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orzzzzzzzzz
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mathMagicOPS, Jan 9, 2025, 3:40 AM
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this css is sus
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ihatemath123, Aug 14, 2024, 1:53 AM
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391345 views moment
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ryanbear, May 9, 2023, 6:10 AM
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We need virgil nicula to return to aops, this blog is top 10 all time.
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OlympusHero, Sep 14, 2022, 4:44 AM
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blog
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tigerzhang, Aug 1, 2021, 12:02 AM
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Amazing blog.
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OlympusHero, May 13, 2021, 10:23 PM
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the visits tho
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GoogleNebula, Apr 14, 2021, 5:25 AM
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Bro this blog is ripped
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samrocksnature, Apr 14, 2021, 5:16 AM
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Holy- Darn this is good. shame it's inactive now
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the_mathmagician, Jan 17, 2021, 7:43 PM
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godly blog. opopop
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OlympusHero, Dec 30, 2020, 6:08 PM
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long blog
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MrMustache, Nov 11, 2020, 4:52 PM
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372554 views!
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mrmath0720, Sep 28, 2020, 1:11 AM
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wow... i am lost.
369302 views!
-piphi
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piphi, Jun 10, 2020, 11:44 PM
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That was a lot! But, really good solutions and format! Nice blog!!!!
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CSPAL, May 27, 2020, 4:17 PM
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impressive
awesome. 358,000 visits?????
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OlympusHero, May 14, 2020, 8:43 PM
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Nice blog.
The element of Mathematics is dense here.
I love the style of your posts.
Thanks for writing the blog! I can learn Geometry from here too.
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epsilonist, Jun 27, 2011, 6:00 AM
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Thank you very much for this blog.
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Affine, May 7, 2011, 6:20 PM
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You seem to have solved every existing problem.
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Goutham, May 2, 2011, 10:59 AM
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I love your posts, man (those which I understand)! It's interesting how you punish integrals with recurrence relations and sometimes make seemingly unrelated integrals evaluate each other!
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Caspar, Apr 11, 2011, 2:36 PM
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Thank you a lot dear Virgil for an interesting and instructive blog.
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vittasko, Mar 12, 2011, 9:42 AM
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Interesting Blog
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ShahinBJK, Mar 12, 2011, 5:07 AM
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Interesant, d-le Nicula.
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silent_control, Mar 3, 2011, 6:00 PM
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Yours is the largest blog full of math equations as far as I have known! Congrats!
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Goutham, Jan 30, 2011, 2:06 PM
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15001th view.
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fries4guys, Jan 20, 2011, 4:57 PM
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Thanks to all !
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Virgil Nicula, Dec 11, 2010, 8:32 PM
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El_Ectric:
You can either copy some parts from the blog post, paste it at "Search Blogs" (which is lower on the page, after "Blog Stats") and find the whole post.
You're welcome.
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Thomas_, Dec 8, 2010, 6:42 PM
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Thanks Thomas_.
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El_Ectric, Dec 8, 2010, 4:21 PM
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El_Ectric: Try to open this blog with Internet Explorer (browser). If it still doesn't work, just copy the post you're interested in, and than paste it into Word Document, or almost any kind of reading documents and you'll have all the details.
Nice blog!
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Thomas_, Nov 24, 2010, 4:59 PM
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Dear Virgil !
Congratulations for this blog !
Babis stergiou - Greece
by
stergiu, Nov 12, 2010, 6:15 AM
by Virgil Nicula, Nov 5, 2011, 9:58 AM 
and the points 
, 
, 
so that 
. Prove that ![$\frac {[PQR]}{[ABC]}=\frac {m^3+n^3}{(m+n)^3}$](//latex.artofproblemsolving.com/7/4/6/74629ddaf757059e3117a80d793755a263e04a76.png)
. 
yields ![$\frac{[PQR]}{[ABC]}=\frac{1+\frac{m}{n} \cdot \frac{m}{n} \cdot \frac{m}{n}}{(1+\frac{m}{n})(1+\frac{m}{n})(1+\frac{m}{n})}=\frac{m^3+n^3}{(m+n)^3}$](//latex.artofproblemsolving.com/1/b/e/1bea7496a9444019a039378c1a91a05aa7c8be05.png)
.
be a triangle with ![$S=[ABC]$](//latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
. Then ![$ \left\|\begin{array}{ccc} M\in (BC) & , & \frac {MB}{MC} = m \\
\\
N\in (CA) & , & \frac {NC}{NA} = n \\
\\
P\in (AB) & , & \frac {NA}{NB} = n\end{array}\right\|\ \implies\ [MNP] $](//latex.artofproblemsolving.com/f/a/8/fa88e75ba23f707cdb009ff5ac08d74e8c896e25.png)
.
