Constructive Counting+Tidbits on life
by shiningsunnyday, Apr 24, 2016, 9:10 AM
It's Sunday afternoon again!!
Below are some selected favorites regarding constructive counting, a really fascinating flavor of problems. I will post my (or that of the official solution) thought process for each problem. The theme here is:
Play with small numbers, look for patterns, and prove your results.
Your 'Average' Problem
A Gem
Also, on a completely random note, I convinced my dad to buy me a treadmill!
Also, on another completely random note, I had the coolest/weirdest dream
...and I still haven't touched the endless heaps of schoolwork for the weekend yet (literally 10-20 hours worth of work).
Me yesterday:
Should I start doing the homework? Oh god. Chinese past due recording, vocabulary test studying. English project, and studying for the literary analysis exam this week. APUSH exam countdown, and I'm only at WW II, possibly another socratic seminar coming up. Biology homework? Are you kidding me. PE project? Oh god oh god haven't even started on my journal yet (supposed to have started weeks ago). Loads of AP Calc review worksheets as well as studying for retake/upcoming free response. Still haven't talked to my counselor about AP Chem for next year. Also there's some stuff going on in my math club that need to be done. *Slowly and despairingly heads toward my schoolbag*
Interm CP: Are you going to leave me again? After that amazing night you gave me yesterday?
Yea we definitely made some nice one-to-one correspondences! But...
106: You've been solving a lot of my collinearity problems and for god's sake now you can't think straight?
Menelaus, Ceva, and I totally demolished that problem of yours! But ugh...
108:
Proof: Trivial
Well life's full of inequalities... I wish I could only deal yours...
104: I know your life is composite of many factors, but now is a prime time fer mat h!
Haha nice one. I really wanna oil it up with some Euler's. But wait...
After wrestling with myself for a long time, I watched this: https://www.youtube.com/watch?v=qOaqiCBum2w (great channel, by the way).
Then I spent the rest of the day doing math.
...And now I'm freaking out (it's 5 PM and 0/10-20 hours schoolwork done)
P.S: Do you guys find my personifying of math books weird? If it is I guess I'll keep it to myself in the future...
Me yesterday:
Should I start doing the homework? Oh god. Chinese past due recording, vocabulary test studying. English project, and studying for the literary analysis exam this week. APUSH exam countdown, and I'm only at WW II, possibly another socratic seminar coming up. Biology homework? Are you kidding me. PE project? Oh god oh god haven't even started on my journal yet (supposed to have started weeks ago). Loads of AP Calc review worksheets as well as studying for retake/upcoming free response. Still haven't talked to my counselor about AP Chem for next year. Also there's some stuff going on in my math club that need to be done. *Slowly and despairingly heads toward my schoolbag*
Interm CP: Are you going to leave me again? After that amazing night you gave me yesterday?
Yea we definitely made some nice one-to-one correspondences! But...
106: You've been solving a lot of my collinearity problems and for god's sake now you can't think straight?
Menelaus, Ceva, and I totally demolished that problem of yours! But ugh...
108:
Proof: TrivialWell life's full of inequalities... I wish I could only deal yours...
104: I know your life is composite of many factors, but now is a prime time fer mat h!
Haha nice one. I really wanna oil it up with some Euler's. But wait...
After wrestling with myself for a long time, I watched this: https://www.youtube.com/watch?v=qOaqiCBum2w (great channel, by the way).
Then I spent the rest of the day doing math.
...And now I'm freaking out (it's 5 PM and 0/10-20 hours schoolwork done)
P.S: Do you guys find my personifying of math books weird? If it is I guess I'll keep it to myself in the future...
Below are some selected favorites regarding constructive counting, a really fascinating flavor of problems. I will post my (or that of the official solution) thought process for each problem. The theme here is:
Play with small numbers, look for patterns, and prove your results.
Old Putnam wrote:
Let 
be the number of subsets 
of 
such that the elements of have an integer average. Prove that 
is even.
be the number of subsets 
of 
such that the elements of have an integer average. Prove that 
is even.First I had the crazy idea of actually computing the number of subsets and see if some pattern pops up. Then I realized we only need to prove is even. Let's play around with 
Say we have
The average is 
. Hey! That immediately gives us another set: 
Evenness makes us think of parity, and what better thing to do then try to pair them up? In fact, we'll try to establish an one-to-one correspondence.
