Solving a viral geometry problem from Twitter

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Can you solve for the area of the square and the length of the tangent line? Special thanks this month to Fiona Harper, Kyle, Mike Robertson, Michael Anvari, Robert Zarnke. Thanks to all supporters on Patreon! / mindyourdecisions
Adapted from a puzzle by Twitter @Cshearer41 (Catriona Agg)
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6 окт 2024

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Комментарии : 657

@engelbrecht777 3 года назад

The art of eyeballing told me AB=18. I am happy you confirmed it with this math thing.

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@SpaceCadet4Jesus 3 года назад

Your eye balls are laser beams.

@AaronHamm 2 года назад

You can abuse a symmetry argument to do it via eyeball, which is likely what you were intuitively doing: Since the line passes through the tangent point of the two circles, and the centers of the two circles are on the perimeter of the square, and any square is symmetric about 90 degree rotations, and there is a 90 degree angle between any tangent line and a radial line that passes through the point of tangency, both the line segment and the distance to the two centers of the circles must be the same length. This doesn't work if one of the circles is so much larger than the other that the tangent line doesn't intersect two opposing sides of the square, but that's not the case here lol

@engelbrecht777 2 года назад

@@AaronHamm The ancient art of eyeballing is older than any math expression. Please keep a respectful tone and stop abusing eyeballing with your modern math techniques. We should both be able to live in peace on this beaten up planet.

@ballin1006 2 года назад

Lol

@shravanichawathe1358 3 года назад

It was tricky , but the tools used are helpful to solve many other geometry problems . Thank you 😊

@shalakakhandagale 3 года назад

Are you Maharashtrian?

@shravanichawathe1358 3 года назад

@@shalakakhandagale हो

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@JLvatron 3 года назад

For the 2nd part, I spent a lengthy series of calcs with many decimals. And I got it wrong. I can't believe how easily and cleanly you did it! Very nice!

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@geoninja8971 3 месяца назад

I did the same, with coordinate geometry - ended up with 17.999999.... :) Near enough!

@JLvatron 3 месяца назад

@@geoninja8971 Pretty good!

@יונתןאלוןחלימי 3 года назад

3:23 love your definition of simplify

@Aiden-xn6wo 3 года назад

The second part was genius.

@rohangeorge712 3 года назад

it was indeed.

@LeoStaley 3 года назад

I skipped straight to the second part. I noticed that the line in question was perpendicular to the line connecting the centers, and then realized that because it was in a square, the two perpendicular lines would have to be equal distance. There was no need for most of the first half at all.

@angrytedtalks 3 года назад

I thought that part was obvious.

@DerpyDinoBro 3 года назад

@@LeoStaley I thought it was because the two triangles shared at least two angles and one of the sides were congruent aka the length of the square proofs baby tell me if that’s wrong 😑 plz

@AnotherDuck 3 года назад

@@LeoStaley Well, the second part was the question in the thumbnail, so I pretty much just had to watch to confirm that it was a square and not a rectangle. Then I got side tracked with trying to figure out the area, since that's what he focused on. The thing I noticed first when we got back to AB was that the angles have to be the same, which leads to the same result. There are a few different ways to reach that conclusion, so it's not that hard of a problem. The area is harder, since it requires some equation solving.

@khiemngo1098 3 года назад

I felt great solving both parts of the problem after five hours or so. For me, part 2 (finding AB) is pretty quick though. It's a great geometry contest problem!

@mranonyms5901 2 года назад

Damn, you have impressive persistence

@antoinedelepine6144 2 года назад

5 hours what the first part took me like at most 10min (and I couldn't be bothered for the second part), so how did you just not use a calculator and compute all numbers on your own or smth? and even that would take like 5 min I don't wanna sound too mean i just can't believe you actually passed 5 hours on a problem like that. be honest, you added a few hours for the likes didn't you

@khiemngo1098 2 года назад

@@antoinedelepine6144 For me, finding part 2 was a lot easier if not straightforward. No, contest problems are not meant for calculator! I haven't seen a math contest that allowed calculators before. Otherwise, there would be no fun and the problems become pointless to solve. So, no calculators are where the fun part begins. By the way, I didn't do this for the "likes" as you wrongly thought of me. If you have actually solved both parts of the above problem without calculator in five or ten minutes, perhaps, you are more suitable to solving IMO problems or even proposing problems for the USAMO or IMO. I didn't solve the problem in one session. I spent a bit of time on it but got something else more important to do, and then came back to it when I have the time. I didn't know exactly how long it took me to solve the problem. If I report that I did it in 3 hours, and in reality, I spent more than 3 hours, that would make me a liar. So, I felt that it's best to report a higher number, say five hours.

