A special math puzzle for Valentine's Day

[@elkiechan6682]

Just like any Valentine problem, this cannot be solved without breaking the heart.

[@TomFromMars]

Underrated comment

[@鋒澤最棒]

Can be solved by calculas without breaking the heart

[@adityavardhanreddy8332]

@@鋒澤最棒 then it ends up becoming complicated.

[@keinsent]

@@adityavardhanreddy8332 Nothing is complicated if you understand it.

[@jamestanny849]

This broke my brain cells bit by bit

[@kingsleyzhong677]

Lonely Mathematicians ------->

[@bitanmandal977]

First I read it Lovely Mathematicians 😂

[@arnauab25_03]

@@bitanmandal977 I wish it was that...

[@IS-py3dk]

LOL..... 😂😂

[@x714n0____]

🙃✌🏻

[@disappointmint7797]

dont attack me like that

[@Ikiguyy]

Spending my Valentines day doing love related maths problems 👍🏼

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....

[@drp7751]

Hello *darkness* my old friend...

[@blizzard7615]

Perfect question for the perfect day .... atleast for them (Sad noises*)

[@richardgratton7557]

Yeah, in lieu of having a romantic relationship?😋

[@felizianosole896]

This is love

[@ultracapitalistutopia3550]

Lesson of today: Just remember the law of cosine

[@youssefabdallah1157]

I have never been in a situation in my Normal life and needed to use sin, cos or tan , however in other situations it must be used

[@FirstnameLastname-hg5gt]

@@youssefabdallah1157 My father once (working on a construction) knew the lengths of the three sides and the length of one diagonal of a quadrilateral. He needed the length of the other diagonal and he asked me. Of course I made the job using cosine law and sine law.

[@youssefabdallah1157]

Well in this situation it's very useful

[@carcaperu4041]

You actually do not need it. There is a simpler solution. 1) Get the cartisian coordinates of the center of the 2 circles (sqrt(2)/4, 3xSqrt(2)/4) and (0,R), since the distance between them is R-1/2 that gives you the solution.

[@youssefabdallah1157]

There are hundreds of ways that can be used to solve any geometric problem in many different ways

[@typingcat]

It's Valentine's Day? Oh, crap; I forgot to get a girlfriend, again.

[@hacker1oo173]

Lol

[@justanonverifiedyoutubechannel]

Underrated

[@chandranichaki9580]

kzbin.info/www/bejne/bYqulZuji9eClas

[@rafaelliman8167]

After that triangle shaded in green, I wanna avoid sin, cos, tan whenever I can. So I draw an altitude perpendicular to the sqrt(2)-r side, dividing it to 2 right triangles. The upper triangle will be a 45-45-90 triangle and since the hypotenuse of that triangle is 1/2, its legs will each be sqrt(2)/4. We'll focus on the lower triangle, whose hypotenuse is 1/2-r. One of its 2 legs will also be sqrt(2)/4, and the other leg will be sqrt(2)-r-sqrt(2)/4=3sqrt(2)/4-r. Based on the right triangle theorem, the equation will be: (3sqrt(2)/4-r)^2+(sqrt(2)/4)^2=(1/2-r)^2 r^2-3sqrt(2)r/2+18/16+2/16=r^2-r+1/4 (1-3sqrt(2)/2)r=-1 r=1/(3sqrt(2)/2-1)=(3sqrt(2)/2+1)/(7/2)=(3sqrt(2)+2)/7 And that's how you solve without sin, cos, tan. And if Presh went this route, he would have to mention his 'favorite right triangle theorem' again. One has to wonder if he considered this before deciding to go the 'al-kashi' route.

[@andreasproteus1465]

It is simpler to solve the equation of the circle through the 3 known points : (0,0), (1.5, 0.5) and (0.5, 1.5). You get the coordinates of the centre for free too.

[@tanguy_7382]

@@andreasproteus1465 Could you explain how exactly do you get those coordinates ?

[@Мопс_001]

@@andreasproteus1465 I don't understand as well. To get coordinates of center we need to know coordinates of circle which is not really possible on my mind (especially because we don't know where those two halfs of mini circles intersect with big one)

[@andreasproteus1465]

@@Мопс_001 Since the question asks for a numerical answer and not for a formula we can see from the diagram that each of the two tangent points is very close to the apex of the mini circle. So we assume that the big circle passes through these points and the origin. If you solve the three circle equations equations you will get R = 0.884 which is only 0.008 smaller than his answer. Note that because one of the points is (0,0) you do not have to solve a quadratic so the solution is very simple.

