Hmm so I guess I haven't posted here in a while
by djmathman, Jan 19, 2014, 4:50 AM
So... how's January been going for me? Well, interestingly, to say the least. yes that's an adverb, it's grammatically correct... darn you SAT Writing section
I guess this is a bit late, but I placed in Science Olympiad regionals! At the NJIT region, I received a 3rd place medal for Circuit Lab (that was my only event, as Astronomy conflicted) and our team placed sixth overall. The medal came as a complete shock (hehe, get it, electricity, hehe), since my teammate and I totally thought we failed that test. But I guess everyone else failed too, as it was probably a medium State level test, so it's ok. The sixth place finish means that it's our school's first year of qualifying for the State competition without the only-one-team-per-school rule! We're probably going to get killed in States, since powerhouses like West-Windsor Plainsboro, Montgomery, and Princeton weren't there, but hey, there's nothing like getting false hope beforehand, right?
School's been pretty hard so far. All my science courses, such as Chemistry and Physics, have been really easy so far, but all my humanities courses, such as Spanish, AP Lang, and APUSH II, have been incredibly difficult. I just got pummeled by an AP Lang Quarter-Exam as well D:. (Think of it as a college midterm of sorts; that's how hard it can get.)
As for math... idk. I've been continuing to work from 107 Geometry Problems and scattering through the fora/Contest pages for interesting problems. I'm still not at the level of USAMO yet; hopefully I will be when the time comes, of course assuming I qualify which is a nontrivial assumption.
So what specific math problems have I completed during these twenty days of not posting any math on my blog?
I've also decided to go back and look at the British Olympiad Round 1 from 2008 that I printed out a while ago. I was looking through some stuff when I found it, and I wanted to see how much better I've gotten at math since I first found it... maybe two years ago?
I guess this is a bit late, but I placed in Science Olympiad regionals! At the NJIT region, I received a 3rd place medal for Circuit Lab (that was my only event, as Astronomy conflicted) and our team placed sixth overall. The medal came as a complete shock (hehe, get it, electricity, hehe), since my teammate and I totally thought we failed that test. But I guess everyone else failed too, as it was probably a medium State level test, so it's ok. The sixth place finish means that it's our school's first year of qualifying for the State competition without the only-one-team-per-school rule! We're probably going to get killed in States, since powerhouses like West-Windsor Plainsboro, Montgomery, and Princeton weren't there, but hey, there's nothing like getting false hope beforehand, right?
School's been pretty hard so far. All my science courses, such as Chemistry and Physics, have been really easy so far, but all my humanities courses, such as Spanish, AP Lang, and APUSH II, have been incredibly difficult. I just got pummeled by an AP Lang Quarter-Exam as well D:. (Think of it as a college midterm of sorts; that's how hard it can get.)
As for math... idk. I've been continuing to work from 107 Geometry Problems and scattering through the fora/Contest pages for interesting problems. I'm still not at the level of USAMO yet; hopefully I will be when the time comes, of course assuming I qualify which is a nontrivial assumption.
So what specific math problems have I completed during these twenty days of not posting any math on my blog?
All-Russian MO 2001 wrote:
Let the circle 
be internally tangent to another circle 
at 
.Take a point 
on and draw a tangent 
which intersects at 
and 
. Let 
be the midpoint of the arc which is on the opposite side of . Prove that the circumradius of the 
doesnt depend on the choice of .
be internally tangent to another circle 
at 
.Take a point 
on and draw a tangent 
which intersects at 
and 
. Let 
be the midpoint of the arc which is on the opposite side of . Prove that the circumradius of the 
doesnt depend on the choice of .
bisects 
.
to 
sends 
.
,
.
be the ratio of similitude of the two triangles. We wish to show that 
is constant the circumradius of 
will follow to be constant.
.
Thus, we have 
However, note that the homothety centered at 
sending 
to 
.
,
Winter NIMO 2014 wrote:
Let 
be an acute triangle with orthocenter 
and let be the midpoint of 
. (The orthocenter is the point at the intersection of the three altitudes.) Denote by 
the circle passing through 
, , and , and denote by 
the circle passing through 
, , and . Lines 
and 
meet and again at 
and 
, respectively. Rays 
and 
meet and again at 
and 
, respectively. Show that 
and 
have the same area.
be an acute triangle with orthocenter 
and let be the midpoint of 
. (The orthocenter is the point at the intersection of the three altitudes.) Denote by 
the circle passing through 
, , and , and denote by 
the circle passing through 
, , and . Lines 
and 
meet and again at 
and 
, respectively. Rays 
and 
meet and again at 
and 
, respectively. Show that 
and 
have the same area.![[asy]
import olympiad;
size(300);
defaultpen(linewidth(0.8));
pair A=(3,7),B=origin,C=(10,0),Hb=foot(B,A,C),Hc=foot(C,A,B),H=orthocenter(A,B,C),M=C/2;
path wb=circumcircle(H,B,M),wc=circumcircle(H,C,M);
pair P=intersectionpoint(wb,B--A),Q=intersectionpoint(wc,C--A);
pair Rx=10*H-9*P,Rs=2*H-P,R=intersectionpoint(wc,Rs--Rx);
pair Sx=10*H-9*Q,Ss=2*H-Q,S=intersectionpoint(wb,Ss--Sx);
draw(A--B--C--cycle^^C--Hc^^B--Hb^^wb^^wc^^P--R--C^^Q--S--B^^M--H);
draw(S--R,linetype("4 4"));
draw(rightanglemark(B,foot(B,A,C),C)^^rightanglemark(C,foot(C,A,B),B));
pair G=centroid(A,B,C);
label("$A$",A,dir(G--A));
label("$B$",B,dir(G--B));
label("$C$",C,dir(G--C));
label("$M$",M,2*dir(285));
label("$H$",H,2.5*dir(125));
label("$P$",P,dir(170));
label("$Q$",Q,dir(90));
label("$R$",R,dir(H--R));
label("$S$",S,dir(H--S));
[/asy]](http://latex.artofproblemsolving.com/0/8/2/082b60f18a58805f78a5b4aff29bfbf91ea3339c.png)
and 
are both cyclic pentagons, whence![\[\angle APH=\angle BSH=\angle BMH=\angle HQC=\angle HRC.\]](http://latex.artofproblemsolving.com/8/1/7/817d673631382cef8753ee7290eecad188fc25b2.png)
Next, remark that 
implies that 
,
(here 
is the projection of 
,
.
sending 
to 
.