@antoinedelepine6144 2 года назад

@@khiemngo1098 hm. Well, i'm still quite weirded out by 3 hours but i'm guessing i got lucky by getting the idea right first try... but your answer was pretty respectul, so i'll just shut up, this discussion was started by a whim of mine anyways... and it's true that 5min for calculations was a bit of an understatement on my part, 10-20 or more minutes added would probably be more right for those square roots. (I wouldnt know since i did use a calculator for the quadratic formula)

@piman9280 3 года назад

"The Quadratic Formula" - Presh actually said it!

@johnjordan3552 3 года назад

I am going to cry!😂

@maheshpatel7691 3 года назад

Its brahmagupta formula

@IS-py3dk 3 года назад

@@maheshpatel7691 Yes LOL 😂😂😂😂

@epikherolol8189 3 года назад

Lol ye

@advaitpetiwale9596 3 года назад

Or dear old Gougu. He didn't even call it "my favorite right angled triangle theorem". Straight up "We know that..." XD

@geoninja8971 3 года назад

Loved it - I couldn't get the second part though, but was glad to solve the first to exact figures too....

@rebellio619 3 года назад

6 days ago how?

@Sanatan_saarthi_1729 3 года назад

How?? It is just published on RU-vid 🤯😵😶😯😮😲😲

@norukamo 3 года назад

hi time traveler

@mrudulsankhesara6940 3 года назад

I think mind your decision should solve this problem.

@aahaanchawla5393 3 года назад

for those wondering this video was private some time ago and he somehow got access to the link

@BossRoss13 3 года назад

After years of pushing the idea of Gougu Theorem, he just finally gave up.

@e1woqf 3 года назад

👍

@tomdekler9280 Год назад

Both parts were absolutely inspired. I feel silly for not realizing both of the missing corner lengths would logically be equal to the other.

@gracyvuna7848 3 года назад

I like the way you play with you geometry tricks to solve such problems! Thank you master!!!!

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@feelshowdy 3 месяца назад

The way you solved the second part was beautiful.

@mathexin2520 3 года назад

I solved the second part in a different way. Suppose a line passes through the centre of the square and is parallel to the first line (the one formed by the two radii). This line would have a length of 18. Now draw a line perpendicular to this line. As the square is rotationally symmetrical, this new line would still have a length of 18. Lastly, as this line is parallel to AB, it follows that AB has a length of 18.

@Batista77387 2 года назад

but thats the same way he did it

@CeritaSerial 3 года назад

THANK YOU VERY MUCH.I WAIT THIS MOMENT

@jumaidahhakiki9150 3 года назад

Me too

@nitawijaya1516 3 года назад

Yeay

@anitasari9328 3 года назад

I love your channel too cerita serial

@eritamasentosa9700 3 года назад

Square yeay

@rr-tf6hx 3 года назад

I love you too cerita💕

@yassineattia6679 3 года назад

Holy, one of the first geometry problem I can figure out. Your vids make me stronger and stronger in those exercises :D. Keep it up

@randomdude9135 3 года назад

Ikr. Same for me too. 1st time I've solved one of his problems like these 😃

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@Tasarran 3 года назад

I got the first part; stumped on how to do the second part. I suspected the answer from just eyeballing the lengths, and had started focusing on the triangle with the hypotenuse on the right side of the square, but I got stuck there. Genius solution.

@hans-rudigerdrzimmermann 3 года назад

I calculated AB using the rule that perpendicular line slopes obey m1 x m2 = - 1 . The slope of AB is then used using Phytagoras theorem on the right angle triangle to find AB to be 18

@timc5768 2 года назад

Me too, but in doing so, I have (without realising it) implied the existence of the triangle whose congruence with the other would have allowed the simpler solution!

@SogehtMathe 3 года назад

Nice problem. Everytime I see a Math-problem like this one I'm once again impressed of Math. Of course it's a neat video too. 👍

@italixgaming915 3 года назад

Terrible problem, at the beginning he assumes that he can draw an half circle and a quarter of circle inside a square and that these two circles will be tangent. THERE IS ABSOLUTELY NO PROOF THAT THIS IS POSSIBLE. WE MUST DEMONSTRATE IT BEFORE SOLVING THE REST OF THE PROBLEM.

@jarretberenson1214 3 года назад

@@italixgaming915 That information is given at the beginning of the problem. You do not need to prove it.

@italixgaming915 3 года назад

@@jarretberenson1214 You need because otherwise you don't know that drawing the picture and having a square is even possible. If you call alpha the angle between the vertical line and the line between the two centers, you have a condition on alpha where drawing the rectangle is possible, and there is only one specific value of alpha where the rectangle becomes a square (the value where sin(alpha)-cos(alpha)=1/3). If you don't demonstrate that such an angle exists, your whole demonstration is completely INVALID. The proof is not difficult at all but the author of the video was lazy and skipped that essential part of the demonstration.