Say we have a set with an integer average
WLOG assume its average, 
isn't in the set yet, note that: 
This means that every set with an integer average can be paired up with another set with the average removed.
Therefore, must be even, right? Wait! We operated under the assumption that the average wasn't contained in the set, but what if the set only has one element? That's just 
ways, which is odd when is odd, even when is even. So we're done.
Remark
is even. Let's play around with 
Say we have
The average is 
. Hey! That immediately gives us another set: 
Evenness makes us think of parity, and what better thing to do then try to pair them up? In fact, we'll try to establish an one-to-one correspondence.Say we have a set with an integer average
WLOG assume its average, 
isn't in the set yet, note that: 
This means that every set with an integer average can be paired up with another set with the average removed.
Therefore,
must be even, right? Wait! We operated under the assumption that the average wasn't contained in the set, but what if the set only has one element? That's just 
ways, which is odd when is odd, even when is even. So we're done.Remark
This reminded me of the inductive proof for 
interesting how multi-disciplinary ideas come up often.
interesting how multi-disciplinary ideas come up often.Old Canada MO wrote:
How many parallelograms can be counted in an equilateral triangle grid of length 
Below is an example for 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.00cm);
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 = -4.3, xmax = 11.48, ymin = -3.7, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((4.16,5.187818472546175)--(8.46,-2.26));
draw((8.46,-2.26)--(-0.14,-2.26));
draw((-0.14,-2.26)--(4.16,5.187818472546175));
draw((3.085,3.325863854409631)--(5.235,3.325863854409631));
draw((2.01,1.4639092362730874)--(6.31,1.4639092362730874));
draw((0.935,-0.39804538186345617)--(7.385,-0.39804538186345617));
draw((5.235,3.325863854409631)--(2.01,-2.26));
draw((6.31,1.4639092362730874)--(4.16,-2.26));
draw((7.385,-0.39804538186345617)--(6.31,-2.26));
draw((3.085,3.325863854409631)--(6.31,-2.26));
draw((2.01,1.4639092362730874)--(4.16,-2.26));
draw((0.935,-0.39804538186345617)--(2.01,-2.26));
/* dots and labels */
dot((-0.14,-2.26),linewidth(3.pt) + dotstyle);
dot((8.46,-2.26),dotstyle);
dot((4.16,5.187818472546175),linewidth(3.pt) + dotstyle);
dot((2.01,1.4639092362730874),linewidth(3.pt) + dotstyle);
dot((4.16,-2.26),dotstyle);
dot((6.31,1.4639092362730874),linewidth(3.pt) + dotstyle);
dot((3.085,3.325863854409631),linewidth(3.pt) + dotstyle);
dot((5.235,3.325863854409631),linewidth(3.pt) + dotstyle);
dot((7.385,-0.39804538186345617),linewidth(3.pt) + dotstyle);
dot((6.31,-2.26),linewidth(3.pt) + dotstyle);
dot((2.01,-2.26),linewidth(3.pt) + dotstyle);
dot((0.935,-0.39804538186345617),linewidth(3.pt) + dotstyle);
dot((3.085,-0.3980453818634561),dotstyle);
dot((4.16,1.463909236273087),dotstyle);
dot((5.235,-0.3980453818634561),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](//latex.artofproblemsolving.com/8/c/f/8cf5cd06d76a01cf5b634fef14253931b0a04ae5.png)
First, an easier problem
The trick to that problem is to choose two of
and two of the 
vertices. When you connect them, you will find a correspondence between 
rectangles. And that cracks the problem.
How about this? My first intuition was to see if we can select points on the exterior in a similar way to form a parallelogram. How about something like this?
