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jlacosta   0
Monday at 3:57 PM
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0 replies
jlacosta
Monday at 3:57 PM
0 replies
f this \8char
v4913   30
N 22 minutes ago by eg4334
Source: EGMO 2022/2
Let $\mathbb{N}=\{1, 2, 3, \dots\}$ be the set of all positive integers. Find all functions $f : \mathbb{N} \rightarrow \mathbb{N}$ such that for any positive integers $a$ and $b$, the following two conditions hold:
(1) $f(ab) = f(a)f(b)$, and
(2) at least two of the numbers $f(a)$, $f(b)$, and $f(a+b)$ are equal.
30 replies
v4913
Apr 9, 2022
eg4334
22 minutes ago
Weird length condition
Taco12   16
N an hour ago by lpieleanu
Source: USA January Team Selection Test for EGMO 2023, Problem 4
Let $ABC$ be a triangle with $AB+AC=3BC$. The $B$-excircle touches side $AC$ and line $BC$ at $E$ and $D$, respectively. The $C$-excircle touches side $AB$ at $F$. Let lines $CF$ and $DE$ meet at $P$. Prove that $\angle PBC = 90^{\circ}$.

Ray Li
16 replies
Taco12
Jan 16, 2023
lpieleanu
an hour ago
ABC is similar to XYZ
Amir Hossein   56
N an hour ago by lksb
Source: China TST 2011 - Quiz 2 - D2 - P1
Let $AA',BB',CC'$ be three diameters of the circumcircle of an acute triangle $ABC$. Let $P$ be an arbitrary point in the interior of $\triangle ABC$, and let $D,E,F$ be the orthogonal projection of $P$ on $BC,CA,AB$, respectively. Let $X$ be the point such that $D$ is the midpoint of $A'X$, let $Y$ be the point such that $E$ is the midpoint of $B'Y$, and similarly let $Z$ be the point such that $F$ is the midpoint of $C'Z$. Prove that triangle $XYZ$ is similar to triangle $ABC$.
56 replies
Amir Hossein
May 20, 2011
lksb
an hour ago
Cubes and squares
y-is-the-best-_   61
N 2 hours ago by ezpotd
Source: IMO 2019 SL N2
Find all triples $(a, b, c)$ of positive integers such that $a^3 + b^3 + c^3 = (abc)^2$.
61 replies
y-is-the-best-_
Sep 22, 2020
ezpotd
2 hours ago
No more topics!
USAMO 2003 Problem 4
MithsApprentice   72
N May 27, 2025 by endless_abyss
Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$.
72 replies
MithsApprentice
Sep 27, 2005
endless_abyss
May 27, 2025
USAMO 2003 Problem 4
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MithsApprentice
2390 posts
#1 • 10 Y
Y by pifinity, Adventure10, megarnie, Mango247, ohiorizzler1434, ItsBesi, buddyram, cubres, and 2 other users
Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$.
This post has been edited 1 time. Last edited by MithsApprentice, Sep 27, 2005, 10:00 PM
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MithsApprentice
2390 posts
#2 • 2 Y
Y by Adventure10, Mango247
When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Please use LaTeX for posting solutions. Thanks.
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darij grinberg
6555 posts
#3 • 5 Y
Y by Adventure10, megarnie, and 3 other users
By the Ceva theorem, applied to the triangle BCF and the concurrent cevians BM, CA and FE (in fact, these cevians concur at the point D), we have $\dfrac{MF}{MC}\cdot\dfrac{EC}{EB}\cdot\dfrac{AB}{AF}=1$ (we work with non-directed segments here). Hence, $\dfrac{MF}{MC}=\dfrac{AF}{AB}\cdot\dfrac{EB}{EC}=\dfrac{AF}{AB}: \dfrac{EC}{EB}$. Thus, MF = MC holds if and only if $\dfrac{AF}{AB}=\dfrac{EC}{EB}$. But by Thales, $\dfrac{AF}{AB}=\dfrac{EC}{EB}$ is equivalent to AE || FC, and obviously we have AE || FC if and only if < EAC = < ACF. Now, since the points A, B, D and E lie on one circle, < EAD = < EBD, what rewrites as < EAC = < CBM; on the other hand, trivially < ACF = < DCM. Thus, we have < EAC = < ACF if and only if < CBM = < DCM. Now, as it is clear that < CMB = < DMC, we have < CBM = < DCM if and only if the triangles CMB and DMC are similar (in fact, you can see it this way: the triangles CMB and DMC already have one pair of equal angles, namely < CMB = < DMC, and thus they are similar if and only if they have one more pair of equal angles). On the other hand, the triangles CMB and DMC are similar if and only if MB : MC = MC : MD (in fact, the triangles CMB and DMC have one pair of equal angles, namely < CMB = < DMC, and thus they are similar if and only if the sides adjacent to these angles are in the same ratio). Finally, it is clear that MB : MC = MC : MD is equivalent to $MB\cdot MD=MC^2$.