@jarretberenson1214 3 года назад

@@italixgaming915 You know that it is possible because it is given in the problem. You sound like you are a lot of fun at parties. You could be doing something more productive than complaining about a RU-vid video.

@italixgaming915 3 года назад

@@jarretberenson1214 If I give you 2+2=5 in a problem, will you be able to tell me the result of 3+4? I'm sorry but the fact that we CAN draw a square is not obvious at all, we must DEMONSTRATE IT. In most geometrical problems, the "building" of the picture is explained step by step so you don't need to do that, but in this specific case, the first step is to verify that there is a way to obtain the picture shown with a square. I'm complaining for a very right reason, the author of the video was extremely lazy and skipped an asbolutely necessary part of the demonstration. That ruins the entire problem. The silliest thing is that the proof is not complicated. I wouldn't have been so annoyed if we were in the general case where we have a rectangle because the proof that we can draw a rectangle for certain values of alpha can be considered as obvious enough to be skipped but this the proof that we can find alpha in order to have a square and not a simple rectangle is NOT obvious. Not very difficullt, but not avoidable at all.

@krzysztofmazurkiewicz5270 3 года назад

Woke up... started the video... saw the puzzle... thought this is impossible.... 5 minutes later i stand corrected... And not a single law of cosign was used!

@peterwright5311 2 года назад

If AB is tangential to both circles, then you can draw a straight line connecting the centres of the circles through the point of tangent. The length is the sum of the radii, 6+12 =18. This must be at 90 degrees to AB, therefore AB = 18 as well.

@Stofferjib 3 года назад

Fresh Skywalker's pronunciation of "Kwor-tir sir-cull" is far more impressive than the math he displays

@rohitmehta3953 3 года назад

This was the comment I came looking for

@dcterr1 3 года назад

Geez, I wasn't even close on this one! I spent about half an hour doing a bunch of really complicated calculations without seeing the geometric tricks presented in this video. Geometry was never my strong area of math I'm afraid. Great derivation!

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@BaljeetSingh0718 3 года назад

Good question

@leodaric5447 2 года назад

You can shift AB up so that B is hitting the top right corner of the box. The length is preserved and the angles of the right triangle formed match the angles of the triangle formed by connecting the two centers of the circles because the two hypotenuses are orthogonal, so the triangles are congruent. Therefore length of AB is just the sum of the 2 radii.

@gamingmusicandjokesandabit1240 3 года назад

My day is ruined and my disappointment is immeasurable: You didn't say the quadratic formula was Brahmagupta's 😂

@advaitpetiwale9596 3 года назад

Or Gougu. Presh is avoiding touchy terminologies. 😂

@mayankgoel3798 3 года назад

It is called sri dharacharya formula not brahmagupta

@advaitpetiwale9596 3 года назад

@@mayankgoel3798 new to the channel?

@snoxh2187 3 года назад

its in the wrong order ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

@mayankgoel3798 3 года назад

@@advaitpetiwale9596 yes

@SuperVertraulich 3 года назад

I did it a bit different: it was clear the first triangle had a hypothenuse of 6 + 12 and the othersides of the triange had a length of x and x - 6 so i did a simple pythagorean: (x - 6)² + x² = 18² -> x = 3 + sqrt(153) and the area of the square is x² = (3 + sqrt(153))² = 162 + 6 sqrt(153) ~= 236.216 The length of AB was obvious the same as the hypothenuse of the first triangle so it had to be 18.

@pedroloures3310 3 года назад

Thanks for the problem and for the beautiful solution .

@leickrobinson5186 3 года назад

That’s funny! The 2nd half is unexpectedly *much* easier! Since AB is perpendicular to the line segment that connects the centers of the two circles (and which must also pass through the point of tangency), it intersects the side of the square at the same angle, and thus both line segments must have the same length. Since that other line segment connects the centers of the two circles, then the length of both that line segment and AB must be 18. :-D

@krishnakumarverma4115 3 года назад

Only works till the lines are touching opposite sides of the square and not adjacent ones. Though I used the same method 🤘

@englishlife5838 3 года назад

I don’t see how your method works.

@aashsyed1277 3 года назад

hello! leick! i also watch michael penn& sybermath!

@italixgaming915 3 года назад

The 0th part of the demonstration is missing anyway: he assumes from the beginning that he can draw an half circle and a quarter of circle inside a square and that these two circles will be tangent but THERE IS NO PROOF OF THAT. WE MUST DEMONSTRATE IT.