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.00cm);
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 = -4.3, xmax = 11.48, ymin = -3.7, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((3.98,5.907818472546172)--(8.28,-1.54));
draw((8.28,-1.54)--(-0.32,-1.54));
draw((-0.32,-1.54)--(3.98,5.907818472546172));
draw((2.905,4.045863854409629)--(5.055,4.045863854409629));
draw((1.83,2.183909236273086)--(6.13,2.183909236273086));
draw((0.755,0.3219546181365429)--(7.205,0.3219546181365429));
draw((5.055,4.045863854409629)--(1.83,-1.54), blue);
draw((6.13,2.183909236273086)--(3.98,-1.54), blue);
draw((7.205,0.3219546181365429)--(6.13,-1.54));
draw((2.905,4.045863854409629)--(6.13,-1.54), green);
draw((1.83,2.183909236273086)--(3.98,-1.54), green);
draw((0.755,0.3219546181365429)--(1.83,-1.54));
/* dots and labels */
dot((-0.32,-1.54),dotstyle);
dot((8.28,-1.54),dotstyle);
dot((3.98,5.907818472546172),linewidth(3.pt) + dotstyle);
dot((1.83,2.183909236273086),linewidth(3.pt) + dotstyle);
dot((3.98,-1.54),linewidth(3.pt) + dotstyle);
dot((6.13,2.183909236273086),linewidth(3.pt) + dotstyle);
dot((2.905,4.045863854409629),linewidth(3.pt) + dotstyle);
dot((5.055,4.045863854409629),linewidth(3.pt) + dotstyle);
dot((7.205,0.3219546181365429),linewidth(3.pt) + dotstyle);
dot((6.13,-1.54),linewidth(3.pt) + dotstyle);
dot((1.83,-1.54),linewidth(3.pt) + dotstyle);
dot((0.755,0.3219546181365429),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](//latex.artofproblemsolving.com/8/0/5/8052329c73c0dacd43e3ac13c7b2073d20fd2c29.png)
The intuition behind such a choosing is that first, a parallelogram has four sides, so it probably mandates choosing four points. Also, because we need two pairs of parallel sides, one way we can accomplish that is by extending lines from two points each on two sides.
Sure, picking two points on each of two sides might work, but what if this situation occurs?
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.00cm);
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 = -4.3, xmax = 11.48, ymin = -3.7, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((4.38,6.7478184725461725)--(8.68,-0.7));
draw((8.68,-0.7)--(0.08,-0.7));
draw((0.08,-0.7)--(4.38,6.7478184725461725));
draw((3.305,4.88586385440963)--(5.455,4.88586385440963));
draw((2.23,3.023909236273086)--(6.53,3.023909236273086), green);
draw((1.155,1.161954618136543)--(7.605,1.161954618136543), green);
draw((5.455,4.88586385440963)--(2.23,-0.7), blue);
draw((6.53,3.023909236273086)--(4.38,-0.7));
draw((7.605,1.161954618136543)--(6.53,-0.7), blue);
draw((3.305,4.88586385440963)--(6.53,-0.7));
draw((2.23,3.023909236273086)--(4.38,-0.7));
draw((1.155,1.161954618136543)--(2.23,-0.7));
draw((6.53,3.023909236273086)--(8.68,3.02390923627309), linetype("4 4") + green);
draw((8.68,3.02390923627309)--(7.605,1.161954618136543), linetype("4 4") + blue);
/* dots and labels */
dot((0.08,-0.7),dotstyle);
dot((8.68,-0.7),dotstyle);
dot((4.38,6.7478184725461725),linewidth(3.pt) + dotstyle);
dot((2.23,3.023909236273086),linewidth(3.pt) + dotstyle);
dot((4.38,-0.7),linewidth(3.pt) + dotstyle);
dot((6.53,3.023909236273086),linewidth(3.pt) + dotstyle);
dot((3.305,4.88586385440963),linewidth(3.pt) + dotstyle);
dot((5.455,4.88586385440963),linewidth(3.pt) + dotstyle);
dot((7.605,1.161954618136543),linewidth(3.pt) + dotstyle);
dot((6.53,-0.7),linewidth(3.pt) + dotstyle);
dot((2.23,-0.7),linewidth(3.pt) + dotstyle);
dot((1.155,1.161954618136543),linewidth(3.pt) + dotstyle);
dot((8.68,3.02390923627309),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](//latex.artofproblemsolving.com/7/7/2/7721ca3f91a4b58ff89287a170c2428ccf5429a1.png)
Uh-oh. This leads to the parallelogram extending outside the equilateral triangle, which we can't have. In addition, note that the green lines extend to the right as opposed to down, like above. There're simply so many restrictions we have to incorporate if we want to keep going down this road. Picking points on two sides can get nasty. What if we can make a parallelogram by only picking points on one side?