Combining all these equivalences, we see that MF = MC holds if and only if $MB\cdot MD=MC^2$.
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Bourne
31 posts
#4 • 8 Y
Y by Mobashereh, Kanep, Adventure10, mst.4921, sabkx, LeYohan, and 2 other users
It's such a easy problem.Why you use such a difficult method to solve it.
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dhernandez
88 posts
#5 • 3 Y
Y by Adventure10 and 2 other users
MithsApprentice wrote:
Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E$, respectively. Lines $AB$ and $DE$ intersect at $F$, while lines $BD$ and $CF$ intersect at $M$. Prove that $MF = MC$ if and only if $MB\cdot MD = MC^2$.

Assume $MF = MC$.
1) Draw parallel to $AD$ from $F$ and parallel to $ED$ from $C$. Let them intersect at $Y$.
2) $Y$ must be on line $BM$ because $FDCY$ is a parallelogram and its diagonals must intersect at their midpoints.
3) $\angle FYC = \angle ADE$ and $\frac{AD}{FY} = \frac{BD}{BY} = \frac{DE}{CY}$ therefore triangles $FYC$ and $ADE$ are similar and $FC \parallel AE$
4) $\angle MCA = \angle YFC = \angle DAE = \angle DBE$ hence the circumcircle of triangle $BDC$ is tangent to line $MC$ therefore $MC^2 = MD \cdot DB$

Assume $MC^2 = MD \cdot DB$
1) Then $\angle MCD = \angle DBE = \angle DAE$ hence $FC \parallel AE$
2) $\angle MFE = \angle DEA = \angle DBA$ hence the circumcircle of triangle $BDF$ tangent to line $MF$
3) Hence $MF^2 = MD \cdot MB = MC^2 \Rightarrow MF = MC$

Daniel
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giancarlo
27 posts
#6 • 2 Y
Y by Adventure10, Mango247
nice solution :)
dhernandez wrote:
3) $\angle FYC = \angle ADE$ and $\frac{AD}{FY} = \frac{BD}{BY} = \frac{DE}{CY}$ therefore triangles $FYC$ and $ADE$ are similar and $FC \parallel AE$
4) $\angle MCA = \angle YFC = \angle DAE = \angle DBE$ hence the circumcircle of triangle $BDC$ is tangent to line $MC$ therefore $MC^2 = MD \cdot DB$

