The 1/7 (of area) triangle theorem.
How knowing *about* Linear Algebra makes this an exciting and doable exercise.
A problem exploration thread (1/14).
How knowing *about* Linear Algebra makes this an exciting and doable exercise.
A problem exploration thread (1/14).
(2/14) The Problem:
Take triangle ABC. Let A’ be the point in BC that’s twice as far from C as from B.
Similarly, B’ ... on CA twice as far from A as C
C’....on AB 2ce as far from B as A
Connect AA’, BB’, CC’. Prove the middle triangle has 1/7 the area of the original, ABC.
Take triangle ABC. Let A’ be the point in BC that’s twice as far from C as from B.
Similarly, B’ ... on CA twice as far from A as C
C’....on AB 2ce as far from B as A
Connect AA’, BB’, CC’. Prove the middle triangle has 1/7 the area of the original, ABC.
(3/14) If you doubt the value of the “Linear Algebra attitude,” try it first, now.
The *only* fact from geometry is that:
The area of a triangle is (1/2)base(height).
This fact will be used freely without mention.
Now I break to shovel 5 million pounds of snow (2 million kg).
The *only* fact from geometry is that:
The area of a triangle is (1/2)base(height).
This fact will be used freely without mention.
Now I break to shovel 5 million pounds of snow (2 million kg).
(4/14) Notation (1st attempt):
(sA) is the area of the small triangle that has A as a vertex. (“s” is for “small”—this helps me at the beginning).
qA is the area of the quadrilateral with A.
Our featured triangle has area T.
(sA) is the area of the small triangle that has A as a vertex. (“s” is for “small”—this helps me at the beginning).
qA is the area of the quadrilateral with A.
Our featured triangle has area T.
(5/14) The Linear Algebra Attitude:
-write down connections
-see if they’re enough to achieve the goal
-if possible, find independent connections
1st connection: 2sA+2sB+2qB =
T+sC+qA+qC
There are 2 other similar equations of the same “type”, as shown
-write down connections
-see if they’re enough to achieve the goal
-if possible, find independent connections
1st connection: 2sA+2sB+2qB =
T+sC+qA+qC
There are 2 other similar equations of the same “type”, as shown
(6/14) This is *not* enough, so we draw a line to break up a quadrilateral into 2 new triangles (“n” is for “new”).
nC’ (respectively nC’’) is the area of a new triangle with C’ (respectively C’’).
Thus qB=nC+nC’’
This suggests a complete refinement of notation, without q:
nC’ (respectively nC’’) is the area of a new triangle with C’ (respectively C’’).
Thus qB=nC+nC’’
This suggests a complete refinement of notation, without q:
(7/14) Adjusted notation means we will redo the type 1 equations, but that’s easy and can wait.
The exciting thing is that we get *two* new types of connections.
First, the “type 2” equations. As usual, the highlighted blue triangle has twice the area as the red one beside it:
The exciting thing is that we get *two* new types of connections.
First, the “type 2” equations. As usual, the highlighted blue triangle has twice the area as the red one beside it:
(8/14) the 3rd type, “type 3 equations” feel lucky at first, as we did not need to draw extra lines to see them!
(If you tried this problem before, try it again now! Also, I have to shovel again, a lot of shoveling—and freezing rain is mixed in it—argh!)
(If you tried this problem before, try it again now! Also, I have to shovel again, a lot of shoveling—and freezing rain is mixed in it—argh!)
(9/14) The Goal: proving (understanding) that
7T=sum of all ten variables
However (because linear algebra!) we rewrite as:
6T-(sum of other nine variables)=0
Arranging variables to make the matrix look nice helps. I’ll omit 0s from it, and leave “The Goal” in red at the top:
7T=sum of all ten variables
However (because linear algebra!) we rewrite as:
6T-(sum of other nine variables)=0
Arranging variables to make the matrix look nice helps. I’ll omit 0s from it, and leave “The Goal” in red at the top:
(10/14) Now to rewrite the “type 1” equations with the new notation, and for the matrix.
I’ll do the first one in detail and just summarize the results in the next tweet. It looks really nice!
I’ll do the first one in detail and just summarize the results in the next tweet. It looks really nice!
(11/14) This symmetric matrix has so much symmetry!
The first row (red) is there to remember the goal; then 3 rows of type 1 equations, 3 rows of type 2 equations (rewritten) and 3 rows of type 3 equations.
Adding all type 1 and type 2 equations together gets us *really close*
The first row (red) is there to remember the goal; then 3 rows of type 1 equations, 3 rows of type 2 equations (rewritten) and 3 rows of type 3 equations.
Adding all type 1 and type 2 equations together gets us *really close*
(12/14) So, it’s not hard to get the goal (1st row) from combining the other 9 rows.
Sum of type 1 equations
+ sum of type 2 equations
+ twice the sum of type 3 equations
= The Goal
This is a proof. I prefer to see everything at a glimpse. That’s next.
Sum of type 1 equations
+ sum of type 2 equations
+ twice the sum of type 3 equations
= The Goal
This is a proof. I prefer to see everything at a glimpse. That’s next.
(13/14) This has everything.
I suspect some notation and organizing can be done a bit better, but hopefully one can look at this and see the idea; feel the result!
The triangle in the middle has 1/7 the area of the biggest triangle ABC! summary of proof picture:
I suspect some notation and organizing can be done a bit better, but hopefully one can look at this and see the idea; feel the result!
The triangle in the middle has 1/7 the area of the biggest triangle ABC! summary of proof picture:
(14/14) The learning of Linear Algebra allows one to develop a sense of how much information is needed; this, together with the organization and the structure helps makes this (otherwise surprising) result intuitive.
Linear algebra FTW!
End
Linear algebra FTW!
End