,
.
and 
,![$[BSM]=[MSC]$](http://latex.artofproblemsolving.com/f/c/a/fca3f61ab6468bfd6562d1caec6d1d2899d9492c.png)
and ![$[BRM]=[MRC]$](http://latex.artofproblemsolving.com/8/7/6/8768ad740dc23cd630672c116d9c23dc98e49e39.png)
.![$[BRS]=[CRS]$](http://latex.artofproblemsolving.com/d/9/b/d9b3bfa24e20aa4e37b82e4d7c64ffdfea7d9ecd.png)
,I've also decided to go back and look at the British Olympiad Round 1 from 2008 that I printed out a while ago. I was looking through some stuff when I found it, and I wanted to see how much better I've gotten at math since I first found it... maybe two years ago?
Problem 2 wrote:
Find all real values of 
, 
, and 
such that
![\[(x+1)yz=12,\,\,\,(y+1)zx=4,\,\,\,\text{and}\,\,\, (z+1)xy=4.\]](//latex.artofproblemsolving.com/1/0/e/10ee77616816a04e776dc3adbdfc13bf5c941b2c.png)
, 
, and 
such that![\[(x+1)yz=12,\,\,\,(y+1)zx=4,\,\,\,\text{and}\,\,\, (z+1)xy=4.\]](http://latex.artofproblemsolving.com/1/0/e/10ee77616816a04e776dc3adbdfc13bf5c941b2c.png)
![\[\left.\begin{array}{cc} (x+1)xyz&=12x\\(y+1)xyz&=4y\\(z+1)xyz&=4z\end{array}\right\}\implies \dfrac{3x}{x+1}=\dfrac{y}{y+1}=\dfrac{z}{z+1}.\]](http://latex.artofproblemsolving.com/5/9/a/59ab9539a0b3f02d5387f92570c454387e4c4f0a.png)
Let all three expressions equal 
which immediately implies that 
.
and 
.![\[\dfrac{(x+1)y}{x(y+1)}=3\implies x=\dfrac y{2y+3}.\]](http://latex.artofproblemsolving.com/f/a/9/fa92efde888f82789c98e3f6a3a249676a103b0a.png)
Now substituting this expression for ![\[y^2\left(\dfrac y{2y+3}+1\right)=12\implies (y-3)(y+2)^2=0.\]](http://latex.artofproblemsolving.com/2/d/b/2dbe80576567b025948b1b596fb71e6778bccede.png)
Therefore 
or 
.
and 
respectively, so we have the solutions 
.Problem 3 wrote:
Let 
be a parallelogram such that is an acute-angled triangle. The circumcircle of triangle meets the line 
again at . Prove that 
if, and only if, 
. The circumcircle of a triangle is the circle which passes through its vertices.
be a parallelogram such that is an acute-angled triangle. The circumcircle of triangle meets the line 
again at . Prove that 
if, and only if, 
. The circumcircle of a triangle is the circle which passes through its vertices.
is a cyclic quadrilateral, and since 
,
and 
,
,
.
,
.
,
,Problem 6 wrote:
The obtuse-angled triangle has sides of length 
and 
opposite the angles 
respectively. Prove that
![\[a^3\cos A+b^3\cos B+c^3\cos C<abc.\]](//latex.artofproblemsolving.com/f/e/e/feef5a9a5fbf120385956cfc6de7813ce174a4dd.png)
has sides of length 
and 
opposite the angles 
respectively. Prove that![\[a^3\cos A+b^3\cos B+c^3\cos C<abc.\]](http://latex.artofproblemsolving.com/f/e/e/feef5a9a5fbf120385956cfc6de7813ce174a4dd.png)
.![\[\sum a^3\left(\dfrac{b^2+c^2-a^2}{2bc}\right)<abc\implies \sum a^4(b^2+c^2-a^2)<2a^2b^2c^2.\]](http://latex.artofproblemsolving.com/1/d/d/1dd59cec067f1e79179da16ed57ca25da2f2f032.png)
Let 
.![\[\sum x^2(y+z-x)<2xyz\implies \sum x^2y-\sum x^3<2xyz.\]](http://latex.artofproblemsolving.com/5/4/d/54dbd34cce917d68aa51e148c03e27421bccdbfe.png)
If we can prove this, then we are done.
.
,
Next, I claim that 
.![\begin{align*}z^3-3z(x^2+y^2)+2x^3+2y^2&=z[z^2-3(x^2+y^2)]+2x^3+2y^2\\&>(x+y)[(x+y)^2-3(x^2+y^2)]+2x^3+2y^3\\&=(x+y)[-2x^2+2xy-2y^2]+2x^3+2y^3=0.\end{align*}](http://latex.artofproblemsolving.com/a/7/d/a7d83d03b9bec3e27b26022d03cd1f50cd91678d.png)
Finally, adding 
to the nasty inequality just above gives
Dividing both sides by three gives us the desired inequality, and we are done. This post has been edited 4 times. Last edited by djmathman, Apr 29, 2016, 4:06 AM
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