You should try. For me it made the vids 10x more fun. If you can do it. It's somehow very satisfying seeing him do your exact steps. If you fail, you get to learn and see exactly where you went the wrong direction. You can even pause again and continue like you got a hint.
In two minutes you found what I found using algebra (second degree equations) and the Pythagorean theorem with quite annoying calculations, although conceptually not difficult; in practice I found that the two parts into which the tangent segment is divided from the point of tangency measure (5*sqrt(313)+65)/18 (the one on the left) and (169-5*sqrt(313))/18 (the one on the right); and in fact the sum of these values gives 13. And preliminarily I had to find the length of the side of the square.
My 'intuitive' approach: the diagonal through the radii is 13 and crosses the two horizontal parallel lines of the square. Since the black tangent line is perpendicular to the radii and crosses the vertical lines of the square, that also must be 13 units long.
How do you get that if it's perpendicular to the radius then it must be the same length? If all the end points of both lines lay on a circle then that is true by intersecting cords theorem but this is a square.
@@maxhagenauer24 If both lines end at the opposite side (and not the adjacent side), or a corner, then they always have exactly the same length. Like in this problem. (I'm quite sure you can't adjust the position and sizes of the circles so that the tangent line to the two circles wouldn't do that, with the condition that the semicircle must be in the corner, but I don't know how to see or prove that.)
@@synchronos1 Is that a theorem? I don't know why that works or how someone would know that besidesmaybesomethingwith symmetry and right triangles. Obviously if the lines are going from corner to corner then they are equal or if they go from mid point of the side to mid point then they are equal, but I didn't know about anywhere else on the side length. Also I don't think you are correct that it would always be the case no matter the size of the circles because image if the quarter circle in the corner got larger which would make the semi circle smaller. You should see that there would eventually be a point in its size where the tangent line is connecting adjacent sides and not opposites. Like images the quarter circle is the biggest it can be where it's radius equal to the side length of the square, the tangent line would definitely be past the corner to corner mark and would connect adjacent sides of the square.
A fun one! I got it, but I didn't notice the SAS congruency so it took a WAY more steps for me. I used the diagonal and the law of sines, and I ended up approximating a bit much and got about 12.996, but I was confortable rounding the answer up to 13 due to my approximations and honestly just being familiar with the solutions from these questions. My area ended up being ~(11.35)^2 though which I had less faith in, so on to part two!
I got this one pretty intuitively by noticing that the diagonals are congruent due to their equivalent angles and the fact that they both go from one side of the square to the other!
(during a test for which I have not studied) i have no fucking idea how to do this whatsoever but the radius of that semicircle is 5 and 8 + 5 = 13 so with the power of divination I'm guessing that one
I have a question. How are you sure that the 2 triangles are congruent. The 2 θs may not be equal. For example it may be 40 ° in one triangle and 30 ° in another. You had no right to declare that the 2 triangles are congruent based on ASA.
Great job sharing mathmatical thinking in a fun way! To see how a free, browser-based tool solves similar questions: ruclips.net/video/6l8HMIfRA9M/видео.html or ruclips.net/video/BLDpTQ5ediM/видео.html.
Gotta be honest, I never pause and try it on my own.😢
You should try. For me it made the vids 10x more fun.
If you can do it. It's somehow very satisfying seeing him do your exact steps.
If you fail, you get to learn and see exactly where you went the wrong direction.
You can even pause again and continue like you got a hint.
ngl grudaren ur so real for that 💀 me too
I tried this one without pausing and got it right :D
I have no idea on how to solve these. So it’s a learning experience on how to do math.
If I were to pause every math video when they ask to try it on my own, I would have hundreds of eternally paused videos.
"Every triangle is a love triangle when you love triangles" - Pythagoras
In two minutes you found what I found using algebra (second degree equations) and the Pythagorean theorem with quite annoying calculations, although conceptually not difficult; in practice I found that the two parts into which the tangent segment is divided from the point of tangency measure (5*sqrt(313)+65)/18 (the one on the left) and (169-5*sqrt(313))/18 (the one on the right); and in fact the sum of these values gives 13. And preliminarily I had to find the length of the side of the square.
You gave the subscribers the treasure of knowledge.....
Excellent solution !!👍
My 'intuitive' approach: the diagonal through the radii is 13 and crosses the two horizontal parallel lines of the square. Since the black tangent line is perpendicular to the radii and crosses the vertical lines of the square, that also must be 13 units long.
How do you get that if it's perpendicular to the radius then it must be the same length? If all the end points of both lines lay on a circle then that is true by intersecting cords theorem but this is a square.
@@maxhagenauer24 precisely because it's a square and 2 perpendicular lines.
@@rubicon24 But how does that imply they are the same length? You can definitely have 2 perpendicular lines in a square that are not the sane length.
@@maxhagenauer24 If both lines end at the opposite side (and not the adjacent side), or a corner, then they always have exactly the same length. Like in this problem. (I'm quite sure you can't adjust the position and sizes of the circles so that the tangent line to the two circles wouldn't do that, with the condition that the semicircle must be in the corner, but I don't know how to see or prove that.)
@@synchronos1 Is that a theorem? I don't know why that works or how someone would know that besidesmaybesomethingwith symmetry and right triangles. Obviously if the lines are going from corner to corner then they are equal or if they go from mid point of the side to mid point then they are equal, but I didn't know about anywhere else on the side length. Also I don't think you are correct that it would always be the case no matter the size of the circles because image if the quarter circle in the corner got larger which would make the semi circle smaller. You should see that there would eventually be a point in its size where the tangent line is connecting adjacent sides and not opposites. Like images the quarter circle is the biggest it can be where it's radius equal to the side length of the square, the tangent line would definitely be past the corner to corner mark and would connect adjacent sides of the square.
Solved part 2 very quickly. This one took me a bit more time hehe 😅. Fun stuff as always!
How exciting!
A fun one! I got it, but I didn't notice the SAS congruency so it took a WAY more steps for me. I used the diagonal and the law of sines, and I ended up approximating a bit much and got about 12.996, but I was confortable rounding the answer up to 13 due to my approximations and honestly just being familiar with the solutions from these questions.
My area ended up being ~(11.35)^2 though which I had less faith in, so on to part two!
I got this one pretty intuitively by noticing that the diagonals are congruent due to their equivalent angles and the fact that they both go from one side of the square to the other!
Got it correct again 😊
next time can we have a problem that doesn’t look like a fun one
Oooo nice
Man, i got the answer right but i couldnt prove my answer, great work man..
How exiting 👍😅
Oh my bad
Exciting
(during a test for which I have not studied) i have no fucking idea how to do this whatsoever but the radius of that semicircle is 5 and 8 + 5 = 13 so with the power of divination I'm guessing that one
Красавчик!
zoom in picture of you rubbing one out with your fingers
I have a question. How are you sure that the 2 triangles are congruent. The 2 θs may not be equal. For example it may be 40 ° in one triangle and 30 ° in another. You had no right to declare that the 2 triangles are congruent based on ASA.
Bro S was just 10. The question states it is a square.
Not true. Look as the semicircle with diameter 10. It does not fill s
Great job sharing mathmatical thinking in a fun way! To see how a free, browser-based tool solves similar questions: ruclips.net/video/6l8HMIfRA9M/видео.html or ruclips.net/video/BLDpTQ5ediM/видео.html.