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@sanchezking6188 3 года назад

Ive been ambitious here and tried to find a generalized solution for any square with half and quarter circles in it. However, my ambitions were not crowned by success, but I find this solution here much more elegant anyway.

@mrcookie97 2 года назад

Elegant solution for second part!

@mohdfaik 3 года назад

For a more general solution: l (side of square) = 2 *r +(sqrt(17)*r-3*r)/2 whereby r is the radius of the smaller circle; approx. side of square l=2.56155 r

@kimba381 3 года назад

Good solution. I was a little more direct:: x^2 +(x-6)^2=18^2 solve for x. and find x^2. And I am afraid that the second result seemed obvious to me.

@poshun9407 3 года назад

Can you explain how you got your formula please?

@stillliving7670 3 года назад

@@poshun9407 I did it the same way: x is the length of the side of square. The right side of the triangle is therefore just x and the upper part ist x-6 (you subtract the left part of the semi-circle). Hope it explained it and sorry for my crappy english.

@poshun9407 3 года назад

@@stillliving7670 Thanks, I see it now. My confusion was that I was using x + 12 to represent the square's side and assumed you both did as well 😁

@kimba381 3 года назад

@@stillliving7670 Thanks for that. Good English, too.

@avuaronar6815 2 года назад

I'm so proud of myself cuz as soon as I looked at the thumbnail, I figured out that AB would be equal to the 2 radiuses added up, though I couldn't figure out how. So then I surprised myself EVEN MORE and found the side of a square (exactly the way he did it!) and the area of the square. Then I was stuck on proving how AB=6+12. Had to share my joy haha

@JD-nz9dj 3 года назад

I solved this using algebra but the calculation turns out to be kinda messy and complicated. Even so, I'd like to share my method here: Take that corner of the square which lies on the semi circle to be O(0,0), the corner nearest to the semi circle which does not lie on the semicircle to be P [This is the corner on which the right triangle was made in the video] and the corner nearest to the quarter circle but not lying on it (other than P) to be Q. The positive x direction is OP and positive y direction is OQ. So the center of the small circle is at A(6,0). Let us assume there is a point T(a,b) on the small circle and there is another point B collinear to AT such that T divides AB in the ratio 1:2. [The large quarter circle in the diagram is centered at B with BT as the radius and T being the tangent point between the circles, hence the specific ratio of division]. Now we can find the value of B in terms of a, b as B(3a-12,3b). Now P lies on x axis hence its y coordinate is 0, and as OP is perpendicular to BP, x coordinate of P and B is equal. Thus P is (3a-12, 0). So, the length of the square side is 3a-12=3b => b=a-4. Thus T is expressed as (a,a-4). Now we know length of AT = 6, A(6,0), T(a,a-4). We solve for a using distance fomula and quadratic equation and get a=5±√17. Since a-4>0 (due to the way x and y axes are defined), a=5+√17. From this, side length of the square = 3a-12 = 3√17 + 3. Thus the area = (3√17 + 3)² = 162 + 18√17

@SerbAtheist 3 года назад

There is a quicker way: Let C and D be the two centers of circles. B' the corner forming a right-angled triangle with C and D, A' the point such that ABB'A' is a parallelogram and E corner point on the smaller circle. We have that AB is perpendicular to CD, but then so is A'B', thus

@nethminaxd4414 3 года назад

i am 15 years old and i am so glad that i made that first question

@emilia4355 3 года назад

Great job!

@amrelmowaled5786 3 года назад

Great job !!! , Keep going !

@mwysocz 3 года назад

I solved the second part easier I think - Just notice that AB and the 6+12 are perpendicular. Also they are both constrained by two sides of a square each and these pairs of sides are also perpendicular. So if you rotate AB and the sides that constrain it by 90 deg you will get AB parallel to 6+12 and they will be both constrained by sides that are the same distance away

@Ch.Hemachander-03 3 года назад

2:48 You can even more simplify the equation by dividing "2" on both sides. Firstly, I don't understand How to solve that question but after watching this I got an idea. Thanks for this❤️🔥. Love from Andhra Pradesh, India.

@amartyarandive6078 3 года назад

Yes, numbers will be smaller

@__dane__ 2 года назад

Every video I have seen from this channel has reminded me that if two circles are tangent then their centers and the tangent point are colinear.

@regibson23 3 года назад

This video was right above a clip from It's Always Sunny in Philadelphia. And I really had to decide which to watch first. RU-vid's algorithm is pretty damn good.

@artnotes 3 года назад

Since AB is a tangent line and it is on the parallel edge of the square so that it's the same as the line between two center of circle. that means it's 18.