[@Currywurst4444]

@@andreasproteus1465 It doesnt matter if it is close to the real answer. If there is any difference from the right answer, no matter how small, the solution is wrong.

[@brodysdaddy]

Searching the comments for all the, "easy, did this one in my head" people.....they must be taking the day off.

[@DharmveerSingh-ko3nd]

If you want more hard maths loving questions then I will highly recommend you this channel's latest videos #onlymathlovers ...

[@simonsez6200]

or, they're on a date

[@gravysnake78]

Bro they get all the bitches 😫

[@michaelroditis1952]

Hahaha I usually am of those people but when it gets trigonometrical..

[@Tiqerboy]

As it turns out, it looks like one of those "I think can do this in my head" problems but it turns out it's not that easy. The four triangles with the diagonal slice is another example that is harder than it looks. Same with the ladder leaning against the wall with a cubic box below it

[@universalyoutuber195]

Mingles- ❤️ Singles - ✍️

[@kurumi394]

Hmm yes, let me make my date solve this question, I'm sure she'll love it lol

[@archeopterixneuroza4715]

Maybe

[@prajwals2793]

Fastest way to break up. (Unless she's a mathematician ofc) but if you solve it in a satisfying way she'd love it.. If my partner were to pull out something like this and solve it satisfactorily, I'd never let go 😂

[@ediguzelis8034]

I am so lonely that I tought you were being sarcastic

[@fionalee6744]

ಠ∀ಠ

[@tompirovano9755]

@@ediguzelis8034 I feel you man

[@ahsanbulbul8512]

Imagine you propose him/her and they pull out this math as a condition for acceptance ........ 😂

[@yeyogapriya5883]

Of cource, this is a right way

[@jellyfish0311]

Well, if the suitor delivers you know it's the right one

[@Z-eng0]

And tell you to do it without trigonometry too

[@supu8599]

Y=X²+(Y-X⅔)² makes a heart in graph.😄😄💕💕

[@Susp7]

Oh yeah it does

[@Tiqerboy]

One of the rare times I solved this EXACTLY how Presh did it. I was thinking of maybe solving it using co-ordinate geometry but thinking that would lead to a messier solution. Happy Valentine's Day to Presh, may he find the love of his life.

[@barunkumarsahu1354]

No coordinate geometry is easier if you choose the perfect origin...

[@vedants.vispute77]

Happy Valentines Day... I am gonna date with integrals tonight ❤️

[@shreyan1362]

I would be dating with Prime Numbers. I had a crush on them and they just agreed yesterday. My day is set 😏😏😏

[@adityasuhane8930]

*i* am ok with my imaginary date and i don't *even* have power to make it real...

[@ManishGupta-hb4iu]

kzbin.info/www/bejne/rIbGf3ePoaenpcU

[@hubb8049]

I hate that your profile pic is technically correct

[@Kingkhan-og8xw]

Me to my calculator: (Me+ You)²= Me²+You²+2Babies My calculator:👁👄👁

[@whothefisyash]

Sending to my GF 😁

[@aayush_7037]

do you have 😂

[@ManishGupta-hb4iu]

kzbin.info/www/bejne/rIbGf3ePoaenpcU

[@prachi4110]

You will edit it after 10 min yelling that you are single cause your girlfriend couldn't solve it

[@whothefisyash]

@@prachi4110 my GF❤ is complex number (as they are imaginary) 😉

[@abdullahmohammed8502]

@@whothefisyash damn, sad. Good luck tho

[@ManaredPickens]

For once I was actually on the right track to solving it lol

[@rishianand153]

Even I but until I reached equation the square roots were scary lol..

[@jus1t-gg]

now i know how to make a heart shape in ms paint

[@arkabanerjee1091]

Your geometry problems are truly out of the world - Simple but elegant

[@mathmaticalproblemandsolution]

at (2:15)my question is ....how exactly the center of big circle and opposite vertex of square lie on a same line ?????

[@Z-eng0]

2:15 For anyone (like me) who may wonder how we know that the centre of the big circle lies ON the square's diagonal, I found a small proof of that. Start with the assumption that the centre is NOT on the diagonal, draw the diagonal, then draw the centre anywhere away from that line, connect the centre to the point of intersection with the square below, Notice that you can draw a circle surrounding the square with its diameter being the diagonal [search up cyclic quadrilaterals and circum circle of a right triangle], Notice that the new circle and the original big circle both have to intersect at the bottom point (point where the square edge and the circle originally meet), Draw the tangent to both circles there, Notice that it makes 90deg angle with both the diagonal AND the radius of the big circle, Since these are 2 different lines meeting at a common point AND having the same angle with the horizontal, it means they MUST BE COLINEAR (meaning they lie on the same line) [btw, note that 2 non parallel lines can intersect only once, and 2 parallel lines having same inclination angles can NEVER MEET, if they DO meet, that means they meet on infinite points and are the same line]

[@carcaperu4041]

I solve it in two different ways. 1) I got the cartisian coordinates of the center of the 2 circles (sqrt(2)/4, 3xSqrt(2)/4) and (0,R), since the distance between them is R-1/2 .That gives me the solution. 2)In a similar way using the cartisian coordinates found the point x,y were they intersect and the derivatives are the same. That is algebraically more complicated but that method can be used for any type of figures (for example ellipses).

