Solve This Crazy Area of the Intersection Between a Circle and a Square | Geometry 

Hi, could you possibly briefly explain how you know the line at 1:55 is a diameter? Thanks!

I was trying to demonstrate geometricaly that the round shaded parts on the bottom and the left were the missing part of the shaded area respectively on the top and on the right. I think I lacked a symetry argument (or something to apply Thales) to show the circle cuts the square at the same place on each side. You ended up demonstrating what I wanted but with subtracting shapes. Clever, I think.

how do u have only 287 subs wtffffff

Excellent way of manipulation of areas

Hi, what an interesting lesson !!! Havn‘t yet heard of the surprising Formular 4r^2 😳 A mathematical treasure 👍🏽😊👍🏽

Nice!

At 2:05 how do you know the purple triangle is isoceles?

Points to be labeled: Center of the circle: O Vertices of the square (clockwise from top left): ABCD. Circle/square intersection/tangent points: M (tangent to AB) N (tangent to BC) P (intersects CD) Q (intersects DA) Let r be the radius of the circle. Draw diagonal DB. Let the point on the circumference that it passes through be T. As the figure is symmetrical about DB, it will pass through O, thus OD = OT = r. Draw QP. As ponts Q, P, and D are all points on the circumference of a circle, and ∠PDQ = 90° (as the vertex of a square), then QP must be a diameter of the circle, by Thales' Theorem. Draw chords QT and TP. As DT is already known to be a diameter of the circle, and QT and TP are chords and thus perpendicular to any radii passing through their midpoints, they are parallel to tangents AB and BC respectively and DQTP is a square. As diagonals DT and QP are diameters of circle O, their lengths are 2r. Thus the side length s of DQTP is 2r/√2 or √2r. This means that the thicknesses of the circular segments subtended by arcPD and ard DQ, and by symmetry the distance from QT to AB and TP to BC, is (2r-√2r)/2 = (2-√2)r/2. Thus the side length of the larger square is √2r plus this amount. 1 = √2r + (2-√2)r/2 1 = √2r + r - r/√2 1 = (2+√2-1)r/√2 r = √2/(√2+1) r = √2(√2-1)/(√2+1)(√2-1) r = (2-√2)/(2-1) r = 2 - √2 s = √2r = √2(2-√2) = 2√2 - 2 The area of the shaded region is equal ro the difference between the areas of the two squares, as the chords QT and TP are the same distance from the center of the circle as the chords PD and DQ that form the two orange circular segments. Area = 1² - (2√2-2)² A = 1 - (8-8√2+4) A = 8√2 - 11 ≈ 0.314

Con centro el del círculo, giramos 180º los dos segmentos circulares; en la nueva posición, sus cuerdas delimitan un nuevo cuadrado inscrito en el círculo, cuyo vértice inferior izquierdo y los lados y diagonal correspondiente se superponen a los del cuadrado de lado 1 ud → En el trazado resultante, los dos segmentos que unen el centro del círculo con los puntos de tangencia y el que lo une con el vértice común de ambos cuadrados tienen una longitud igual al radio "r" del círculo.→ r+r√2=1*√2→ r(1+√2)=√2→ r=2-√2 → Área roja =Diferencia entre las áreas de ambos cuadrados → 1²-[(2r)²/2] =1-2r² =1-2(2-√2)² =8√2 -11. Gracias y un saludo cordial.

was expecting pi, left dissapoinyed. but great work!

I love your channel's math problems but I feel weird with your voice. Can you speak louder and get a better mic?