@gibbogle 3 года назад

That's really doing it the hard way. Let P be the point where the circles touch. Draw a horizontal line through P. The angle between this line and AB equals the angle that the line through the centres and P makes with the right side of the square. Call this angle beta. Let d be the length of the side of the square, then cos(beta) = d/18. But AB = d/cos(beta) = 18.

@pro-nav 3 года назад

First part of the question is a 8th grade question in India, it is also solved example in RD Sharma. 2nd part was definitely a easier approach but I tried to apply trigonometry for this easy question. smile_with_tear_drop_emoji

@jeemonjose 3 года назад

I realized the length of AB is 18 at rhe beginning. Because I have this law in my mind that if two line segments perpendicular to each other, and inside a square, they're of equal length, if each line touches two opposite sides of the square. Just like here, points A and B touch opposite sides, and center points of two circles touch opposite sides(even though one point is in a corner, it does touch one side. I don't have a mathematical proof for this, I've just had many instances of seeing this, and thought it to be true, maybe there is a proof or someone can make one. First I thought only the perpendicular lines going through the center point of the square has this property. But, more trail and examples proved this idea. And I think the formation of triangles from the lines, and comparing them like Presh has done here, is a proof of this concept.

@Flare_0219 2 года назад

There's an area in the square where this works. I'm not 100% sure, but I believe the point has to be somewhere in that circle for it to work. If we slide the line segment that was made by the circles to the right, other than being useless to the problem, it would disprove the equivalency of the lines

@midhatmowla8820 2 года назад

Now that I think of it I think you are right...

@midhatmowla8820 2 года назад

Its like...you KNOW its right but without any proof XD

@andydorray9093 2 года назад

this was my logic as well. I'm sure a mathematical proof can be easily developed

@trantorino7030 3 года назад

There's one little bit missing to your proof which is showing that vertices A and B do actually lie on the opposite sides of the square. This you can get from your congruence argument at 4:47. B is at 2/3(12+x) + 1/3(6+x) vertically which is less than 12+x, and A is at 2/3(12+x) - 2/3(6+x) vertically, which is above zero.

@TechyMage 3 года назад

I left at x^3+18x-72 because i thought u will tell more elegant way to solve it. And 2nd part was genius

@יוכבדשםבדוי 2 года назад

Intuitively i knew AB is equal to their radii, but i struggled to prove it. The solution is so elegant and brilliant

@bibhuprasadmahananda6986 3 года назад

Fantastic solution ....

@bibhuprasadmahananda6986 3 года назад

I used coordinate Geometry to find the solution..... Very tedious and cumbersome...... I let the lower left bottom be the origin... And the side length be a.... Then wrote the circle equations and line equations(by differentiation) ha! .... But Finally found it out....

@josephlupton8196 3 года назад

this was a pretty easy problem to solve. Just basic quadratics for the first part and geometry for the second. The way I looked at the second part was since the segment connecting the centers of the circles is perpendicular to the tangent, that means the right triangles with each one as a hypotenuse would have to be at least similar. This is because perpendicular lines have slopes that are negative reciprocals of each other, which means the ratio of the larger leg to the smaller leg in each triangle would have to be the same. And since each triangle has a larger leg with length of the side of the square, they are congruent.

@fuadalali2162 3 года назад

2x^2+36x-144=0 Divide both sides by 2 X^2+18x-72=0 (x-3)(x+24)=0 x=3 or x=-24 (rejected) So each side of the square is 15 Area=225

@anandk9220 3 года назад

This is a really interesting problem. I solved part 1 exactly the same simplest way as in the video but solved part 2 using trigonometry. Here's how I did it. Let 'x' be the remaining side on the top and right edges of the square. x = 3√17 - 9 = 3.36931687685 Part 2 : Applying sine theta to the base acute angle of right triangle at upper right side of square (i.e. triangle formed by joining centers of semicircle and quarter circle), Base acute angle = theta =asin(15.36931687685/18) = 58.63302232706 Angle formed by AB with horizontal = 90° - theta = 31.3669776729° So, AB = 15.36931687685 ÷ (sin 31.3669776729°) = 17.9994... = 18 units

@mohammedal-haddad2652 3 года назад

Beautiful!