[@geoninja8971]

That was a satisfying problem.... loved it! good to see some engaging geometry again Presh...

[@abhi36292]

u first :0

[@jagratishrivastava7504]

How is your comment 5 days ago when the video is just uploaded?!!!

[@sakshamdobriyal9952]

It's been like a minute 😂

[@popeye3522]

Lol 5 days ago... How it's possible. ?

[@hansondesa188]

Patreon support They get vid before

[@PetraKann]

The Radius is a term that relates to circular geometries such as a 2D circle or a 3D sphere. There are terms in engineering such as the Hydraulic diameter or equivalent particle size for irregular shaped particles but these are used as averages in order to estimate other properties. For example, if you have laminar flow in a circular geometry, it's relatively simple to determine the flow characteristics, pressure drop etc. The same fluid flowing in a pipe with a peanut shaped cross section, at the same average volumetric flow rate, can be characterised by using the hydraulic diameter as the dimension within the flow. [The Hydraulic diameter formula is given by 4(A/p) where A=cross sectional area of pipe or duct & p= wetted perimeter of pipe or duct. In this case, the area of the heart and its perimeter need to be known or can be estimated indirectly.]

[@Z-eng0]

There's a non-trigonometric way of solving it as well (and also proving that the center lies ON the vertical diagonal of the square without saying it's just symmetry, but I'll ignore that for now and make that assumption like in the vid). So first I construct both the square's diagonals, then construct a line from the left semi-circle intersection with the big circle through the semi-circles center (which is the midpoint of the square's side) and extend it to meet with the vertical diagonal, and do the same for the left semi-circle (they should both intersect on the diagonal if drawn correctly), then construct a smaller square by connecting the center (diagonal intersection point) of the big square with the 2 semi-circle centers, this would make a small square half the deminsions of the big one, construct it's diagonals as well, the distance form the big circle center to the left and right semi-circle centers is r-1/2, those 2 lines along with the diagonal of the small square form an isosceles triangle which we'll redraw outside our whole figure to analyze, the whole base is 2^0.5/2 and the bisector from the big circle center divides that into 2^0.5/4, we now also have 2 right triangles with only 1 side missing, looking at the original figure that side would be half the small diagonal minus distance from big square center to the big circle center, which is r-(2)^0.5/2 (whole radius minus half big diagonal), making the missing triangle side equal (2^0.5)/4 - (r-(2^0.5)/2), now we have a right triangle with 2 legs of 2^0.5/4 and 3*2^0.5/4-r and a hypotenuse of r-1/2, using Pythagorean theorem we can simplify and solve for r (the r² terms will cancel anyway), and we get the radius desired

[@thricegreatart]

"We can solve this using Al-Kashi's Law of Cosines. It's a familiar formula." No... no it's not.

[@MarieAnne.]

We can also use coordinate geometry to solve. Let bottom vertex of square be at origin (0,0). Then top vertex is at (0, √2) and left/right vertices are at (±√2/2, √2/2). Centres of semi-circles are halfway between top vertex and left/right vertex and have coordinates (±√2/4, 3√2/4). Square and large circle are symmetric about y-axis, so centre of circle with radius r is on y-axis and a distance of r units from origin. Centre of large circle = (0, r). Just like Presh did in video, I found that distance between center of either semi-circle and larger circle = r − 1/2. But we can also use distance formula to find this, since we have coordinates of centres of large circle and semi-circles: r − 1/2 = Distance between (0, r) and (±√2/4, 3√2/4) (r − 1/2)² = (0 − (±√2/4))² + (r - 3√2/4)² It's fairly straightforward from there to find r = (3√2 + 2) / 7

[@marienbad2]

Beautiful solution. I got in a mess of angles and stuff as I overthought it. I did think about tangents but missed the obvious colinearity of them inside the square!