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.004261515405575cm);
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 = -3.585906140754743, xmax = 11.018355374650833, ymin = -2.696739164831635, ymax = 6.5581794127384425; /* image dimensions */
/* draw figures */
draw((4.,5.547818472546173)--(8.3,-1.9));
draw((8.3,-1.9)--(-0.3,-1.9), blue);
draw((-0.3,-1.9)--(4.,5.547818472546173), green);
draw((2.925,3.68586385440963)--(5.075,3.68586385440963));
draw((1.85,1.8239092362730867)--(6.15,1.8239092362730867));
draw((0.775,-0.03804538186345663)--(7.225,-0.03804538186345663), blue);
draw((5.075,3.68586385440963)--(1.85,-1.9), green);
draw((6.15,1.8239092362730867)--(4.,-1.9));
draw((7.225,-0.03804538186345663)--(6.15,-1.9));
draw((2.925,3.68586385440963)--(6.15,-1.9));
draw((1.85,1.8239092362730867)--(4.,-1.9));
draw((0.775,-0.03804538186345663)--(1.85,-1.9));
/* dots and labels */
dot((-0.3,-1.9),dotstyle);
dot((8.3,-1.9),dotstyle);
label("$D$", (8.371448661465791,-1.715717795609207), NE * labelscalefactor);
dot((4.,5.547818472546173),linewidth(3.pt) + dotstyle);
label("$A$", (4.095676278628418,5.743746577912276), NE * labelscalefactor);
dot((1.85,1.8239092362730867),linewidth(3.pt) + dotstyle);
dot((4.,-1.9),linewidth(3.pt) + dotstyle);
dot((6.15,1.8239092362730867),linewidth(3.pt) + dotstyle);
dot((2.925,3.68586385440963),linewidth(3.pt) + dotstyle);
dot((5.075,3.68586385440963),linewidth(3.pt) + dotstyle);
label("$B$", (5.1877566707816865,3.9297825367085406), NE * labelscalefactor);
dot((7.225,-0.03804538186345663),linewidth(3.pt) + dotstyle);
label("$C$", (7.297878106467664,0.07973640843938816), NE * labelscalefactor);
dot((6.15,-1.9),linewidth(3.pt) + dotstyle);
dot((1.85,-1.9),dotstyle);
dot((0.775,-0.03804538186345663),dotstyle);
dot((2.925,-0.03804538186345663),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](//latex.artofproblemsolving.com/8/1/c/81c5181b7b84ca68c832626bd033c093fd4f97e9.png)
Aha! This is looking good. WLOG let's count only the case involving one of the three sides (we can multiply by three at the end). Note that whenever we have four points, we can always use the first two points to extend downwards, and the last two points to extend leftwards, forming a parallelogram. So is the answer simply
(Note it's because there're vertices) What if this happens?