. Observe that ![$ \frac {[MNP]}{[ABC]} = \frac {[ABC] - [APN] - [BMP] - [CNM]}{[ABC]} =$](//latex.artofproblemsolving.com/b/d/a/bdad1b627dde889e91f146be1e832c84d13a01af.png)
.
(sideline !) and 
(sideline !), then ![$ \frac {[AXY]}{[ABC]} = \frac {AX}{AB}\cdot\frac {AY}{AC}$](//latex.artofproblemsolving.com/b/d/0/bd096649d5a13d33df1ea33b229243803bda69b5.png)
. If 
![$\frac {[MNP]}{[ABC]} = \frac {1 + m^3}{(1 + m)^3} = \frac {r^3 + s^3}{(r + s)^3}$](//latex.artofproblemsolving.com/d/8/1/d81e9f2d08aaf2f502058ce334ee721fd1e49bdb.png)
(the proposed problem). Otherwise. 
![$[APN]=[BMP]=$](//latex.artofproblemsolving.com/9/6/d/96d0ec390f12acf1a2954429f1704f9dea7e637f.png)
![$[CNM]=\frac {rs}{(r+s)^2}\cdot [ABC]\ \implies$](//latex.artofproblemsolving.com/9/0/9/909b4d50d4f2417aa358c88fca3ea2b9052c8baa.png)
![$ \frac {[MNP]}{[ABC]}=1-\frac {3rs}{(r+s)^2}=$](//latex.artofproblemsolving.com/9/d/d/9dd0be17ee7d114ede5fc63d56219111cf928232.png)
![$ \frac {r^2+s^2-rs}{(r+s)^2}\ \implies\ [MNP]=$](//latex.artofproblemsolving.com/6/c/9/6c91031814502272221bc8f47fdf053728d0e1d6.png)
![$ \frac {r^3+s^3}{(r+s)^3}\cdot [ABC]$](//latex.artofproblemsolving.com/f/f/6/ff68931d513cec1a3d762eb007a17b488239738f.png)
.
![$ \boxed {\ [MNP] =\frac {(1 + xyz)(1 + mnpxyz)}{[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}\cdot [ABC]\ }$](//latex.artofproblemsolving.com/9/7/7/9771f549337f8e363bd12611202e4e9eaf1763ed.png)
.![$ x = y = z = 1\ \implies\ 4\cdot [MNP] = [DEF]$](//latex.artofproblemsolving.com/8/1/1/8114ac1cf64a41e63b333b1e419bec6005a8dcde.png)
(the Pohoatza's problem).
and![$ [MNP] = \frac {1 + uvw}{(1 + u)(1 + v)(1 + w)}\cdot [XYZ] =$](//latex.artofproblemsolving.com/b/a/a/baa427bae61dcc6addc19153e06969af4c520bdb.png)
![$ \frac {1 + uvw}{(1 + u)(1 + v)(1 + w)}\cdot\frac {1 + xyz}{(1 + x)(1 + y)(1 + z)}\cdot [ABC] \ \implies$](//latex.artofproblemsolving.com/6/b/1/6b180810c9ca902942153a0774ce162484ddc0a3.png)
![$ [MNP] =$](//latex.artofproblemsolving.com/9/8/8/9885bae70fa4e7708d7c76e61005bb0e13119504.png)
![$ \frac {1 + \frac {1}{mnpxyz}}{\frac {[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}{mnpxyz(1 + x)(1 + y)(1 + z)}}\cdot\frac {1 + xyz}{(1 + x)(1 + y)(1 + z)}\ \implies$](//latex.artofproblemsolving.com/1/d/6/1d6f2855170e0e7ea6d94057e1c1886fbe3d08b5.png)
![$ \frac {(1 + xyz)(1 + mnpxyz)}{[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}\cdot [ABC]$](//latex.artofproblemsolving.com/c/9/e/c9e78704f394fce961e02f5449db757144a55535.png)
.
and the points 
, 
, 
so that 
. Prove that ![$\frac {[PQR]}{[ABC]}=\frac {m^3+n^3}{(m+n)^3}$](http://latex.artofproblemsolving.com/7/4/6/74629ddaf757059e3117a80d793755a263e04a76.png)
. 
yields ![$\frac{[PQR]}{[ABC]}=\frac{1+\frac{m}{n} \cdot \frac{m}{n} \cdot \frac{m}{n}}{(1+\frac{m}{n})(1+\frac{m}{n})(1+\frac{m}{n})}=\frac{m^3+n^3}{(m+n)^3}$](http://latex.artofproblemsolving.com/1/b/e/1bea7496a9444019a039378c1a91a05aa7c8be05.png)
.
be a triangle with ![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
. Then ![$ \left\|\begin{array}{ccc} M\in (BC) & , & \frac {MB}{MC} = m \\
\\
N\in (CA) & , & \frac {NC}{NA} = n \\
\\
P\in (AB) & , & \frac {NA}{NB} = n\end{array}\right\|\ \implies\ [MNP] $](http://latex.artofproblemsolving.com/f/a/8/fa88e75ba23f707cdb009ff5ac08d74e8c896e25.png)
.