[@dicandeo9698]

I solved it a different way: Suppose the center of the square is the origin of a cartesian coordinate system. A circle with center (a;b) and radius r is given by the equation (x - a)^2 + (y - b)^2 = r^2. The center of the large circle has to have 0 as its x-coordinate, so a = 0. Since the bottom point of the square has the coordinates (0; -1/sqrt(2)), we get the equation (I): 1/2 + 2b/sqrt(2) + b^2 = r^2 Since the circles are touching each other from the inside, the distance between the two centers must be equal to the difference between the radii. The center of the half circle to the right is at (sqrt(2)/4; sqrt(2)/4). The distance between that center and the center of the large circle (0; b) is equal to sqrt((sqrt(2)/4)^2 + (sqrt(2)/4 - b)^2). Simplyfing this, we get that the distance between the two centers is equal to sqrt(b^2 - b/sqrt(2) + 1/4). Since this must be equal to the difference between the radii, we get (II): sqrt(b^2 - b/sqrt(2) + 1/4) = r - 1/2 Solving this system of two equations, we get that b = (-sqrt(2)+4)/14 and r = sqrt(22+12*sqrt(2))/7 = (3*sqrt(2)+2)/7, which is approximately 0.891805812 or 0.892 if rounded to the thousandths. Generalising this problem for squares with a side length of s, we get that the center of the large circle has the coordinates (0; s * (-sqrt(2)+4)/14) and a radius r = s * (3 * sqrt(2) + 2)/7

[@yoapps137]

Honestly the easiest way to solve this is by knowing thatthe 3 points that the big circle touches creates an isosceles triangle whose circumcenter can be calculated fairly easily because already know one of the perpendicular bisectors is the squares diagonal!!!

[@carlosmingorance2110]

I just traced an AB chord joining both tangent points. That's the hypotenuse of a rect triangle with a third corner in the center of the square, and both legs had a length of 1, so its length was easy to calculate. Then I traced the diameter, of length equal to the diagonal to the square plus x. Intersecting chords theorem then resolves: x times the diagonal (square root of two) is equal to the square of half the square root of two (both halves of the first chord multiplied). For some reason, it seemed way simpler to me.

[@denvercheddie]

I was actually able to solve this using only geometry (no trigs). Complete one of the smaller circles, noting that it passes through the center of the square. We can then observe that there are two intersecting chords in this smaller circle. These chords are 1) half the diagonal of the square, and 2) the normal drawn from the point of tangency of the 2 circles. Using the chord intersection theorem, we need to solve r*(1-r) = (r - sqrt(2) / 2) * (sqrt(2) - r) to get the same result as in the video.

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....

[@o_enamuel]

fun fact: in Brazil, Valentine's Day is on June 12th (Which is my b-day too)

[@I_am_Anugrah]

Happy Birthday bro Just come here and see this message on June 12th🤣

[@o_enamuel]

@@I_am_Anugrah Ty

[@I_am_Anugrah]

@@o_enamuel You are supposed to thank me on June 12th😅

[@o_enamuel]

@@I_am_Anugrah is that there is a risk that I will forget

[@I_am_Anugrah]

@@o_enamuel Okay Okay I accept this😅 It was really an awesome experience to talk to you .....😊

[@ФилиппЛыков-д8е]

Great solution! I solved it applying your favourite right triangle theorem to the triangle that is the lower part of the one you have considered in your solution.

[@PHYSICSSIRJEE]

Nice valentine's one Presh ❤️ . Spreading the love through mathematics 👌😊

[@DharmveerSingh-ko3nd]

If you want more hard maths loving questions then I will highly recommend you this channel's latest videos #onlymathlovers ...

[@mohamedachrafnahid8822]

Your channel is wonderful!!!Thank you so hard for these valuable informations!!have a super mathematical day

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....

[@mrpumaknight]

Happy valentine! ❤🌹 I misunderstood something! How are you sure that the line segment between two centers (center of large circle and the semicircle) will pass the tangent point?! 🤔

[@polyacov_yury]

That's according to the theorem in the beginning of the video. Since large circle and the semi-circle are tangent - their centers and the tangent point are on the same line.

[@RavikumarG]

Normal to a circle always passes through center of that circle. (You can verify that by finding general equation of normal to circle centered at 0,0 it would come out as in form : y = m*x. This is WLOG as you can bring any circle to 0,0 by shifting axis. ) Since both curves (arc of semicircle and big circle) are meeting at a single point, they are tangential to each other, so they share a common tangent and a common normal. Now that they share a common normal, this common normal should pass through point of intersection of curves and both the centers, making them collinear.

[@mathmaticalproblemandsolution]

at (2:15)my question is ....how exactly the center of big circle and opposite vertex of square lie on a same line ?????

