@nsato7.bsky.social
Hence, (10P - 3A)/7 = (4O + 3B)/7. The coefficients on both sides add up to 1, so this common vector lies on both lines AP and OB. Therefore, the common vector coincides with point N, which means N = (4O + 3B)/7. This gives us ON:NB = 3:4.
January 23, 2026 at 1:43 AM
Hence, (10P - 3A)/7 = (4O + 3B)/7. The coefficients on both sides add up to 1, so this common vector lies on both lines AP and OB. Therefore, the common vector coincides with point N, which means N = (4O + 3B)/7. This gives us ON:NB = 3:4.
From the given, M = (A + B)/2 and P = (2O + 3M)/5. Substituting, we get P = (2O + 3(A + B)/2)/5 = (4O + 3A + 3B)/10. Then 10P = 4O + 3A + 3B, so 10P - 3A = 4O + 3B.
January 23, 2026 at 1:43 AM
From the given, M = (A + B)/2 and P = (2O + 3M)/5. Substituting, we get P = (2O + 3(A + B)/2)/5 = (4O + 3A + 3B)/10. Then 10P = 4O + 3A + 3B, so 10P - 3A = 4O + 3B.
Similarly, a = -1 and b = 6 implies a^2 + ab + b^2 = 31, and a = 1 and b = 5 implies a^2 + ab + b^2 = 31. This suggests that taking a or b negative does not give us any additional values.
January 21, 2026 at 4:31 AM
Similarly, a = -1 and b = 6 implies a^2 + ab + b^2 = 31, and a = 1 and b = 5 implies a^2 + ab + b^2 = 31. This suggests that taking a or b negative does not give us any additional values.
That's an interesting point. For example, if we take a = -3 and b = 5, then we get a^2 + ab + b^2 = (-3)^2 + (-3)(5) + 5^2 = 19. But if a = 2 and b = 3, then a^2 + ab + b^2 = 2^2 + (2)(3) + 3^2 = 19.
January 21, 2026 at 4:31 AM
That's an interesting point. For example, if we take a = -3 and b = 5, then we get a^2 + ab + b^2 = (-3)^2 + (-3)(5) + 5^2 = 19. But if a = 2 and b = 3, then a^2 + ab + b^2 = 2^2 + (2)(3) + 3^2 = 19.
Your table leaves out the values of a^2 + ab + b^2 where a or b is 0 (which are the perfect squares).
January 19, 2026 at 2:16 AM
Your table leaves out the values of a^2 + ab + b^2 where a or b is 0 (which are the perfect squares).
Given triangle ABC and point P, let AP, BP, CP intersect BC, AC, AB at D, E, F. Then there exists a unique conic that is tangent to the sides of triangle at D, E, F (known as an inconic). The point P is known as the Brianchon point (also perspector) of the inconic.
January 13, 2026 at 3:48 PM
Given triangle ABC and point P, let AP, BP, CP intersect BC, AC, AB at D, E, F. Then there exists a unique conic that is tangent to the sides of triangle at D, E, F (known as an inconic). The point P is known as the Brianchon point (also perspector) of the inconic.
This ties nicely to your recent posts on difference of squares.
We can write 91 = 100 - 9. Then by difference of squares, 100 - 9 = (10 - 3)(10 + 3) = 7*13.
We can write 91 = 100 - 9. Then by difference of squares, 100 - 9 = (10 - 3)(10 + 3) = 7*13.
January 11, 2026 at 4:24 PM
This ties nicely to your recent posts on difference of squares.
We can write 91 = 100 - 9. Then by difference of squares, 100 - 9 = (10 - 3)(10 + 3) = 7*13.
We can write 91 = 100 - 9. Then by difference of squares, 100 - 9 = (10 - 3)(10 + 3) = 7*13.
As @diffgeom.bsky.social points out, inversion takes care of the problem nicely. (And many of these Sangaku problems were solved by inversion.)
The three arcs invert to three lines. There are two circles that are tangent to these three lines; we want the "top" one.
The three arcs invert to three lines. There are two circles that are tangent to these three lines; we want the "top" one.
August 5, 2025 at 3:08 PM
As @diffgeom.bsky.social points out, inversion takes care of the problem nicely. (And many of these Sangaku problems were solved by inversion.)
The three arcs invert to three lines. There are two circles that are tangent to these three lines; we want the "top" one.
The three arcs invert to three lines. There are two circles that are tangent to these three lines; we want the "top" one.
Here is a graph of a regular 2n-gon using absolute value. (You can adjust the value of n using the slider.) www.desmos.com/calculator/c...
July 21, 2025 at 4:10 AM
Here is a graph of a regular 2n-gon using absolute value. (You can adjust the value of n using the slider.) www.desmos.com/calculator/c...
I have confirmed that your answer is correct.
June 21, 2025 at 1:29 AM
I have confirmed that your answer is correct.
The model of Dandelin spheres works for cones and cylinders: en.wikipedia.org/wiki/Dandeli...
Dandelin spheres - Wikipedia
en.wikipedia.org
March 13, 2025 at 4:15 AM
The model of Dandelin spheres works for cones and cylinders: en.wikipedia.org/wiki/Dandeli...
Reposted
Trig is like an acquaintance who introduces themselves as being about triangles.
When you really get to know them, their real identity is uncovered: they are about circles (more specifically about triangles inside circles w/base a radius from center edge)
#iTeachMath
When you really get to know them, their real identity is uncovered: they are about circles (more specifically about triangles inside circles w/base a radius from center edge)
#iTeachMath
March 8, 2025 at 3:54 AM
Trig is like an acquaintance who introduces themselves as being about triangles.
When you really get to know them, their real identity is uncovered: they are about circles (more specifically about triangles inside circles w/base a radius from center edge)
#iTeachMath
When you really get to know them, their real identity is uncovered: they are about circles (more specifically about triangles inside circles w/base a radius from center edge)
#iTeachMath
Nice! This is known as the Leaning tower illusion (en.wikipedia.org/wiki/Leaning...)
I have used an example like this, to illustrate to students that you should not trust geometric diagrams too much, e.g. lines that look parallel may not actually be parallel.
I have used an example like this, to illustrate to students that you should not trust geometric diagrams too much, e.g. lines that look parallel may not actually be parallel.
February 25, 2025 at 3:18 AM
Nice! This is known as the Leaning tower illusion (en.wikipedia.org/wiki/Leaning...)
I have used an example like this, to illustrate to students that you should not trust geometric diagrams too much, e.g. lines that look parallel may not actually be parallel.
I have used an example like this, to illustrate to students that you should not trust geometric diagrams too much, e.g. lines that look parallel may not actually be parallel.
I found this: www.youtube.com/watch?v=LjjA...
A Packing Puzzle Requiring Some Math Skills
YouTube video by Dom Puzzles
www.youtube.com
February 23, 2025 at 4:51 AM
I found this: www.youtube.com/watch?v=LjjA...
This is a consequence of the Incenter-Excenter lemma (web.evanchen.cc/handouts/Fac...): In triangle ABC, if I is the incenter, I_A is the A-excenter, and M is the midpoint of arc BC, then MB = MC = MI = MI_A.
February 21, 2025 at 9:27 PM
This is a consequence of the Incenter-Excenter lemma (web.evanchen.cc/handouts/Fac...): In triangle ABC, if I is the incenter, I_A is the A-excenter, and M is the midpoint of arc BC, then MB = MC = MI = MI_A.