BC, \gamma$ at $D,K$ respectively. Let $CE$ be the second tangent from $C$ to $\omega.$ Assume that $(KEC)$ meets $BC$ again in $Q.$ Show that $BQ=BD.$
Let $M$ be the midpoint of arc $BC$ not containing $K$ in $\gamma.$ By a well known lemma (shooting lemma), $K,D,M$ are collinear. (proof is by homothety taking $\omega$ to $\gamma$). Let $(M,MC)$ be the circle at $M$ with radius $MC.$

We start off by the following key lemma:

Lemma: The points $ED, CM$ meet at $X,$ where $X$ lies on $(M,MC)$
Proof: Let $O$ be the center of $\omega.$ Let $OC \cap ED=N.$ Now consider the inversion about $(M,MC).$ It is not too hard to see that $\omega$ is fixed under this inversion (since it is tangent to $BC, \gamma,$ both of which are swapped under this inversion). Hence $\omega, (M,MC)$ are orthogonal.

Thus, $ON \cdot OC=OD^2$ is the power of $O$ with respect to $(M,MC).$ Hence, $N$ lies on $(M,MC).$ Since $\angle CND=\pi/2,$ hence this implies the lemma. $\square$
[asy]
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[/asy]
Claim: Now, we claim that $X$ also lies on $(KEC).$
Proof: Indeed,
$$\measuredangle KEX=\measuredangle KED=\measuredangle KDB=\text{arc}(BK)+\text{arc}(MC)=\text{arc}(KM)$$(where the last part since $M$ is the arc midpoint.) But also $$\measuredangle KCX=\measuredangle KCM=\text{arc}(KM)$$Hence $\measuredangle KEX=\measuredangle KCX$ giving that $X$ lies on $(KEC).$ $\square$

To finish, see that $\measuredangle XQC=\measuredangle XEC$ by $(KEC).$ But also $\measuredangle XEC=\measuredangle DEC=\measuredangle CDE=\measuredangle QDX$ and so $XQ=XD.$ But $XB \perp BC$ as $X \in (M,MC)$ and so $B$ is the midpoint of $QD.$ $\blacksquare$
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vivoloh
59 posts
#10 • 2 Y
Y by amar_04, Adventure10
I have a solution which involve inversion about point $K$ and a little bit of lengthy angle chase.
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BOBTHEGR8
272 posts
#11 • 1 Y
Y by Adventure10
Shweta_16 wrote:
In triangle $\triangle{ABC}$ with incenter $I$, the incircle $\omega$ touches sides $AC$ and $AB$ at points $E$ and $F$, respectively. A circle passing through $B$ and $C$ touches $\omega$ at point $K$. The circumcircle of $\triangle{KEC}$ meets $BC$ at $Q \neq C$. Prove that $FQ$ is parallel to $BI$.

Proposed by Anant Mudgal

Solution-
Let tangent to $\omega$ at $K$ intersect $BC$ at $T$ and $AC$ at $R$. Let $KE$ intersect $BC$ at $S$.
In $\Delta TRC$ , $ K,E,D$ are the incircle touch points and hence $R(T,C,S,D)=-1 \implies R(D,C,S,T)=1-(-1)=2$
Hence $\frac{DS}{SC}=2\frac{DT}{TC} \implies \frac{DS}{SC}(SD-SC)=2\frac{DT}{TC}(TC-TD)\implies \frac{SD^2}{CS}-DS=2(DT-\frac{TD^2}{CT})$
But $SD^2=SE\cdot SK=SC\cdot SQ \implies \frac{SD^2}{SC}=QS$ and $TD^2=TK^2=TB\cdot TC \implies \frac {TD^2}{CT}=BT$
So we have $QS-DS=2(DT-BT) \implies QD=2BD \implies BQ=BD=BF$ and hence we have $\angle QFD=90 $
But $BI\perp FD \implies BI \parallel QF$
Hence proved !!!
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mmathss
282 posts
#12 • 2 Y
Y by GeoMetrix, Adventure10
GeoMetrix wrote:
[asy]
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[/asy]
IMO the title doesnt suit the prob.
Proof: Let $I_A$ be the $A$-excentre. Its well known(ISL 2002 G7) that $K=I_AD \cap \omega$. Now we'll use a phantom approach . Let $\ell$ be the line through $F$ parallel to $BI$. and let $\ell \cap BC=Q'$ We just need to show that $Q' \in \odot(KEG)$. Firstly define $\ell \cap AI=L$. Notice that $\angle LQ'C=\angle IBC=\angle IKC=\angle LKC \implies LQ'I_AC$ cyclic. Now notice that $\Delta FAL \cong \Delta EAL \implies \angle AFL=\angle AEL$. But notice that $\angle AEL=\angle AFL=\angle ABI=\angle LQ'C\implies LEQ'C$ cyclic. This combined with the fact that $LQ'I_AC$ is cyclic $\implies LQ'CI_AE$ is cyclic. So now we just need to show that $K\in \odot(LQ'CI_AE)$ But notice that $\angle I_AEQ'=\angle I_ACQ'=90-\frac{\angle C}{2}$ and $\angle Q'I_AE=\angle Q'CE=\angle C \implies I_AQ=I_AE$. FInally $\angle EKD =\angle CDE=\angle I_ACD=\angle I_AEQ'=\angle I_AQ'E\implies K \in \odot(I_AQ'E)$ as desired. $\blacksquare$

Well there was no need of phantom point approach
Here's how you can finish it easily:
$\triangle AEI_A\equiv \triangle AFI_A\Rightarrow I_AE=I_AF\Rightarrow I_A$ is circumcenter of $QFE$.Since $\angle QI_AE=C$ we get $\angle QFE=180-\frac {C}{2}$ and we are done.
This post has been edited 2 times. Last edited by mmathss, Feb 1, 2020, 7:01 PM
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ihategeo_1969
235 posts
#13
Y by
We will introduce some new points.

$\bullet$ Let $D$ be $A$-intouch point.
$\bullet$ Let $I_A$ be $A$-excenter and $D'$ be midpoint of $\overline{DI_A}$.

Then by IMO Shortlist 2002/G7 and RMM 2012/6, we get $K \in \overline{DI_A}$ (it is the $D$-Why pointof $\triangle DEF$ btw) and $D' \in (BKC)$ respectively.

Claim: $I_A \in (KEC)$.
Proof: Now $\overline{DE} \parallel \overline{I_AC}$ and so \[\measuredangle KI_AC=\measuredangle DI_AC=\measuredangle KDE=\measuredangle KEA=\measuredangle KEC\]And done. $\square$

Now by PoP we get \[DQ \cdot DC=DK \cdot DI_A=2DD' \cdot DK=2DB \cdot DC \iff DQ=2DB\]So we get $B$ is midpoint of $\overline{QD}$ and hence $\frac 12$ homothety at $D$ sends $\overline{QF}$ to $\overline{BI}$ and done.
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