[@RavikumarG]

​@@mathmaticalproblemandsolution Well, you can always draw a line joining two points :P I get the question is "How the center of big circle falls on square's vertical diagonal" (say, VD = vertical diagonal) The heart shown should be symmetrical about VD, Like VD is a mirror. Also the big circle can be thought of as circumcircle of triangle formed with square's bottom vertex & two points of contact with semi-circles and big circle. circumcenter gotta be symmetrical about VD, but it's unique, so it will lie on VD.

[@mathmaticalproblemandsolution]

@@RavikumarG bro your every point is about assumption can we show geometrically that it form a diameter ....as presh did not tell about that also in the video so may be the question is incomplete or prove it with geometrical properties from the given data .....

[@philipkudrna5643]

In the end, it almost always comes down to al-quasi and his law of cosines...!

[@jaylenlockhart6467]

I never was great with geometry but doing this question was fun! I got stuck around 3:01 and had to look at the solution.

[@rishiaman2]

pro

[@maverick0149]

Another simple method to solve the problem: Use a Cartesian reference with its origin in the down corner of the square and its axes along the sides of the square. Let A = (0.5 , 1) the centre of the semicircle on the left C = (c , c) the centre of the larger circle R: radius of the larger circle, being R = 2^0.5 x c r : radius of the semicircle, being r = 0.5 a : distance between A and C Now it is: a = R - r that is a = 2^0.5 x c - 0.5 and also a^2 = (0.5 - c)^2 + (1 - c )^2. Solving the simple equation (2^0.5 x c - 0.5)^2 = (0.5 - c)^2 + (1 - c )^2 we can find c and then R.

[@Someone-vj2cz]

Such a great way to spend my Single Awareness Day

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers ........

[@shreyan1362]

Maybe there's a reason why this video is *420* long

[@slolilols]

Yes. Weed

[@ManishGupta-hb4iu]

kzbin.info/www/bejne/rIbGf3ePoaenpcU

[@sharduljoshi5679]

HAAHAHAHHAHA xd

[@jackhandma1011]

Nice

[@nelser1160]

Haha weed number OMG you're so funny with your original jokes redditor 😁😂🤣🤣🤣😂😎😎😎

[@ЕркебуланМолдагали-э7ц]

What if we solve it by multiplication of chords, like 0.5×(2)^1/2×0.5×(2)^1/2=(2)^1/2×t t=0.5/(2)^1/2 R=[(2)^1/2+0.5/(2)^1/2]:2 R=[5×(2)^1/2]8

[@marceloalves7627]

Hi Presh. I really love your videos and I use to show many of them to my students who love as well. I have a question that you probably have hear before. Which software or softwares do you use to make your videos? I would love to make a few videos for my classes using this technique but I have no idea how to. If you help me I am sure my students will be thrilled. I’m looking forward to do that.

[@dcterr1]

Very clever solution. I paused the video and tried to solve it using another method, but I came up with 5√2/8, which is incorrect.

[@lamgiakhang5712]

Me: Can't solve the problem and keep thinking:Where is the center of the big circle

[@shauryabhansali3731]

But if we divide the diagonal of square and calculate the half part , it can also find out the radius......then it's radius is coming ...... 1/√2

[@trueriver1950]

2:56 in Al Kashi's law of Cosines, by setting theta to a right angle it's one way of getting an argument going about the name of that special case

[@richardgratton7557]

Yes, we know. Don’t start something...it’s a nice Sunday morning, and it’s Valentines Day. Gogou for a walk or something!😉

[@nemoumbra0]

1:11 It's quite funny... You consider the fact that points A, T and B are collinear to be more complicated and less known than the fact that the tangent lines to these circles are, in fact, one and the same!

[@vinayakpandey1032]

How can you say that diameter of the large circle will pas through the diagonal of square.

[@mathmaticalproblemandsolution]

yes the same question i have at (2:15)my question is ....how exactly the center of big circle and opposite vertex of square lie on a same line ?????

[@dlangeberg]

Dang, we all really out here circumcising circles on Valentine's day

[@Qermaq]

Hope we make bris work of it!

[@FireStormOOO_]

Oddly, when solving based on the large inscribed isosceles triangle, I'm getting 5/8*sqrt(2), which is within 1% of the answer given in the video but definitely not equal.

[@janfloh9549]

You are right. I have also your solution.

[@gatisgraubins7359]

I came to same result solving this in 2 different ways

[@Caomhanach]

Same, using coordinate geometry.

[@Gwunderi25]

Same here

[@krishnachopda1062]

...circumference of the circle...an angle 45 degrees... and that’s our answer!!

