A Simple Solution to an Otherwise Challenging Problem - Algebra and Geometry Working Together!
Vložit
- čas přidán 23. 03. 2024
- Complete playlist: • The MOST Fundamental C...
grinfeld.org
/ pavelgrinfeld
grinfeld.org/books/An-Introdu...
grinfeld.org/books/A-Tensor-D...
I really like your teaching: clear and fun!
Thank you - much appreciated!
Great teachers "turn the lights on"; this is what Pr. Pavel does here, and elsewhere too!
Thank you for your kind words - it's much appreciated!
I like seeing multiple proofs - there’s always something to learn from each one.
Here’s an outline of my approach: note that the median divides the original triangle into two with equal areas. Keep using that idea with other sub-triangles and you can eventually see that many occur in equal-area pairs, but especially the top two. Then use equal areas of the top two to show their bases must be equal (because same altitude and areas). That doesn’t yet prove their base lines are parallel to the bottom line, so assume they aren’t and draw a line from end of one side of that line that is parallel to the bottom of the original triangle. Use similar triangle ratios to get a contradiction, showing that base of those upper triangles must be parallel to the bottom after all. It works, but like Fermat said, showing all the steps would be too long for this comment 🤔. My takeaway was using the equal bases for one set of triangle to show equal areas, then eventually using it backwards to show equal areas implies equal bases for the top set.
Thank you - what an interesting approach!
This is solved almost trivially with Ceva's theorem. Label the triangle ABC clockwise from the top and the landing points of the segments X, Y, Z from the bottom, so AX is the median.
Ceva's theorem says that three cevians (segments from one vertex of a triangle to the opposite side) concur at a single point if and only if (AZ/ZB)·(BX/XC)·(CY/YA)=1. Since BX/XC=1, that fraction disappears, and simple rearrangement gives AZ/ZB=YA/CY, which is exactly the condition we need to prove that YZ is parallel to CB.
This is easy in one way, but it obviously requires knowledge of Ceva's Theorem, which might not be so well-known.
Great insight!
On utilise théorème de Thalès excellent travail merci professeur
Thank you!
Pick an arbitrary point P outside the triangle. Draw the vectors from that point through each of the interesting points of the triangle. Continue these vectors below point A (lower left). Then draw a horizontal vector from A to the right, and note where it intersects the vectors coming from the point P. You can prove that the projection from (A+B)/2 is still the midpoint of A->B' using a similar triangle argument. Then draw a vertical from the projection of the midpoint to find the projection of the top of the triangle, where it intersects the line from P. Now you have a projective transform, basically a list of numbers like gamma, and the projected triangle is easy to solve by symmetry. You have to show that projective transforms preserve proportions, though, or draw lots more similar triangles.
Thank you for that!
α+ 👍
By the way I was a little (but pleasantly) surprised when you pronounced φ to rhyme with ‘high’. That’s the way we pronounce it in England but I’ve only ever heard Americans pronounce it as ‘fee’.
♫ ♬ Beeta bayta theeta thayta let’s call the whole thing off! ♩♫ 😁
Fee-fi-fo-fum, I smell the bones of an Englishman!
@@MathTheBeautiful Touché!
It’s a zombie Redcoat insur-resurrection.
Watch out they don’t chop down your beanstalk and steal all your golden mathematical nuggets! 😁
Can we use inverse intercept theorem ?
In intercept theorem we know that angle is cut by two parallel lines
Inverse intercept theorem allows us to make conclusion that lines are parallel
Maybe not because we have to have some ratios to use inverse intercept theorem
but it was my first idea
Yes, we are
I cannot help but feel that William of Occam might take exception to this approach. Linear Algebra can serve to solve this problem, for sure. You can also give your beard quite the shave with a samurai sword. "IMPOSING VECTORS" rather than making a simple construction and doing a geometry proof is the "ingenuity" here, I think. LOL. Though it is nice to see that MATH is internally consistent, keeping one's Euclid sharp is probably good advice. 🤪
Entia non sunt multiplicanda praeter necessitatem
"Entities must not be multiplied beyond necessity"
(the simplest explanation is usually the best)
I respect everything you're saying and equally respectfully disagree. I believe that it is best to have as many approaches to the same problem as possible and to discuss their relative strengths and weaknesses.
Different sorts of empowerment for students, perhaps. Which is good. But only "grade school" skills (and a straight edge) are REQUIRED to solve the "challenging problem" with Geometry. Whereas rather advanced (and capable) university students might be challenged applying the Linear Algebra Vectors method. And I tend towards the democratization (and broadening) rather than specialization (and narrowing) of problem solving triumphs. Though - even "little Harry" (the great sculptor-in-waiting from Star Trek TNG, S01E17) needed his "basic skills" right? - czcams.com/video/ETt8GJRbqLc/video.html @@MathTheBeautiful
@@MathTheBeautiful I believe it was Michael Penn who coined (or, perhaps, borrowed where I could see) the phrase "the unreasonable effectiveness of linear algebra". While I enjoy an elegant geometric demonstration as much or more as the next person - when translating a geometric problem to another domain allows a proof without requiring intuition, I regard that also as a demonstration of elegance.
@@pietergeerkens6324 Agreed!
@yank5429 I am also in favour of democratisation. But the linear algebra is also a type of democratisation. Now even us idiots can solve geometry problems. No Euler, Euclid, Archimedes, required.
At 8:00, shouldn't gamma be *minus* alpha / beta?
Yes!