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.058437764831375cm);
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 = -3.783806193687872, xmax = 11.274631571143503, ymin = -2.5097681721962575, ymax = 7.032968053932643; /* image dimensions */
/* draw figures */
draw((3.98,5.907818472546172)--(8.28,-1.54));
draw((8.28,-1.54)--(-0.32,-1.54));
draw((-0.32,-1.54)--(3.98,5.907818472546172));
draw((2.905,4.045863854409629)--(5.055,4.045863854409629));
draw((1.83,2.183909236273086)--(6.13,2.183909236273086), blue);
draw((0.755,0.3219546181365429)--(7.205,0.3219546181365429), blue);
draw((5.055,4.045863854409629)--(1.83,-1.54), green);
draw((6.13,2.183909236273086)--(3.98,-1.54), green);
draw((7.205,0.3219546181365429)--(6.13,-1.54));
draw((2.905,4.045863854409629)--(6.13,-1.54));
draw((1.83,2.183909236273086)--(3.98,-1.54));
draw((0.755,0.3219546181365429)--(1.83,-1.54));
/* dots and labels */
dot((-0.32,-1.54),dotstyle);
dot((8.28,-1.54),dotstyle);
dot((3.98,5.907818472546172),linewidth(3.pt) + dotstyle);
dot((1.83,2.183909236273086),linewidth(3.pt) + dotstyle);
dot((3.98,-1.54),linewidth(3.pt) + dotstyle);
dot((6.13,2.183909236273086),dotstyle);
label("$B$", (6.2551523161997125,2.452454665390771), NE * labelscalefactor);
dot((2.905,4.045863854409629),linewidth(3.pt) + dotstyle);
dot((5.055,4.045863854409629),linewidth(3.pt) + dotstyle);
label("$A$", (5.1672803864210195,4.28466002080752), NE * labelscalefactor);
dot((7.205,0.3219546181365429),linewidth(3.pt) + dotstyle);
label("$C$", (7.323938773526147,0.44848005790370166), NE * labelscalefactor);
dot((6.13,-1.54),linewidth(3.pt) + dotstyle);
dot((1.83,-1.54),linewidth(3.pt) + dotstyle);
dot((0.755,0.3219546181365429),linewidth(3.pt) + dotstyle);
dot((3.98,2.183909236273086),dotstyle);
dot((2.905,0.3219546181365429),dotstyle);
dot((5.055,0.32195461813654286),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](//latex.artofproblemsolving.com/b/7/7/b77d40685a42062a17178eb6a460bd0e3b55b2ab.png)
In this case, we try to bound the parallelogram first with the green lines and blue lines. The problem, though, is that
in this case assumes the role of 
from last diagram. Because this parallelogram touches a side of the equilateral triangle, we only need three points to bound it, the first two extending southwest and the last two extending west.
We can verify such a correspondence is unique, and our answer is
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.00cm);
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 = -4.3, xmax = 11.48, ymin = -3.7, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((4.16,5.187818472546175)--(8.46,-2.26));
draw((8.46,-2.26)--(-0.14,-2.26));
draw((-0.14,-2.26)--(4.16,5.187818472546175));
draw((3.085,3.325863854409631)--(5.235,3.325863854409631));
draw((2.01,1.4639092362730874)--(6.31,1.4639092362730874));
draw((0.935,-0.39804538186345617)--(7.385,-0.39804538186345617));
draw((5.235,3.325863854409631)--(2.01,-2.26));
draw((6.31,1.4639092362730874)--(4.16,-2.26));
draw((7.385,-0.39804538186345617)--(6.31,-2.26));
draw((3.085,3.325863854409631)--(6.31,-2.26));
draw((2.01,1.4639092362730874)--(4.16,-2.26));
draw((0.935,-0.39804538186345617)--(2.01,-2.26));
/* dots and labels */
dot((-0.14,-2.26),linewidth(3.pt) + dotstyle);
dot((8.46,-2.26),dotstyle);
dot((4.16,5.187818472546175),linewidth(3.pt) + dotstyle);
dot((2.01,1.4639092362730874),linewidth(3.pt) + dotstyle);
dot((4.16,-2.26),dotstyle);
dot((6.31,1.4639092362730874),linewidth(3.pt) + dotstyle);
dot((3.085,3.325863854409631),linewidth(3.pt) + dotstyle);
dot((5.235,3.325863854409631),linewidth(3.pt) + dotstyle);
dot((7.385,-0.39804538186345617),linewidth(3.pt) + dotstyle);
dot((6.31,-2.26),linewidth(3.pt) + dotstyle);
dot((2.01,-2.26),linewidth(3.pt) + dotstyle);
dot((0.935,-0.39804538186345617),linewidth(3.pt) + dotstyle);
dot((3.085,-0.3980453818634561),dotstyle);
dot((4.16,1.463909236273087),dotstyle);
dot((5.235,-0.3980453818634561),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/8/c/f/8cf5cd06d76a01cf5b634fef14253931b0a04ae5.png)
First, an easier problem
How many rectangles can be counted among a 
by lattice grid?
by lattice grid?The trick to that problem is to choose two of
and two of the 
vertices. When you connect them, you will find a correspondence between 
rectangles. And that cracks the problem.How about this? My first intuition was to see if we can select points on the exterior in a similar way to form a parallelogram. How about something like this?