[@nibaranghosh2202]

Thank you sir! 😇 ❤

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .......

[@NZC_Meow]

Happy Valentine's Day to Everyone! ❤️❤️❤️ Secret to how that person wrote a comment 5 days ago when the video was uploaded few minutes ago: There are some unlisted videos in the playlist which aren't uploaded yet. Go to a random playlist and sort by "date added (newest)" and then you may find some videos which aren't uploaded yet. They're uploaded for members only, but they're in the playlists. This is why we cant see them, but we can access them through playlists. This is the secret. Now you can time travel too!

[@ManishGupta-hb4iu]

kzbin.info/www/bejne/rIbGf3ePoaenpcU

[@ΓιαννηςΜπουρας-ξ8ν]

I have a question about this problem. How do we know that the point of the left half-circle which touches the circle with its center and with the center of the circle is indeed straight line? Please someone answer this.....

[@manusarda]

That's explained in start of solution by tangent concept. Only difference is that in his question small circle is within large circle but in explaination small circle is outside large circle

[@laniemi2062]

Thanks for the problem. I got the result r=0.89 using only ruler and compass. The midpoint of the circle is easier to find and construct if you seek a circle where radius =r-0.5

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....

[@laniemi2062]

This site is hard enough for me because I am trying to solve the problems in my head only.

[@hmv678]

Nice one for Valentine's day. Thank you for your work. Subscribed.

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....

[@AlthafShameelPP]

So we "nerds" are also celebrating valentines day😎

[@palootto4021]

Just what I need to start my day.

[@DharmveerSingh-ko3nd]

If you want more hard maths loving questions then I will highly recommend you this channel's latest videos #onlymathlovers ...

[@noobmaster-zh9vk]

Time 4:20

[@gamingmusicandjokesandabit1240]

Who loves MindYourDecisions?

[@krissbarnwal8003]

This is how I spend My Valentine's Day 🥳 Truly Mathly....😉

[@mikemanh]

I had tried a different method before i watched your solution. My method is: 1. I chose 3 points on the large circle: where the square and the two semicircles are tangent on the big circle. 2. Side length of the isosceles triangle are a=sqrt(2) and b=sqrt(2,5) 3. I tried to find the circumcircle of this triangle 4. And my result is: 0.8839 Where did i make mistake??

[@janfloh9549]

I think your solution is right. 👍

[@maqsudxorazm]

Thanks very much

[@proferonquito7228]

The large circle does not circumscribe the figure of the heart. The cut

[@eminet.9845]

Other people: Happy Valentine's Days MYD:Nahh, i wanna find the area.

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers ........

[@yokkie2846]

Happy Valentine's day to you to, mathematics.

[@cosinex8361]

Is there a way to solve this without the law of cosines

[@alexhuttunen3631]

Yes you can make a little right-angled triangle adjacent to the one Presh uses. Hyp is (r-1/2) sides 1/ (2rt(2)) and (3/2rt(2)) - r ... then solve using Pythagoras

[@alexhuttunen3631]

I found this by putting the base of the square at (0,0) and asking what co-ordinate the centre of the semi-circle would be.

[@alexhuttunen3631]

x co-ordinate is 1/2rt(2) and y is 3/2rt(2) - you know this because the centre of the semi-circle is the centre of the side of the square

[@uclongle3783]

How to draw that large circle????

[@Wazarrd]

PLEASE SOMEONE HELP!!! I made two triangles. One with vertices at the bottom vertex of the square, the top of the circle and the last where the two circles touch. This is right angled. The other has vertices at the top of the circle, where the two circles meet and the top vertex of the square, this is also right angled. I worked out some side lengths from root(1.5^2 + 0.5^2) = 2.5 and root(2.5^2 - root(2)^2) = root(2)/2 From this you get simultaneous equations (d=diameter) (Root(2)/2)^2 + (d - root(2))^2 = x^2 And (Root(5/2))^2 + x^2 = d^2 I then got d = root((5-2root(2)) Therefore radius is d/2 = 0.737 If anyone can help me know where I went wrong it would be much appreciated. It’s really bothering me now.

