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

Geometry puzzle: Solve this crazy area of the intersection between a circle and a square! This is not your everyday geometry question-it's a brain teaser that will put your mathematical skills to the test!
🔵 What You'll Learn:
- Step-by-step methods to solve this geometry problem
- Key concepts in calculating the areas of intersecting shapes
- Practical tips for approaching and solving advanced math puzzles
🧠 Why Watch This Video?
Whether you're a student aiming to improve your math skills, a teacher looking for engaging problems for your class, or simply a geometry enthusiast, this video is perfect for you. We'll break down the problem in a clear and comprehensive way, making it easy to follow and understand.
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Share your solutions or questions in the comments below!
Join the Chanel 👉 / @thephantomofthemath
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#Geometry #MathChallenge #CircleAndSquare #GeometryProblems #MathPuzzles

[@mathendspas]

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.

[@e23779]

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

[@ThePhantomoftheMath]

Sure: Our purple triangle is a right triangle, meaning it has a 90-degree angle. Since we have inscribed a 90-degree angle, the arc that it subtends must be a 180-degree arc. Therefore, the hypotenuse of the purple triangle must be the diameter of the circle.

[@tylerduncan5908]

​@@ThePhantomoftheMathI was about to ask, but after looking at your explanation it seems so obvious. Thank u

[@Patrik6920]

@@ThePhantomoftheMath thats a falso proof...as the square can get larger or smaller and still tuch the circle at a 90 deg angle, but the line will change position. i assume u refere to the triangle incribed in a circle as diameter-chord-chord theorem, it imply we already know where the center line is... if a square tuchng both sides of a circle and also meet at a 90 deg angle always intersects the circle in a tangent line that intersects it in half i dont know...

[@Patrik6920]

@@ThePhantomoftheMath did a check to see if this is true... (not using inscribed triangle theorem).. and it seems a square with the side length 1 that tuches the circle as in the video indeed intersect the circle at [x,y]= [0, 2(√2-1); 2(√2-1), 0] (it devides the circle at origo and have two intersection points at [x,y]=[1,√2-1; √2-1, 1] this can be checked by the following equations Lines tuching the edge of circle: x=(0)y+1 y=(0)x+1 intersection points: x=y x=-y +2(√2-1) (x-√2-1)² + (y-√2-1)² = (2/(2+√2)² ..and it checks out to be the case... 👍👍👍

[@Askedgd]

how do u have only 287 subs wtffffff

[@ThePhantomoftheMath]

Lol. Thank you for that! :) I only started a couple of months ago, so I'm relatively new to this. But your comment gives me the energy to keep making more and more videos like this.

[@jacquespictet5363]

You could use the red parts of the circle to complete the red parts of the square. Seems quicker.

[@enricolucarelli816]

WOW!!!👏👏👏👏👏👏 Glad I gave up trying to solve it by myself quite soon. I usually don’t do that. But in this case I think It would have taken an eternity for me to solve it.🤗

[@Darisiabgal7573]

The square is side one. The Chord = r * 1.41421, its bisector is 0.7071 Thus 1.707 r = 1 or r = 1/1.7071 The red area = outside the square ARC 90°- chord 90° = (pi/4 - 1/2) r^2 there are two of these areas The area in the corner is ((2/1.7071)^2 - pi (1/1.707)^2)/4 …..

[@vcvartak7111]

Excellent way of manipulation of areas

[@ThePhantomoftheMath]

Thank you!

[@dr.rolandzagler8831]

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

[@ThePhantomoftheMath]

Hi! I'm glad you liked the video. I was also surprised by this formula when I found out about it, so I knew I needed to make a video that includes it. It's truly amazing.

[@JoeCarsto]

Nice!

[@quigonkenny]

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

[@santiagoarosam430]

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.

[@JJ-zp5jz]

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

[@ThePhantomoftheMath]

The purple triangle is isosceles for the following reasons: - The square is inscribed in the circle, meaning the circle passes through all four vertices of the square. Therefore, the diameter of the circle is equal to the diagonal of the square. - The green triangle's vertices are: The top-left corner of the square. The bottom-right corner of the square. The bottom-left corner of the square. - The line from the bottom-left vertex to the top-left vertex of the square is a side of the square, which we can call - a. - The diagonal of the square (which is the hypotenuse of the right triangle formed by the square's sides) has a length of - a root of 2. - In the purple triangle, the two sides from the bottom-left vertex to the top-left vertex and from the bottom-left vertex to the bottom-right vertex are equal in length because they are both sides of the square, each with length - a. Therefore, the purple triangle is isosceles because it has two sides of equal length.

[@rgcriu2530]

👍👍

[@Larsbutb4d]

was expecting pi, left dissapoinyed. but great work!

[@ThePhantomoftheMath]

Sorry for that! Next time will be Pi included 😂

[@Larsbutb4d]

@@ThePhantomoftheMath its ok js expected pi bc yknow... circle

[@Michaelishere-sl1kg]

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

[@ThePhantomoftheMath]

Hi. Thank you for your feedback and support! I'm glad you enjoy the math problems. I appreciate your suggestion about the audio quality. While I can't change my voice and prefer not to use AI tools for voiceovers to keep things authentic, I understand the importance of good audio. I currently use a Razer Siren Pro mic, but I agree it could be better. I'm planning to invest in a higher-quality setup soon. I'm continuously working on improving the channel and value your critiques. Thank you for helping me make this channel better for everyone!

[@Michaelishere-sl1kg]

​@@ThePhantomoftheMath That's fine buddy, keep up your work!