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.00cm);
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 = -4.3, xmax = 11.48, ymin = -3.7, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((3.98,5.907818472546172)--(8.28,-1.54));
draw((8.28,-1.54)--(-0.32,-1.54));
draw((-0.32,-1.54)--(3.98,5.907818472546172));
draw((2.905,4.045863854409629)--(5.055,4.045863854409629));
draw((1.83,2.183909236273086)--(6.13,2.183909236273086));
draw((0.755,0.3219546181365429)--(7.205,0.3219546181365429));
draw((5.055,4.045863854409629)--(1.83,-1.54), blue);
draw((6.13,2.183909236273086)--(3.98,-1.54), blue);
draw((7.205,0.3219546181365429)--(6.13,-1.54));
draw((2.905,4.045863854409629)--(6.13,-1.54), green);
draw((1.83,2.183909236273086)--(3.98,-1.54), green);
draw((0.755,0.3219546181365429)--(1.83,-1.54));
/* dots and labels */
dot((-0.32,-1.54),dotstyle);
dot((8.28,-1.54),dotstyle);
dot((3.98,5.907818472546172),linewidth(3.pt) + dotstyle);
dot((1.83,2.183909236273086),linewidth(3.pt) + dotstyle);
dot((3.98,-1.54),linewidth(3.pt) + dotstyle);
dot((6.13,2.183909236273086),linewidth(3.pt) + dotstyle);
dot((2.905,4.045863854409629),linewidth(3.pt) + dotstyle);
dot((5.055,4.045863854409629),linewidth(3.pt) + dotstyle);
dot((7.205,0.3219546181365429),linewidth(3.pt) + dotstyle);
dot((6.13,-1.54),linewidth(3.pt) + dotstyle);
dot((1.83,-1.54),linewidth(3.pt) + dotstyle);
dot((0.755,0.3219546181365429),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/8/0/5/8052329c73c0dacd43e3ac13c7b2073d20fd2c29.png)
The intuition behind such a choosing is that first, a parallelogram has four sides, so it probably mandates choosing four points. Also, because we need two pairs of parallel sides, one way we can accomplish that is by extending lines from two points each on two sides.
Sure, picking two points on each of two sides might work, but what if this situation occurs?
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.00cm);
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 = -4.3, xmax = 11.48, ymin = -3.7, ymax = 6.3; /* image dimensions */
/* draw figures */
draw((4.38,6.7478184725461725)--(8.68,-0.7));
draw((8.68,-0.7)--(0.08,-0.7));
draw((0.08,-0.7)--(4.38,6.7478184725461725));
draw((3.305,4.88586385440963)--(5.455,4.88586385440963));
draw((2.23,3.023909236273086)--(6.53,3.023909236273086), green);
draw((1.155,1.161954618136543)--(7.605,1.161954618136543), green);
draw((5.455,4.88586385440963)--(2.23,-0.7), blue);
draw((6.53,3.023909236273086)--(4.38,-0.7));
draw((7.605,1.161954618136543)--(6.53,-0.7), blue);
draw((3.305,4.88586385440963)--(6.53,-0.7));
draw((2.23,3.023909236273086)--(4.38,-0.7));
draw((1.155,1.161954618136543)--(2.23,-0.7));
draw((6.53,3.023909236273086)--(8.68,3.02390923627309), linetype("4 4") + green);
draw((8.68,3.02390923627309)--(7.605,1.161954618136543), linetype("4 4") + blue);
/* dots and labels */
dot((0.08,-0.7),dotstyle);
dot((8.68,-0.7),dotstyle);
dot((4.38,6.7478184725461725),linewidth(3.pt) + dotstyle);
dot((2.23,3.023909236273086),linewidth(3.pt) + dotstyle);
dot((4.38,-0.7),linewidth(3.pt) + dotstyle);
dot((6.53,3.023909236273086),linewidth(3.pt) + dotstyle);
dot((3.305,4.88586385440963),linewidth(3.pt) + dotstyle);
dot((5.455,4.88586385440963),linewidth(3.pt) + dotstyle);
dot((7.605,1.161954618136543),linewidth(3.pt) + dotstyle);
dot((6.53,-0.7),linewidth(3.pt) + dotstyle);
dot((2.23,-0.7),linewidth(3.pt) + dotstyle);
dot((1.155,1.161954618136543),linewidth(3.pt) + dotstyle);
dot((8.68,3.02390923627309),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/7/7/2/7721ca3f91a4b58ff89287a170c2428ccf5429a1.png)
Uh-oh. This leads to the parallelogram extending outside the equilateral triangle, which we can't have. In addition, note that the green lines extend to the right as opposed to down, like above. There're simply so many restrictions we have to incorporate if we want to keep going down this road. Picking points on two sides can get nasty. What if we can make a parallelogram by only picking points on one side?