[@PuzzleAdda]

Riddle: I had $2.00. My mom gave me $10.00 while my dad gave me $30.00. My aunt and uncle gave me $100.00. I had another $5.00. How much do I have? Answer: kzbin.info/www/bejne/oJCZoKh7bNt-itE

[@BTECESaiSachithReddy]

In which world did they say that the center of the bigger circle and the semi circle and the point of contact are collinear

[@Abhinav-sx1dz]

Video :22 Seconds ago.. Comment: 4days ago ... Riddle: KZbin

[@vedants.vispute77]

They r patreons

[@ankitlakum1]

Happy Valentine's Day to Everyone! ❤️❤️❤️

[@zdrastvutye]

hi! unlike the "circle given by 3 points" this is a nonlinear equation system, because the tangent point cannot be predicted, there is just a relationship of distances: 10CLS:WINDOW OPEN:WINDOW FULL:n1=4:n2=180:DIM x(n1,n2),y(n1,n2) 20PRINT "umkreis von 'herz' aus quadrat und zwei halbkreisen":wdh=0 30lq=1:xm1=lq:ym1=0.5*lq:xm2=.5*lq:ym2=lq:x1=0:y1=0:x2=lq:y2=0 40x3=0:y3=lq:rh=lq/2:sw=.01:nu=130:OPTION DEGREES:GOTO 130 50druk=ruk/SQR(2):zab11=xm1^2+ym1^2-rh^2 60nab11=2*(xm1-druk):ab11=zab11/nab11:zab12=-2*(ym1-druk) 70nab12=nab11:ab12=zab12/nab12:ab21=(druk-xm1)/(ym1-druk)*druk 80ab21=druk+ab21:ab22=(xm1-druk)/(ym1-druk):yi1=(ab21-ab11)/(ab12-ab22) 90zx11=-2*yi1*(ym1-druk):zx11=zx11+xm1^2+ym1^2-rh^2 100xi1=zx11/nab11:dlu1=(xi1-xm1)^2/rh^2:dlu2=(yi1-ym1)^2/rh^2 110dl=dlu1+dlu2-1:IF wdh=0 THEN ?dl 120RETURN 130ruk=SQR(xm1^2+ym1^2)/2:GOSUB 50 140ruk1=ruk:dl1=dl:ruk=ruk+sw:IF ruk>lq THEN STOP 150ruk2=ruk:GOSUB 50:IF dl1*dl>0 THEN 140 160ruk=(ruk1+ruk2)/2:GOSUB 50:IF dl1*dl>0 THEN ruk1=ruk ELSE ruk2=ruk 170IF ABS(dl)>1e-9 THEN 160 ELSE CLOSE WINDOW 4 180xm=druk:ym=xm:PRINT "der Umkreisradius=";ruk:rukl=ruk 190PRINT "der Mittelpunkt liegt auf=";xm;ym:GOSUB 680 200pro=SQR((xi1-xm)^2+(yi1-ym)^2)-ruk:PRINT "fehler=";pro*100;"%" 210i=0:xu1=x1:xu2=x2:yu1=y1:yu2=y2:GOTO 240 220dxu=(xu2-xu1)/n2:dyu=(yu2-yu1)/n2:FOR a=0 TO n2 230dxa=dxu*a:dya=dyu*a:x(i,a)=xu1+dxa:y(i,a)=yu1+dya:NEXT a:RETURN 240GOSUB 220:i=3:xu2=x3:yu2=y3:GOSUB 220:REM die kreisboegen berechnen 250xmu=xm1:ymu=ym1:wau=270:wae=180:ru=rh:GOTO 290 260dwu=wae/n2:FOR a=0 TO n2:dwa=a*dwu:wa=dwa+wau:dw=360*INT(wa/360) 270wa=wa-dw:dx=ru*COS(wa):dy=ru*SIN(wa):x(i,a)=xmu+dx:y(i,a)=ymu+dy 280NEXT a:RETURN 290i=1:GOSUB 260:i=2:xmu=xm2:ymu=ym2:wau=0:GOSUB 260:i=4:xmu=xm:ymu=ym 300wae=360:ru=ruk:GOSUB 260:GOSUB 350:wvu=45:GOTO 330:REM verdrehen 310xvx=xu*COS(wvu):xvy=yu*SIN(wvu):xv=xvx-xvy 320yvx=xu*SIN(wvu):yvy=yu*COS(wvu):yv=yvx+yvy:RETURN 330FOR a=0 TO n1:FOR b=0 TO n2:xu=x(a,b):yu=y(a,b):GOSUB 310 340x(a,b)=xv:y(a,b)=yv:NEXT b:NEXT a:CLS:GOSUB 350:END 350sumx=0:sumy=0:anz=0:colx=2 360FOR a=0 TO n1:FOR b=0 TO n2:sumx=sumx+x(a,b):sumy=sumy+y(a,b) 370anz=anz+1:NEXT b:NEXT a:mittelx=sumx/anz:mittely=sumy/anz 380xmin=mittelx:xmax=xmin:ymin=mittely:ymax=ymin 390FOR a=0 TO n1:FOR b=0 TO n2 400IF x(a,b)xmax THEN xmax=x(a,b) 420IF y(a,b)ymax THEN ymax=y(a,b) 440NEXT b:NEXT a:IF xmin=xmax OR ymin=ymax THEN ELSE 480 450IF ymin=ymax THEN 470 460mass=YVIRTUAL/(ymax-ymin):GOTO 520 470mass=XVIRTUAL/(xmax-xmin):GOTO 520 480masx=XVIRTUAL/(xmax-xmin):masy=YVIRTUAL/(ymax-ymin) 490IF masx

[@md.ahsanulkabirrana2042]

Why the center of the large circle will lie on the diameter of the square?