. Observe that ![$ \frac {[MNP]}{[ABC]} = \frac {[ABC] - [APN] - [BMP] - [CNM]}{[ABC]} =$](http://latex.artofproblemsolving.com/b/d/a/bdad1b627dde889e91f146be1e832c84d13a01af.png)
.
(sideline !) and 
(sideline !), then ![$ \frac {[AXY]}{[ABC]} = \frac {AX}{AB}\cdot\frac {AY}{AC}$](http://latex.artofproblemsolving.com/b/d/0/bd096649d5a13d33df1ea33b229243803bda69b5.png)
. If 
![$\frac {[MNP]}{[ABC]} = \frac {1 + m^3}{(1 + m)^3} = \frac {r^3 + s^3}{(r + s)^3}$](http://latex.artofproblemsolving.com/d/8/1/d81e9f2d08aaf2f502058ce334ee721fd1e49bdb.png)
(the proposed problem). Otherwise. 
![$[APN]=[BMP]=$](http://latex.artofproblemsolving.com/9/6/d/96d0ec390f12acf1a2954429f1704f9dea7e637f.png)
![$[CNM]=\frac {rs}{(r+s)^2}\cdot [ABC]\ \implies$](http://latex.artofproblemsolving.com/9/0/9/909b4d50d4f2417aa358c88fca3ea2b9052c8baa.png)
![$ \frac {[MNP]}{[ABC]}=1-\frac {3rs}{(r+s)^2}=$](http://latex.artofproblemsolving.com/9/d/d/9dd0be17ee7d114ede5fc63d56219111cf928232.png)
![$ \frac {r^2+s^2-rs}{(r+s)^2}\ \implies\ [MNP]=$](http://latex.artofproblemsolving.com/6/c/9/6c91031814502272221bc8f47fdf053728d0e1d6.png)
![$ \frac {r^3+s^3}{(r+s)^3}\cdot [ABC]$](http://latex.artofproblemsolving.com/f/f/6/ff68931d513cec1a3d762eb007a17b488239738f.png)
.
![$ \boxed {\ [MNP] =\frac {(1 + xyz)(1 + mnpxyz)}{[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}\cdot [ABC]\ }$](http://latex.artofproblemsolving.com/9/7/7/9771f549337f8e363bd12611202e4e9eaf1763ed.png)
.![$ x = y = z = 1\ \implies\ 4\cdot [MNP] = [DEF]$](http://latex.artofproblemsolving.com/8/1/1/8114ac1cf64a41e63b333b1e419bec6005a8dcde.png)
(the Pohoatza's problem).
and![$ [MNP] = \frac {1 + uvw}{(1 + u)(1 + v)(1 + w)}\cdot [XYZ] =$](http://latex.artofproblemsolving.com/b/a/a/baa427bae61dcc6addc19153e06969af4c520bdb.png)
![$ \frac {1 + uvw}{(1 + u)(1 + v)(1 + w)}\cdot\frac {1 + xyz}{(1 + x)(1 + y)(1 + z)}\cdot [ABC] \ \implies$](http://latex.artofproblemsolving.com/6/b/1/6b180810c9ca902942153a0774ce162484ddc0a3.png)
![$ [MNP] =$](http://latex.artofproblemsolving.com/9/8/8/9885bae70fa4e7708d7c76e61005bb0e13119504.png)
![$ \frac {1 + \frac {1}{mnpxyz}}{\frac {[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}{mnpxyz(1 + x)(1 + y)(1 + z)}}\cdot\frac {1 + xyz}{(1 + x)(1 + y)(1 + z)}\ \implies$](http://latex.artofproblemsolving.com/1/d/6/1d6f2855170e0e7ea6d94057e1c1886fbe3d08b5.png)
![$ \frac {(1 + xyz)(1 + mnpxyz)}{[mz(1 + y) + (1 + z)]\cdot [nx(1 + z) + (1 + x)]\cdot [py(1 + x) + (1 + y)]}\cdot [ABC]$](http://latex.artofproblemsolving.com/c/9/e/c9e78704f394fce961e02f5449db757144a55535.png)
.
August 2018
January 2018