You can also say that if FDCY is a parallelogramme, then FBCY is cyclic ( beause $\angle ADE = \angle FDC = \angle FYC$), so power of M: $MC^2= MF \cdot MC = MY \cdot MB = MD \cdot MB$
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calc rulz
1126 posts
#7 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Nice Solution for the Only If Part
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April
1270 posts
#8 • 2 Y
Y by Adventure10, Mango247
MithsApprentice wrote:
Let $ABC$ be a triangle. A circle passing through $A$ and $B$ intersects segments $AC$ and $BC$ at $D$ and $E,$ respectively.
Lines $AB$ and $DE$ intersect at $F,$ while lines $BD$ and $CF$ intersect at $M.$
Prove that $MF = MC$ if and only if $MB\cdot MD = MC^{2}$
Nice problem! And here is a nice proof. :lol:
Proof. Take $G\in BD: \,FG\parallel CD$
We have:
$MF=MC\Longleftrightarrow \textrm{the quadrilateral}\; CDFG\; \textrm{is a parallelogram}\\ \Longleftrightarrow FD\parallel CG\Longleftrightarrow\angle FDA=\angle GCD\Longleftrightarrow\angle FDA+\angle CGF=180^\circ\\ \Longleftrightarrow \angle ABE+\angle CGF=180^\circ\Longleftrightarrow\textrm{the quadrilateral}\;CBGF\;\textrm{is cyclic}\\ \Longleftrightarrow\angle CBM=\angle CBG=\angle CFG=\angle DCF=\angle DCM\\ \Longleftrightarrow\triangle BCM\sim\triangle CDM\Longleftrightarrow MB\cdot MD=MC^{2}$
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windrock
46 posts
#9 • 2 Y
Y by Adventure10, Mango247
If MC=MF. Assume that AE not paralel to CF, call T is intersection of AE and CD. we have (CFMT) = -1, on the other hand. M is midpoint of CD. so nonsense. Hence, CF // AE, easy to finish problem
If MC.MC = MB. MD, easy
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windrock
46 posts
#10 • 3 Y
Y by Adventure10, Mango247, and 1 other user
If $ MC=MF$. Assume that $ AE$ not paralel to $ CF$, call $ T$ is intersection of $ AE$ and $ CD$. we have $ (CFMT) = -1$, on the other hand. M is midpoint of CD. so nonsense. Hence, $ CF // AE$, easy to finish problem
If $ MC^{2} = MB. MD$, easy
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SnowEverywhere
801 posts
#11 • 2 Y
Y by Adventure10, Mango247
Sorry for bringing up an old topic. How did the solvers know to create the parallelogram. Is this just a well known construction? Does my solution work?

Solution

Assume that $ MC=MF$. Now, by Ceva's theorem $ \frac{MC}{MF} \cdot \frac{BE}{CE} \cdot \frac{FA}{BA}=\frac{BE}{CE} \cdot \frac{FA}{BA}=1$. Therefore $ \frac{BE}{CE} = \frac{BA}{FA}$ and by SAS similarity, $ BAE \sim BFC$. Therefore $ AE \| FC$. Now note that $ \angle{MFD}=\angle{DEA}=\angle{DBA}$. Hence by AA similarity, $ MDF \sim MFB$ and it follows that $ \frac{MD}{MF}=\frac{MF}{MB}$. Rearranging yields that $ MF^2=MC^2=MD \cdot MB$.

Assume that $ MC^2=MD \cdot MB$. It follows that $ \frac{MD}{MC}=\frac{MC}{MB}$ which by SAS implies that $ MCD \sim MBC$. Hence $ \angle{MCB}=\angle{MDC}=\angle{ADB}=\angle{AEB}$ and therefore $ FC \| AE$. This implies that $ AEB \sim FCB$ and that $ \frac{BE}{CE} = \frac{BA}{FA}$. By Ceva's theorem, it follows that $ \frac{MC}{MF} \cdot \frac{BE}{CE} \cdot \frac{FA}{BA}=\frac{MC}{MF}=1$. Therefore $ MC=MF$.
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ACCCGS8
326 posts
#12 • 2 Y
Y by Adventure10, Mango247
Quite a nice problem. It is possible to prove both parts simultaneously.