@benjobenjowherry 3 года назад

Finally finished one of these questions and got it right! I did it using a lot of algebra using the change in y and x from the centre of the smaller circle to the intersection of the circles, and called them n and m respectively, then found the gradient of the line going downwards, found the negative reciprocal, then used that to find the 2 points on the line, then used pythagoras' theorem to find the distance, my method was much harder but I'm glad I actually came to an answer, a correct one at that, for one of these questions

@althea9909 3 года назад

I completely skipped the first part because it is pretty obvious that the perpendicular lines are equal, thus you can get the result by adding up radiuses of both circles (6+12=18) If you need a quick answer, then this is the way to go but this method will not always work :)

@pradyutlaha01 3 года назад

The first problem was easy.....but the second one is the tougher one.... Nice solution..👍👍

@piusmathsclass4864 2 года назад

Day by day I am becoming your fan

@robertveith6383 3 года назад

@ MindYourDecisions -- *Do not* apply the Quadratic Formula before dividing both sides of the equation by the greatest common factor to reduce the size of the numbers in doing the arithmetic. In this case, apply the Quadratic Formula to x^2 + 18x - 72 = 0.

@rafaelliman8167 3 года назад

That's just simplifying..... but the result will be the same. The formula is applicable to any values of a, b, c.

@randomdude9135 3 года назад

I'm actually really happy that I managed to solve one of your problems 🤩😃

@huyanhhoang5420 3 года назад

The second solution is flawless!!!

@xyy3211 3 года назад

This question is very easy

@shadrana1 3 года назад

We will need to calculate the side length of the square. (6+u)^2+(12+u)^2=(6+12)^2 36+u^2+12u+u^2+24u+144=36+144+2*72 2u^2+36u=144 u^2+18u-72=0,use quadratic formula to find u, u= (-18+/-sqrt(324+288))/2=(-18+/-sqrt (612))/2 = -9+/-(sqrt(36*17))= -9+/-6(sqrt17)/2 u= -9+3(sqrt17) since u>0 Length of side of square=12+u=3+3sqrt17=3(1+sqrt17) Distance from centre of half circle to NE point of square=6+u =6-9+3sqrt17=3(sqrt17-1) Slope of radial line (CD say)= -3(sqrt17+1)/3(sqrt17-1) = -(sqrt17+1)/(sqrt17-1)*1 = -(sqrt17+1)^2/(sqrt17-1)=(18+2sqrt17)/16 = -(9+sqrt17)/8 (the slope is -ve since the line is sloped downwards from the y-axis to the x-axis) The horizontal co-ordinate of CD=3(sqrt17-1) and vertical co-ordinate of CD=3(sqrt17+1) the slope is the tangent of the angle to the top horizontal of the square. The slope of the common tangent line AB = +8/(9+sqrt17)since lines CD and AB intersect at right angles. The AB triangle has swapped the horizontal and vertical components of the CD triangle and the slope is the tangent of the complement of the CD triangle.This means we have AB and CD with three angles in common and one side in common,the length of the side of the square.This means the triangles attached to AB and CD are equal. Therefore, AB=CD=sqrt( (9(sqrt17+1)^2+9(sqrt17-1)^2)) =sqrt 324 =18 units. Area of square = (side of square)^2=(3(1+sqrt17))^2 =9(18+2sqrt17) =18(9+sqrt17) sq. units

@athirkhan5511 3 года назад

I got the area another way which was harder. I made a triangle centre of the semi circle to the top left corner to the bottom right corner. Top left angle is 45 degrees and we have two sides. Use sine rule twice to find the diagonal. Then find the length of one side by dividing by square root 2. Then find the area. For the length AB I thought I had correct but was wrong.

@TheHatkeHaryanvi 3 года назад

Superb

@lucaslucas191202 3 года назад

Finally a challenging puzzle that I actually managed to solve and not just give up on lol

@dioptre 3 года назад

good job! I gave up.

@italixgaming915 3 года назад

You didn't solve anything I'm afraid because there was a MASSIVE ASSUMPTION at the very beginning: he assumes that he can draw an half circle and a quarter circle inside a square the way he did and that these two circles will be tangent. WHERE IS THE PROOF OF THIS?

@lucaslucas191202 3 года назад

@@italixgaming915 That's part of the question. He also 'assumes' that the radii of the circles are 6 and 12, but you don't complain about that. Had he not said they were tangent you wouldn't be able to solve the question because there would be multiple answers

@italixgaming915 3 года назад

@@lucaslucas191202 You can draw two circles like he did, in a way they are tangent, and you can also draw a rectangle (the fact this is a rectangle is pretty obvious) in the way he did but I'm really really really really really really really really sorry, but there is no proof that this rectangle must be a square. You must demonstrate it as it is obviously part of the problem. This is why I say, and that I maintain, that this problem is a complete garbage.

@lucaslucas191202 3 года назад

@@italixgaming915 What are you on about? I will transcript the start of the video for you: "Suppose a *square* contains a semicircle and a quarter circle as shown. The semicircle and quarter circle are *tangent* to each other. Suppose the semicircle has a diameter equal to 12, and the quarter circle has a radius equal to 12." The basis of the problem is that we have square, and that the two circles in it are tangent. If you don't assume that, then you're not solving this problem! Problems are all about assuming certain things, and then finding other information about the geometry based on the assumptions.