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.004261515405575cm);
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 = -3.585906140754743, xmax = 11.018355374650833, ymin = -2.696739164831635, ymax = 6.5581794127384425; /* image dimensions */
/* draw figures */
draw((4.,5.547818472546173)--(8.3,-1.9));
draw((8.3,-1.9)--(-0.3,-1.9), blue);
draw((-0.3,-1.9)--(4.,5.547818472546173), green);
draw((2.925,3.68586385440963)--(5.075,3.68586385440963));
draw((1.85,1.8239092362730867)--(6.15,1.8239092362730867));
draw((0.775,-0.03804538186345663)--(7.225,-0.03804538186345663), blue);
draw((5.075,3.68586385440963)--(1.85,-1.9), green);
draw((6.15,1.8239092362730867)--(4.,-1.9));
draw((7.225,-0.03804538186345663)--(6.15,-1.9));
draw((2.925,3.68586385440963)--(6.15,-1.9));
draw((1.85,1.8239092362730867)--(4.,-1.9));
draw((0.775,-0.03804538186345663)--(1.85,-1.9));
/* dots and labels */
dot((-0.3,-1.9),dotstyle);
dot((8.3,-1.9),dotstyle);
label("$D$", (8.371448661465791,-1.715717795609207), NE * labelscalefactor);
dot((4.,5.547818472546173),linewidth(3.pt) + dotstyle);
label("$A$", (4.095676278628418,5.743746577912276), NE * labelscalefactor);
dot((1.85,1.8239092362730867),linewidth(3.pt) + dotstyle);
dot((4.,-1.9),linewidth(3.pt) + dotstyle);
dot((6.15,1.8239092362730867),linewidth(3.pt) + dotstyle);
dot((2.925,3.68586385440963),linewidth(3.pt) + dotstyle);
dot((5.075,3.68586385440963),linewidth(3.pt) + dotstyle);
label("$B$", (5.1877566707816865,3.9297825367085406), NE * labelscalefactor);
dot((7.225,-0.03804538186345663),linewidth(3.pt) + dotstyle);
label("$C$", (7.297878106467664,0.07973640843938816), NE * labelscalefactor);
dot((6.15,-1.9),linewidth(3.pt) + dotstyle);
dot((1.85,-1.9),dotstyle);
dot((0.775,-0.03804538186345663),dotstyle);
dot((2.925,-0.03804538186345663),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/8/1/c/81c5181b7b84ca68c832626bd033c093fd4f97e9.png)
Aha! This is looking good. WLOG let's count only the case involving one of the three sides (we can multiply by three at the end). Note that whenever we have four points, we can always use the first two points to extend downwards, and the last two points to extend leftwards, forming a parallelogram. So is the answer simply
(Note it's because there're vertices) What if this happens?![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(6.058437764831375cm);
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 = -3.783806193687872, xmax = 11.274631571143503, ymin = -2.5097681721962575, ymax = 7.032968053932643; /* image dimensions */
/* draw figures */
draw((3.98,5.907818472546172)--(8.28,-1.54));
draw((8.28,-1.54)--(-0.32,-1.54));
draw((-0.32,-1.54)--(3.98,5.907818472546172));
draw((2.905,4.045863854409629)--(5.055,4.045863854409629));
draw((1.83,2.183909236273086)--(6.13,2.183909236273086), blue);
draw((0.755,0.3219546181365429)--(7.205,0.3219546181365429), blue);
draw((5.055,4.045863854409629)--(1.83,-1.54), green);
draw((6.13,2.183909236273086)--(3.98,-1.54), green);
draw((7.205,0.3219546181365429)--(6.13,-1.54));
draw((2.905,4.045863854409629)--(6.13,-1.54));
draw((1.83,2.183909236273086)--(3.98,-1.54));
draw((0.755,0.3219546181365429)--(1.83,-1.54));
/* dots and labels */
dot((-0.32,-1.54),dotstyle);
dot((8.28,-1.54),dotstyle);
dot((3.98,5.907818472546172),linewidth(3.pt) + dotstyle);