By Ceva's Theorem, $\frac{FM}{MC} \cdot \frac{CE}{EB} \cdot \frac{BA}{AF} = 1$ so $MF=MC \Leftrightarrow \frac{CE}{EB} \cdot \frac{BA}{AF} = 1$ $\Leftrightarrow \frac{CE}{EB} = \frac{FA}{AB}$ $\Leftrightarrow EA || CF$ $\Leftrightarrow \angle MCA = \angle DAE$ $\Leftrightarrow \angle MCA = \angle DBE$ $\Leftrightarrow MC^2 = MD \cdot MB$ where the last step follows from the Tangent-Secant Theorem and its converse.
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JSGandora
4216 posts
#13 • 4 Y
Y by tim9099xxzz, acegikmoqsuwy2000, Adventure10, Mango247
First let us prove the if statement. If $MD\cdot MB=MC^2$, then $MC$ is tangent to $\odot DBC$ at $C$ and thus $\angle MCD=\angle MBC$ and since quadrilateral $ABDE$ is cyclic, we have $\angle MBC=\angle FAD$ so $AF||FC$ so $\triangle BAE~\triangle BFC$ thus $\frac{BA}{AF}=\frac{BE}{EC}\implies \frac{BA}{AF}\cdot\frac{EC}{BE}=1$. Now by Ceva's, we have
\[\frac{FA}{AB}\cdot\frac{BE}{EC}\cdot\frac{CM}{MF}=1\]
and using $\frac{BA}{AF}\cdot\frac{EC}{BE}=1$, we have $\frac{CM}{MF}=1\implies CM=MF$ as desired.

Now to prove the only if statement, we begin with $MF=MC$ now by Ceva's we find $\frac{FA}{AB}\cdot\frac{BF}{FC}=1\implies{FA}{AB}=\frac{FC}{BF}$ SO $\triangle BAE~\triangle BFC$ which means $AF||FC$ and so $\angle DBF=\angle FAD=\angle DCM$ so $MC$ is tangent to $\odot BDC$ at point $C$ which means $MC^2=MD\cdot MB$ by Power of a Point.
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TheStrangeCharm
290 posts
#14 • 2 Y
Y by Adventure10, Mango247
We work in the projective plane.
Let $X = AE \cap FC$
First, let us show that if $MF = MC$, then $MC^2 = MD\cdot MB$. Note that $(F,C;M,X)$ is a harmonic division. Because $MF = MC$, we must have that $X$ is the point at infinity on line $FC$, so we have that $FC$ is parallel to $AE$. Thus, we have that $\angle{FCA} = \angle{CAE} = \angle{DBC}$, where the last angle equality follows from cyclic quadrilateral $ADEB$. Thus, we see that $MC$ is tangent to the circumcircle of $\triangle{BDC}$ at $C$, and the result follows.
Now let us show that if $MC^2 = MD\cdot MB$, then $MF = MC$. Note that we must have that $\angle{MCA} = \angle{EBD} = \angle{DAE}$, so we get that $AE$ is parallel to $FC$ again. Thus, $X$ is the point at infinity on line $FC$, but we still have that $(F,C;M,X)$ is harmonic, so we must have that $M$ is the midpoint of $FC$, so we are done. $\blacksquare$



This solution is essentially the same as the ones above, however, a lot of the work using ceva is encapsulated in a well know harmonic division lemma.
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mathbuzz
803 posts
#15 • 2 Y
Y by Adventure10, Mango247
First I do ''only if'' part ----
Note that , in triangle $BCF$ , , we have $BM,EF,AC$ as 3 concurrent cevians. hence , by using ceva's theorem , we get that $AE$ is parallel to $CF$. hence , after a little bit of anngle chasing , we get $\Delta BCM$ is similar to $\Delta CDM$. hence , $MB.MD=MC^2$.

for if part -- $MB.MD=MC^2$ implies that ,
$\Delta BCM$ is similar to $\Delta CDM$ with $\angle CBM=\angle DCM$ , and $\angle CDM= \angle BCM$
hence , a little angle chasing shows that $AE$ is parallel to $BC$.
hence , by Ceva's theorem , we get that $MC=MF$ :D
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