@KnowledgeEducationIndia 3 года назад

I have found the area but finding the length of the tangent is a much difficult task. 😅😅😅

@epikherolol8189 3 года назад

Ikr

@wrOngplan3t 3 года назад

Finally, FINALLY managed one (half) of these problems - part 2, by just looking at the thumbnail. I somehow got AB was the same length as those radiuses, being at right angle to them and ending at opposite edges of the square. That's all :P (maybe I was just lucky).

@Calisthenics_Engineer 3 года назад

Lots of appreciation from India 🇮🇳 🇮🇳

@heytayo2596 2 года назад

On the qestion, you should give the 90 mark degree on the midle of AB line and the radius of the circle. Cause, if its not 90, the way you find AB line cant be used

@michaelxx2x 3 года назад

What I really like about the solving is the using of basic math. Wonderful

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@Vanilla.coke1234 3 года назад

I see the thumbnails for these things and always try to figure them out and then skip to the end to see how far off I was. Can't believe it but I guessed it right this time. What I did was completely different to the video but it worked so I'm happy

@ricardofsantos8258 3 года назад

I spent 3 hours and 9 sheets of paper extending triangles for every side and dealing with enormous roots, until I finally got to 18. Then, Presh says: "draw this line here. You will see that [OBVIOUS ANSWER]". That's why I hate geometry. Because it's so difficult to find those beautiful and simple answers. Ok, I don't hate it.

@maruthasalamoorthiviswanat153 3 года назад

excellent problem and excellent solution. thank you sir

@arkabanerjee1091 3 года назад

Video title was so apt

@meboy5495 3 года назад

I'm agriculture student but i love maths because of you 😍

@kartiklakhotia3861 3 года назад

Nice Problem

@imoj 3 года назад

For the first part I did 2x² + 36x - 144 = 0 Halve it -> x² + 18x - 72 = 0 Complete a² + 2ab + b² -> x² + 2*9*x + 9² - 81 - 72 = 0 So it turns into (x + 9)² - 153 = 0 Move independent part to right and sqrt (x + 9) = ± √153 x + 9 = ±√9*17 = ±3√17 x = -9 ± 3√17 Only use positive because we're measuring distance. x = 3√17 - 9

@JSSTyger 3 года назад

3:14 "...the side of the SQUEER"

@babakabanaras4749 3 года назад

Sir your Questions are really amazing

@ФилиппЛыков-д8е 3 года назад

3:51 First, one needs to prove that the point _B_ lies within the segment that is the vertical side of the square. It might be on the top side while the point _A_ would still be on the left side. Then we would have a totally different configuration. That _A_ is on the vertical side can be proved using the cosine of _alpha_.

@springroll2624 3 года назад

Let F be the center of the semicircle, F' be the center of the quarter circle, C be the point of contact between the semi and quarter circles, and P be the upper right corner of the square. Since F'C-FC=6 and F'P-FP=6, both C and P are on the same hyperbola with F and F' as it's focus. Since FP < F'P, P is above the line AB. (The hyperbola which C and P lie never come closer to F' than the tangent line AB.)

@IS-py3dk 3 года назад

The trick lies in getting that there are 2 right triangles which have lengths of their hypotenyse = 18 and are also congruent to each other

@gelbkehlchen Год назад

Solution: I call the side of the square a. Right triangle: hypotenuse = 12+6 = 18, horizontal leg: a-6 vertical leg: a Pythagoras: (a-6)²+a² = 18² ⟹ a²-12a+36+a² = 324 ⟹ 2a²-12a+36 = 324 |/2 ⟹ a²-6a+18 = 162 |-162 ⟹ a²-6a-144 = 0 |p-q-formula ⟹ a1/2 = 3±√(9+144) = 3±√153 ⟹ a1 = 3+√153 and a2 = 3-√153 [not guilty, because

@viniciusfernandes2303 3 года назад

Thanks for the video!

@TimJSwan 2 года назад

Darn, for the 2nd part, I was going about getting the tangent line then trying to compute the length rather than finding out about proving the symmetry.

@xtraPathshala 3 года назад

Animation is most powerful tool to understand complicated problem

@anjitahlawat 3 года назад

I knew it, but you really deserve a Subscription from my side.

@Anonymous-kw7ls 3 года назад

I solved for the area of square but failed for that length. Btw, the method used for solving the length was Fantastic.🎉 It actually used similarity concept but for three triangles.

