Peter Gallin

Catch the infinity! Normally the exterior of the symmetric white Swiss cross with the dimensions 6 and 7 (see figure) is red. Here the whole infinite exterior red area is reflected on the circle (see figure). What is the area of the red part of the figure.

@viterick Look and see!

@sonukg4india Here my solution without trigonometry!

Regular 12-gon. Find angle x (without trig)

@viterick We construct the blue triangle ABC starting with the red isosceles trapezium CDEF in the regular 18-gon. So we get the equilateral triangle ABH and see that |BE|=a+c+b and that the answer is 10°.

Find ? (By Thiago Durado)

@mnk68636175 @viterick Merci pour les explications. Il serait très utile de voir un mot avant les calculs. Je ne voyait que Pythagore dans votre première ligne.

@GallinPeter @viterick Non ! Dans la démonstration, on a fait aucune référence à une quelconque axe de symétrie. On a utilisé dans un premier temps, le Theorem de Stewart au triangle isocèle. Le Theorem de Stewart ne fait pas nécessairement référence à une quelconque axe de symétrie.

@viterick I cannot find x. Probably we have to find x(R) as function. ???

Find "x"

@viterick Theory of pole and polar: Black points are in harmonic position: c/d = b/(b+c+d) <=> c(b+c+d) = bd Tangent-secant theorem: a² = b(b+c+d) => a²c = bc(b+c+d) =b·bd = b²d QED

@mnk68636175 @viterick This proof assumes that b,c,d is symmetry axis of the figure.

@viterick a²c = b²d

@MathMath901 Look at the grid and see the height 10.8 and the base segment 5 so that the area is 10.8·5/2 = 27.

#math problem 30-05-2026 Geometry puzzle 🔺

@sonukg4india Theory of pole and polar: Black points are in harmonic position: c/d = b/(b+c+d) <=> c(b+c+d) = bd Tangent–secant theorem: a² = b(b+c+d) => a²c = bc(b+c+d) = b·bd = b²d QED

#morning with #math

@TigerGeometerrr The violet segment and the green segment are rotated with -60°. So the red parallelogram is rotated into the blue parallelogram with -60° => answer 60°.

@rdnzus @sonukg4india Here my proof: Look at symmetry axis of pentagon and the parallel connection of the black and white point. So we have 6°=60°–34° and 12°. See then 18°. QED

@GallinPeter @sonukg4india This “well known theorem in the equilateral triangle” needs a separate proof.

Find ?

@rdnzus @sonukg4india Here my proof: Look at symmetry axis of pentagon and the parallel connection of the black and white point. So we have 6°=60°–34° and 12°. See then 18°. QED

@GallinPeter @sonukg4india I happened to have a proof of this "well known theorem":

@viterick The two violet triangles are congruent => answer 45°.

By Thiago Durado.