[@damianbla4469]

I did it in the other way. I got another answer than you got. And I've just figured out what I did wrong. Here is my solution: Part I - Names The square has vertices ABCD (where: A - vertex on the bottom, B - on the right, C - on the top, D - on the left). Center of the square is the point S. The left semicircle is tangent to the large circle in point E. The right semicircle is tangent to the large circle in point F. Part II - The Idea Let's make the triangle AEF. The large circle is circumscribed on the triangle AEF. So radius "R" of the large circle is given by the formula: R = a*b*c / (4*P) where P - is the area of the triangle AEF Part III - Why this Ideai does NOT work :( Because I wrongly assumed that the quadrant points of the left and right semicircles are the points tangent to the large circle. You may see it good on the picture at 02:34.

[@Skandalos]

As always I did it the lazy way using a coordinate system and as usual got a complicated system of equations. It worked but I filled four pages with my calculations.

[@studymakessmart2433]

By looking the diagram it seems like 1.5 will be the radius

[@ThoughtsForShowers]

I guess math will have to be my Valentines again. What you want to torture me with this math problem? You’re the worst Valentines ever. Oh wait, you’re the only Valentines I’ve had, so I guess you’re the best.

[@fivestar5855]

Ah, Mathematician's heart is so small...

[@BiEz78]

That's the case to simply say ... Lovely ;)

[@spiderjerusalem4009]

am i the only one who got 0,883883 as the answer? Or who used the formula of a circle outside triangle?

[@maruthasalamoorthiviswanat153]

Excellent timely creation.

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .......

[@bidish2224]

Sir I have a question. Why there is no exact formula for finding the sum of a definite number terms of a harmonic progressions

[@ostdog9385]

Isnt the denominator 6?

[@alexchou1984]

Great puzzle! Solved the problem through brute force, but trig is definitely more elegant!

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....

[@milaanvigraham8664]

I used a different method and got such a complicated equation I needed Wolfram alpha to solve it. But lo and behold it came out to (2+3√2)/7 !!! Same answer, but such a complicated equation

[@MyXAHOB]

Is it possible to solve this riddle without law of cosines?

[@bombastic_side_eye_007]

Best Valentine's day gift

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....

[@bombastic_side_eye_007]

@@madhukushwaha4578 I guess this hard is enough for me to digest properly

[@tahasami597]

Thank very nice

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .......

[@KamalGamal_KG]

How do you make these videos .

[@JannPoo]

Wait, why are we assuming that the radius of the large circle to the tangent point of the semicircle intersects the square's side exactly in the middle?

[@shameemkm4929]

The center of the small circle is at middle of the side of the square.

[@JannPoo]

@@shameemkm4929 Yeah but why does the radius of the big circle that is perpendicular to the tangent of the small semicircle pass through that specific point?

[@shameemkm4929]

@@JannPoo that is what he explained first the tangent rule.

[@mathmaticalproblemandsolution]

at (2:15)my question is ....how exactly the center of big circle and opposite vertex of square lie on a same line ?????

[@starpawsy]

Being happily married now for 24 years, I dont do Valentine math problems.

[@yusufdenli9363]

Nice problem, perfect solution👍

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....

[@yusufdenli9363]

@@madhukushwaha4578 thanks..

[@ericfuchs123]

Dammit, got roughly 0.853. Put the center of the circle in the wrong spot...

[@IS-py3dk]

SIR I LOVED THIS RIDDLE Happy valentines day 😍 # I LOVE MATH ❤❤

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .......m

[@marina32163]

area is r^2+pi*r^2

[@kihihi4054]

Is there any channel like this but about physics? thanks

[@shambhav9534]

This is exactly how you propose a nerd girl.

[@riyadamin191]

Never thought 1 square and two semicircles give one ❤️

[@Achieve.cookie]

Lovely solution

[@madhukushwaha4578]

Hii! You may have to visit this channel for more harder questions #onlymathlovers .....