Geometry Stumper

[Paul Dale]

(08-16-2018 04:04 PM)Thomas Puettmann Wrote:  He has very strong opinions about that computer math is the "real" math and all what we humans still do ourselves is unimportant because it is of neglegtible complexity.

And the computer's solution to the Riemann Hypothesis is?


Pauli

[brickviking]

(08-17-2018 06:29 AM)Paul Dale Wrote:  

(08-16-2018 04:04 PM)Thomas Puettmann Wrote:  He has very strong opinions about that computer math is the "real" math and all what we humans still do ourselves is unimportant because it is of neglegtible (BV:negligible) complexity.

And the computer's solution to the Riemann Hypothesis is?


Pauli

Sigh. I'm completely lost even when various explanations are provided. I guess I've lost most of the maths understanding I learned when I was at school—that, or I'm particularly weak in geometry. It seems strange that it was my second best subject at school, but that's not saying much, as I only got middling results for my exams.

(Post 269)

[Albert Chan]

Now that we prove CDE is on the same line, here is another challenge:

With DE as diameter of a circle, midpoint K, prove A, B touch the circle.

In other words, prove AK = BK = DK = EK

Geometry talk: the circle touches triangle 2 vertices and 2 excenters.

Edit: this is the prove, try not to peek ... All about Excircles, section 2.2

[Albert Chan]

Just downloaded XCas 1.4.9-57 (win32) today.
I want to see if XCas can handle the tough proof: http://www.hpmuseum.org/forum/thread-112...#pid102245
Prove point D, C, E lies on a straight line (using distance, not slope)

pointE := head(solve([y=ae*x, y=be*(x-1)], [x, y])) //==> [-be/(ae-be), -ae*be/(ae-be)]
pointD := simplify(subst(pointE, [ae=-1/ae, be=-1/be])) //==> [-ae/(ae-be), -1/(ae-be)]
pointC := [k, 1]
dist(x) := sqrt(dot(x, x))

soln := head(solve(dist(pointD - pointC) + dist(pointE - pointC) = dist(pointD - pointE), k))

XCas was unable to solve for k, the mess is too much, *STOP* after 5 minutes ...

To help XCas a bit, scale the distance by abs(ae - be):
(ae, be was slope of line AE, BE. Since A != B, scaling factor > 0)

dist(x) := sqrt(dot(x, x) * (ae - be)^2) // scaled distance

With scaled distance, solving take 2 seconds:

soln = (-ae^2 * be - ae^2 + be^2 - be) / (ae * be^2 - ae^2 * be + ae - be)

Removing the square terms, we get:

simplify(subst(soln, [ae^2=1-2*k*ae, be^2=1-2*(k-1)*be])) //==> k

For any value of k, DCE lies on a straight line, thus DCE is always straight.

[Albert Chan]

Another proof that CDE lies on a straight line, Mathematica, without using Solve:

From analytic geometry thread, learn about properties of excircles.

Rotate the triangle, so BC lies on line x=1, AC on t2 radian line, AB on t3 radian line.
For ABC with inscribed unit circle, these are the coordinates: (d is calculated using symmetry, then rotation)

PHP Code:

c = {1, Tan[t2/2]}; 
e = {1 - Tan[t2/2] Tan[t3/2], Tan[t2/2] + Tan[t3/2]}; (* excircle center *)
d = {{Cos[t2], -Sin[t2]}, {Sin[t2], Cos[t2]}} . (e /. {t2 → t3-t2, t3 → 0-t2}); 


Prove is now trivial:

PHP Code:

Simplify[Det[{d-c, e-c}]] ==> 0 


It can also confirm by distance, with slight help from our end (avoid square root):
If CDE on straight line, distance wise, EC + CD = ED
Square both side: EC² + CD² + 2 √(EC² CD²) = ED²
Rearrange, square again: 4 EC² CD² = (ED² - EC² - CD²)²

PHP Code:

dSqr[pt_] := Simplify[Dot[pt, pt]]
{ec2, cd2, ed2} = Map[dSqr, {e-c, c-d, e-d}];
Simplify[4 ec2 cd2 == (ed2 - ec2 - cd2)^2] ==> True 


[Albert Chan]

PHP Code:

c = {1, Tan[t2/2]}; 
e = {1 - Tan[t2/2] Tan[t3/2], Tan[t2/2] + Tan[t3/2]}; (* excircle center *) 


This prove is even shorter !

Dot[c, e-c] = 1 * -tan(t2/2) tan(t3/2) + tan(t2/2) * tan(t3/2) = 0

-> OCE is right triangle -> by symmetry, OCD also right triangle -> CDE is straight line