@smchoi9948 2 года назад

Simple way to know "|AB| = 18": [General fact] Given a pair of // lines separated by a fixed distance (say, L₁//L₂), while another line L intersects L₁ and L₂ at P and Q respectively. |PQ| could be solely determined by the acute angle between L and L₁ (or L₂). Parallelograms always have equal opposite sides, true? [This problem] Name the centres concerned O and P. Since OP⊥AB (common normal of circles⊥common tangent of circles), the acute angle between OP and the horizontal square sides is the same as that between AB and the vertical square sides, so |OP| = |AB|. Sketch of a direct co-geom. approach: [1] Let C₁ be the circle completed from the blue quarter-sector & C₂ be that completed from the green semicircle. [2] Align the whole figure on the Cartesian plane so that the centre of C₁ is at O(0,0) while OB lies on +ve x-axis. [3] Let s be the length of a square side. The centre of C₂, w/ radius 6, is at P(s,s-6). [4] As |OP| = radii sum of C₁ & C₂ due to external tangency, s² + (s-6)² = 18² ...($). [5] ($) & s>0 ⇒ s = 3+3√(17) ⇒ area of square = s² = 162 + 18√(17) [6] Let T(u,v) be where C₁ touches C₂. As T divides OP internally by 12:6 = 2:1, by the Section Formula, u = 2s/3 & v= 2(s-6)/3. [7] OP⊥AB ⇒ slope of AB = -1/slope of OP = -1/[(s-6)/s] = s/(6-s) [8] Eqn. of the line containing AB is y = [s/(6-s)] x + c, where 2(s-6)/3 = [s/(6-s)] (2s/3) + c (as T is on it) ⇔ c = 4(s²-6s+18)/3(s-6), i.e. y = [s/(6-s)] x + 4(s²-6s+18)/3(s-6) ...(*). [8] Put y=0 into (*), 0 = [s/(6-s)] x + 4(s²-6s+18)/3(s-6) ⇔ x = 4(s²-6s+18)/3s, so B is at (4(s²-6s+18)/3s,0). [9] Put y=s into (*), s = [s/(6-s)] x + 4(s²-6s+18)/3(s-6) ⇔ x = (s²-6s+72)/3s, so A is at ((s²-6s+72)/3s,s). [10] |AB| = √( { [4(s²-6s+18)/3s] - [(s²-6s+72)/3s] }² + (0 - s)² ) = √[(s-6)² + s²] = 18 (by ($))

@Schrodinger8509 3 года назад

Oh my God, very good approach

@dirklaubusch5659 3 года назад

First: Thanx for the question, it was fun to be able to solve it after so many years out of school. BUT: Way to easy; use the quadratic formula, and u have it. How about: " Find the smallest quadrat + circles with sidelenghts / radius [ Element of N ] that fit that geometrical drawing ". ( Sorry for my bad english; i hope u get my point ) That sqrt(17) in the solution makes it just looking ugly :)

@aniketb2010 2 года назад

Nothing is Impossible. Thanks PT.

@anmolraina5650 2 года назад

I solved this it was very easy than other problems 👍👍👍👍💟

@pesockek 2 года назад

Ive seen the answer before clicking on video for explanation and proving “am i right?” Yes Solved in 5 seconds in mind

@karlhendrikse 3 года назад

Haven't watched the video, but it's pretty easy. The distance from the center of one circle to the center of the other is just the sum of the radii = 18. The line AB is perpendicular to that line, and is the same length because it's as wide as the other line is tall.

@mustafaaa69 3 года назад

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@Nikioko 3 года назад

In a square, all angles are 90° and all sides are of equal length. Therefore, any perpendicular line segments connecting opposite sides have the same lengths.

@debonandonchetia8633 3 года назад

The 2nd one was mind blowing

@tubbdoose 3 года назад

After staring at the thumbnail for a while I was just like “oh the lines are parallel and since they are in a square they are probably the same length?”

@spacescopex 3 года назад

See my solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4E1ZSVU13eE.html

@Rahee11 3 года назад

An Interesting Math For You: If a shop purchases 2 lollipops brands A and B. Shop purchased lollipop A 4 for each dollar and B 6 for each dollar. If the shop purchased equal quantities of each brand sells them combined for 5 for each dollar, what is their profit /loss percentage?

@siperglx5051 3 года назад

Wow amazing

@JimsEvilTwin2 3 месяца назад

Love this exercise. But I started solving it based solely on the thumbnail and did not realize that it was given that the containing box is a square. 😂

@ejrupp9555 2 года назад

The 2nd part ... lines that are perpendicular and touch all 4 sides are equal. Like in the circle theorem except the caveat is 4 points must touch all 4 sides.