dot((1.83,2.183909236273086),linewidth(3.pt) + dotstyle);
dot((3.98,-1.54),linewidth(3.pt) + dotstyle);
dot((6.13,2.183909236273086),dotstyle);
label("$B$", (6.2551523161997125,2.452454665390771), NE * labelscalefactor);
dot((2.905,4.045863854409629),linewidth(3.pt) + dotstyle);
dot((5.055,4.045863854409629),linewidth(3.pt) + dotstyle);
label("$A$", (5.1672803864210195,4.28466002080752), NE * labelscalefactor);
dot((7.205,0.3219546181365429),linewidth(3.pt) + dotstyle);
label("$C$", (7.323938773526147,0.44848005790370166), NE * labelscalefactor);
dot((6.13,-1.54),linewidth(3.pt) + dotstyle);
dot((1.83,-1.54),linewidth(3.pt) + dotstyle);
dot((0.755,0.3219546181365429),linewidth(3.pt) + dotstyle);
dot((3.98,2.183909236273086),dotstyle);
dot((2.905,0.3219546181365429),dotstyle);
dot((5.055,0.32195461813654286),dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]](http://latex.artofproblemsolving.com/b/7/7/b77d40685a42062a17178eb6a460bd0e3b55b2ab.png)
In this case, we try to bound the parallelogram first with the green lines and blue lines. The problem, though, is that
in this case assumes the role of 
from last diagram. Because this parallelogram touches a side of the equilateral triangle, we only need three points to bound it, the first two extending southwest and the last two extending west.We can verify such a correspondence is unique, and our answer is
Also, on a completely random note, I convinced my dad to buy me a treadmill!
Wait that sounds familiar
Now time to shed some of that fat...Also, on another completely random note, I had the coolest/weirdest dream
So I was going on an airplane, and I somehow managed to get into business class. All the business class passengers sat in a circle room with a one single strip of window along the circumference of the room. I found my seat, saw a really mean woman in it, and she yelled at me, despite it being my seat. Then I realized the room was actually missing a seat, so I yelled furiously at the flight attendant. "Do you want me to sit on the ground?" Then I got promoted to first class, and I was so happy. So I squeezed myself through the super-narrow stairs leading into first class, and it is actually a circle-shaped cavern that's infinitely deep downwards. So everyone was standing around the circumference of the edge of the cliff, in a circle, surrounding the abyss. Then a music came on and all the first-class passengers began dancing. I found myself dancing next to ALL of my old crushes, and I had the best time of my life, dancing and getting loved. I danced so hard I fell into a lake out of nowhere along with everyone else. Everything suddenly was so perfect. Then I woke up. It was noon, and I had slept for 13 hours, realizing I hadn't done anything this weekend, so much work to be done...
People say dreams reflect your thoughts. What does this dream say about me? Hm...
Sorry, I had no idea why I posted this.
last night.People say dreams reflect your thoughts. What does this dream say about me? Hm...
Sorry, I had no idea why I posted this.
Attachments:
This post has been edited 3 times. Last edited by shiningsunnyday, Apr 25, 2016, 5:29 AM
The thing with 1-1 correspondence problems is that they are very hard to figure out but have the most mind-blowing solutions.
May 2018
April 2018