4 Squares 1 Circle

i completely forgot about coordinate circle solving! very clever!

this was one of the best ones so far. I absolutely loved the idea of putting it on an axis - first time seeing a solution like this. how exciting

The double box was very exciting :)

how exciting

Using coordinated plane here is really clever

Your channel made me like math and look at it differently. More like fun puzzles.

This was really cool to do first and then watch you do it. I took a more geometric approach that gave me similar equations to solve. Ignoring all of the stuff I drew that was unhelpful, here’s what worked for me: All of the squares have the same area of 16. That means their side lengths are sqrt(16)=4. Three points define a unique circle. The center must be equidistant from all three points. That distance is the radius. I drew an approximate center with radii out to each point. Since the two points on the right are exactly vertical to each other, this center would be vertically equidistant between the two, meaning it was vertically 4/2=2 units higher than the lower right point. I also extended the “grid” formed by the squares, specifically horizontal line segments from the bottom-left square over to the right and across the bottoms of the two vertically middle squares. Then, I dropped a vertical line from the center down to the lowest horizontal line segment I drew. This formed two right-angle triangles, both with radii as the hypotenuses. The triangle on the left had a height of 6 and an unknown width of x. The triangle on the right had a height of 2 and an unknown width of 16-x. Using the Pythagorean Theorem, I was able to set r^2 equal to two different statements containing x. r^2=(6^2)+(x^2) r^2=(2^2)+((16-x)^2) I then set those equal to each other and solved for x. (6^2)+(x^2)=(2^2)+((16-x)^2) 36+(x^2)=4+256-32x+(x^2) -224=-32x x=7 I plugged x back into one of the equations to find r^2. r^2=(6^2)+(7^2) r^2=36+49 r^2=85 And then the final answer! A=85(pi) square units

I took the surveyor approach (being a surveyor and surveying instructor). Essentially took the bottom side of the bottom left square, extended it to right to intersect right side of rightmost square extended down. Established perpendicular bisector to left of rightmost square. From this, established right trapezoid with bottom base length of 16, right side perpendicular to base with length of 6, side from right side bisector left to radius point with a length sqrt(R^2-4), and a diagonal left side with distance of R. From radius point, construct line perpendicular to bottom base. This leads to a right triangle with hypotenuse of R, one side of 6, and another side of 16-sqrt(R^2-4). Plug into Pythagorean Theorem and end up with R^2=85. Area=85π.

If you start with a little geometry and draw the perpendicular bisector of two of the circle's chords, you find the center of the circle very easily and from there a little use of Pythagorean theorem gives you the radius squared. Multiply by Pi and voilà :)

really fun doing honestly

I love how the operations are in a different color. Visually makes things easier to follow, thanks!

The rightmost square is centred symmetrically on the horizontal diameter of the circle (this has to be so since the chord it makes with the circle is vertical, but can be made clearer by drawing additional mirror flips of the other squares.) Given this symmetry, k is obviously 2 by inspection.

I used the law of sines to solve it! View the three points touching the circle as vertices to a triangle. Then the circle is the circumcircle of the triangle. Let A be the top right vertex, B be bottom right, and C be bottom left. By law of sines, a/sin(A)=2R where R is the radius of the circumcircle and a=length of BC. sin(A) is easy to calculate: imagine a right triangle by extending AB downwards by another 4 units to point D. Then ADC is a right triangle, and we get AD=8, DC=16, so by Pythagorean AC=8sqrt(5), so sin(A)=opp/hyp=DC/AC=2/sqrt(5). Now we need a=BC, but this can be solved by applying Pythagorean on triangle BDC, since BD=4, DC=16, we have a=BC=4sqrt(17). Combining all these information together, we have R=a/2sin(A)=4sqrt(17)/(4/sqrt(5))=sqrt(85). The area of the circle is this πR^2=85π. Using this method I was able to mental math it in a few minutes 🤓

Mind Blown. It's been so long but you still make it look so easy. This is one of my favorites. Rock On

This one took me a while. I took a totally different approach, which was very simple in setup, but took quite a bit of working out. I used trig. It's kind of hard to explain, but basically, using the coordinate system you use (though the way I did it, I didn't need a coordinate system), I'm looking for x, the distance from (0, 2) to the center. I can also find alpha, the angle between (0, 2) to (0, -14) and (0, 2) to (4, -14). I also note the unknown angle theta, the angle between the center to (0, 2) and the center to (0, 0). From these relationships, it's not too hard to see tan(alpha) = 1/4 and tan(theta) = 2/x. The last piece of the pie is noticing (through some angle relationships) that the angle between the center to (0, -14) and the center to (4, -14) is 2alpha + theta. This gives tan(2alpha + theta) = 6/(16 - x). I then used the tan sum and tan double angle identities to get the last equation in terms of tan(alpha), tan(theta), and x. Substituting, I can get an equation totally in terms of x. After a lot of algebra, I got 0 = 8x^2 - 8x - 576. Using our old friend the quadratic formula, there are two solutions: -8 and 9. It's obviously not -8, since it's a length, so it's 9! From that, I could get the radius with Pythagorus: 9^2 + 2^2 = r^2.

How spicy! 🌶️

I did this defining the bottom left point on the bottom left square as the origin, finding the equations of the lines from that point to the other 2 points on the circumference, then finding the equations of 2 lines perpendicular to those 2 lines that pass through the midpoint (basically the perpendicular bisectors), then by finding the intersection point (7,6) i can calculate the radius by calculating the distance to the origin, since the origin is on the circumference of the circle (pythagoras) and then use that to calculate the area. Very exciting

Well.. I way overcomplicated this problem. I drew an inscribed triangle, used the distance formula to find each side length, used law of cosines to find one of the angles in the triangle, used law of sines to find the radius, and finally calculated the area. I got the same answer but I did wayyyy more work lol

Here's how I would solve it: connect the three points to get a triangle of base 4 and a height of 16. The other two sidelengths can be calculated by the Pythagorean Theorem. From there, the area of the triangle 1/2 • base • height. But another formula for the area of a triangle is the product of the sidelengts divided by the radius of the exocircle. Substituting in the values of the area and the sidelengths gives us the radius and from there the area is just r²π.

That's a spicy assignment.