A Japanese Temple Geometry Problem from 1800.
Рет қаралды 161,825
Күн бұрынWe look at a nice geometry problem involving equilateral triangles inscribed in a circle.
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@russellthorburn9297 3 жыл бұрын
I confess that, while it was easy to follow, I'd never have come up with it on my own.
@LeonardoGualchieri 3 жыл бұрын
an interesting solution, considering that by collecting the square root of 3 over 2, the golden ratio comes out (the inverse of the golden ratio precisely). R * sqrt(3) / 2 * ( (sqrt(5)-1)/2 ) = R * sqrt(3) / 2 * ( 1/PHI ). Moreover that sqrt(3) / 2 can be seen as sin(60°), the peculiar angle of the equilateral triangle. Fascinating.
@johnnyfonseca354 3 жыл бұрын
Moreover, R*sin(60) is also half the side length of the larger triangle. If we denote the side length of the larger triangle by Q, then in his notation q = Q/(2*phi).
@SuperSeagull12 3 жыл бұрын
I wonder if you tried to solve this on a curved surface, like on a sphere, if you would get a similar result but with another factor instead of sqrt(3)/2 that was related to the equilateral angle in that curved space
@AeroZeppelin-rb4pt 3 жыл бұрын
Also if the factor was added by the golden ratio (the inverse of the golden ratio precisely) then R *& 396 (/3÷8)(/_8L it can be seen as my ligma recently . Fascination is real.
@vkarpinsky 3 жыл бұрын
@Johnny Fonseca So, the ratio of the sides of triangles: Q / q = 2φ = 3.236067…
@RockBrentwood 3 жыл бұрын
That's the same thing as 2R sin(18°) sin(60°) = R (cos(42°) - cos(78°)) = R (sin(48°) - sin(12°)). You weren't fascinated enough, because you missed all of that! That's right: sin(48°) - sin(12°) = ¼ (√15 - √3). Now, you can be fascinated.
@marienbad2 3 жыл бұрын
Such an elegant solution utilising geometry, trigonometry, quadratic formula, and root operations. I had the angles but missed the red triangle. Also the angle at the middle is twice 60 degrees at top of circle (inscribed angle), so it was easier to work out the middle angles that way. Beautiful stuff.
@makmidov 3 жыл бұрын
There are so much beautiful Sangaku problems and I hope you will include more of them.
@DoomSpellSkadi 3 жыл бұрын
Next question: Keep making the Equilateral Triangle at the midpoint of last triangle and find the total sum of all triangle.
@paulcleary9107 3 жыл бұрын
I worked that out to be (6*sqrt3)/(5+Sqrt5) or 1.43618120749. The area's of the two triangles drawn are (large to smallest) (3*Sqrt3)/4 and 3/32*(sqrt3*(3-sqrt5)) as a ratio that comes to 1/8*(3-sqrt5). Summing to infinity gives the answer shown. Oh and if there were triangles at the midpoint of all 3 sides the sum to infinity would be (6*Sqrt3)/(3*sqrt5-1) or 1.82059102448
@FineDesignVideos 3 жыл бұрын
Interestingly, if you want the sum of the sides of the triangles after keeping on drawing more triangles, this question is MUCH easier than the question in the video. This is because you can see it as you are drawing a line from the top of the circle going downwards, but at a slope of 30 degrees (this is the right side of the big triangle). When you reach the edge of the circle, move your pen back to the center horizontally and continue (now you'll get the right side of the second triangle), and continue forever. So essentially what you have done is you have drawn a line at a slope of 30 degrees. The line extends for a length of 2r vertically. Using cos(30)=sqrt(3)/2, we can say that the length of the line is 4r/sqrt(3).
@kevin326520 3 жыл бұрын
I do it by invoking the cosine rule: Notice that the purple "R", "x" , and one edge of the orange triangle forms a triangle, where the purple edge corresponding to an angle of 150 degree. It's obvious that x = R/2 and cos(150 degree) = -sqrt(3)/2. Then, invoking the cosine rule gives you the equation found in 7:42.
@panagiotisapostolidis6424 3 жыл бұрын
this is exactly what i was looking for
@starhe 3 жыл бұрын
By symmetry and Intersecting chords theorem, it is easy to get (sqrt(3)R/2)^2 = q*(q+sqrt(3)R/2)
@srijanbhowmick9570 3 жыл бұрын
Loved the problem & the solution!!!
@rhythmmandal3377 3 жыл бұрын
i just used the relation between circle and eq. triangle to get it's height,h and side length,a and then do q = sqrt(r^2-(r-h/2)^2)-a/4, worked out pretty well.
