Tricky Geometry Challenge

For more geometry challenges, try the above page!

[yobrogobrrrrrrrr._.]

This guy is so chill while teaching, it's almost he's playing a game .😊

[samuelking]

I know it’s just geometry, but this guy does a really good job of explaining how he gets from one step to another in a way that anyone can understand.

[Boulder_Bill]

I've learned more about math from a YouTube channel than 4 years of college. How exciting.

[mathmachine]

The great/frustrating thing about geometry is, if you can't think of a clever solution, you can always just turn it into a bunch of vector equations and solve it that way. It won't be as elegant as the easier "intended" solution, but not all real world problems have an easy solutions, so in some cases, you're better off just not trying to look for an elegant way to do it, and just plugging everything into vectors. ¯\_(ツ)_/¯

In this case, you'd do that by solving the position of the bottom most corner (C), and the two corners that intersect the circumference (A and B). C•ĵ=0, |B-C|²=4|A-C|², |A|²=|B|²=1, then when you find all 3, just compute ((A-C)×(B-C))•k, or just 2|A-C|², or just |B-C|²/4.

Still, I like your method better. :)

[animeedits]

I really love his way of teaching, you can see his love for maths through it, he also makes it look easy and lovable for others. Keep up !

[billsmith]

Cool! I've done about 10 of these now and it's a blast having all of this coming back to me. I've only figured out 2 of them, but I'm 67, so I'm feeling a bit cocky. I'm pretty sure I've already figured out a way to save 10 minutes of time mowing my yard more efficiently. I'm finally taking control of my life with Mathematics. I was beginning to lose hope. Thanks for the videos!

[txikitofandango]

Very cool problem! I did it with coordinate geometry. Three of the corners of the rectangle are at (-s, s), (0, s), and (s, 0). The circle has equation (x-h)² + (y-k)² = 25 and goes through the three listed points. Therefore you get three equations with three unknowns:
(s-h)² + k² = 25
(s+h)² + (s-k)² = 25
h² + (s-k)² = 25
These are easy to solve by elimination. You get s = √10, and so area = 20.

[pace_]

i like how you use simple algebra and concepts to solve these.

[JeffreyCH]

Find Area of Blue Rectangle Speedrun (2:53, any%, WR)

[mkarakurt]

This guy really enjoys math. I watch the videos just to witness his joy. Good work man!!

[Zaygone]

I normally dont interact with channels but man you need to keep making these.

[AngryEgg]

Found it in an easier way by finding the slant of the rectangle then define 1 smaller edge as x. Then you can pass a line equal to x in the middle of the rectangle cutting it into 2 squares then you will see a right triangle with x, x/2 and 5sqrt(2)/2. Then use Pythagorean theorem: x^2 + (x/2)^2 = (5sqrt(2)/2)^2 => x^2 = 10
Area = x*(x+x) = x*2x = 2x^2 = 2*10 = 20
This is briefly explained so sorry if it’s unclear what I did.

[jobaecker]

I was an honors math student (including geometry) back in the 1970's. For the life of me I don't ever recall learning the subtended angle on a circle thing. Ever.
Head exploded. How exciting.

[bradballinger]

Lovely solution.

I approached it by drawing a coordinate system aligned with the rectangle. I gave the rectangle's vertices the coordinates (0, 0), (0, 2a), (4a, 2a), and (4a, 0).

We know that the circle passes through (0, 2a), (2a, 2a), and (4a, 0). The first two of these have perpendicular bisector x=a, while the last two of these have perpendicular bisector x-y=2a. These lines meet at (a, -a), which must therefore be the center of the circle.

Now pick any of the three points of contact; its distance from that center is a*sqrt(10) by the Pythagorean Theorem, which must match the radius of 5. Therefore 10a^2=25, so 2a^2=5, so 8a^2=20. That's the area.

[HoSza]

Again, this problem can be solved more easily with coordinate geometry. Choose the coordinate system such that its origin is at marked corner of the rectangle and the x and y axes contain the long and short edges of it respectively. Now we can write 3 equations describing the points that lie on the circle. Those points have the coordinates: (0, x), (x, x) and (2x, 0). Lets denote the center of the circle as (u, v), then the system of equations is:
u²+(x-v)² = r²
(x-u)²+(x-v)² = r²
(2x-u)²+v² = r².
Solve it for x, u, v (r=10 is known):
u = -v = r√10/10
x = r√10/20.
From this, the blue area is:
A = 2x² = r²/5 = 20.

[cristina.valencia]

What software or app do you use for these? like for the graphics and demonstrations? or is it just editing? love your vids ❤

[the_maker]

I totally was not taught inscribed angles in all my math career and this was a great concept to learn. Got another tool in my kit thank you kind sir

[barryomahony]

I struggled with this a bit until I realized that the 3 points of contact with the circle define it, and thus its radius. That the lower left corner is coincident with the diameter chord is irrelevant and a distraction and doesn't affect the answer. I plotted the 3 points on the x-y plane at (-x, x), (0, x), and (x, 0) and solved the simultaneous equations for the radius r. With r=5, the area 2x^2 is then 20.

[maxkhunglo]

1:36 thank you for not losing me there. I'm not smart, but I love to learn to some degree. It really keeps my attention when you make every explanation visible and not imaginative. Thank you sir, I wish I had you as my math teacher.

[Torrle]

That 'cool property' about the relationship between inscribed angles and arcs was new to me! Thank you for teaching me something new!