@kelly4187 3 жыл бұрын
I did it with trig, ending up with the sin of an arccos which is nicely sqrt(1-x^2), which then substituting the lengths of the triangle joining the origin to the where the small triangle touches the circle, the origin, and where the two triangle touch, gives the same answer in one expression.
@anon6514 2 жыл бұрын
I enjoyed solving this one. Did not expect the golden ratio to come up!
@felipemartinezbarrientos1291 3 жыл бұрын
good solution, I arrived at the same using the string theorem in a circle. Regards
@shambosaha9727 3 жыл бұрын
Wow, this is the first time I have seen someone make a video solving a Sangaku problem. Amazing!
@miserepoignee9594 3 жыл бұрын
In fact, he made a video of another Sangaku problem just a few days ago. kzfaq.info/get/bejne/oKqkgLqCmbKZnJs.html
@shambosaha9727 3 жыл бұрын
@@miserepoignee9594 Is it Sangaku?
@miserepoignee9594 3 жыл бұрын
@@shambosaha9727 He doesn't refer to it as such, but Wikipedia says it is.
@shambosaha9727 3 жыл бұрын
@@miserepoignee9594 Oh, that's nice.
@anilsharma-ev2my 3 жыл бұрын
Sangaku Sanganak Saathi Saath Sapta Sata
@isaacchen23 Жыл бұрын
Nice, I did this in my head!
@CTJ2619 Жыл бұрын
I always like these geometry problems
@jarisipilainen3875 3 жыл бұрын
1:33 continue draw line using q line to back first R line. you will have triangle sqrt(R*R-R/4*R/4) draw line from centre straight down till smaller pyramid you have triangle sqrt(R/2*R/2-R/4*R/4) extract last one from first you have answer q. dats how my brain works not need trimming and worping when it will fail some point
@stevenmellemans7215 2 жыл бұрын
Intersecting cords : (Q\2j^2=q(q+Q/2) with Q =side of large triangle and Q = sqrt(3)R
@frankk8080 3 жыл бұрын
Alternative proof using analytic geometry: Let x^2+y^2=R^2 be the equation of the circle. Note the top vertex of the smaller triangle is at (0,-R/2) and the angle of inclination of right side of the equilateral triangle is 120º, so its slope is - Sqrt[3]. The equation of the line along the right side of the smaller triangle is y = -R/2 - x*Sqrt[3]. Substitute this into the equation for the circle gives x^2 + (-R/2 - x*Sqrt[3])^2 = R^2, which simplifies to 4x^2 + (R*Sqrt[3])x -(3/4)R^2 = 0. The positive solution is the x-coordinate of the lower right vertex of the small triangle. This is x = (Sqrt[15] - Sqrt[3])*R/8. Since this is half the length of the lower side, the answer is (Sqrt[15] - Sqrt[3])*R/4.
@goodplacetostop2973 3 жыл бұрын
9:34
@mafprivate8841 3 жыл бұрын
It should be 9:35
@aakashchakrabarty7714 3 жыл бұрын
Are u online all the time how u write comment instantly after uploading of video
@goodplacetostart9099 3 жыл бұрын
Good place to start at 0:00
@jeremycai5870 3 жыл бұрын
dedication
@goodplacetostop2973 3 жыл бұрын
Aakash Chakrabarty Michael uploads his videos at 8am and 8pm EST. So I just have to login few minutes before and wait until the video appears. It’s just that simple
@tabris1135 3 жыл бұрын
I used a different approach. using the Fact that in an equilateral Hexagon inscribed in a circle, the sides lengths are equal to the radius, i get 2 side lengts of an triangle and a known angle (120°). with this I can use the cosine function to get the side length of the bigger triangle in relation to r. after that, it can easily be seen that q is half that value. I don't know how to formally write that down in a comment, but can easily show it on paper. done
@SingaporeSkaterSam 2 жыл бұрын
Threw my intuition for a loop as I “saw” q rotating to touch the circle again at the lowest point. It doesn’t!
@javierdeloera8384 3 жыл бұрын
How do you know that the segment x and q are contained in the same line? I guess by showing that the lower side of both triangles are parallel, but I don't see it. Thanks in advance
@zeno6991 3 жыл бұрын
Interesting solution. I spent the entire day solving this problem using proportions...
@jongyon7192p 3 жыл бұрын
after showing x=r/2 (in the video), I made two functions [x^2+(y+R/2)^2=R^2 & y=sqrt(3)x], then found the intersection between the two. Then q=2x. Simple, right?
@Tiqerboy 3 жыл бұрын
Here is another question. Draw a third equilateral triangle in the space below the second (orange) triangle. Find the ratio of the sides of the large triangle to the orange triangle to the third smallest triangle. Then continue with a 4th triangle and so on. I'm guessing it should be a geometric progression.
@benhouari5021 3 жыл бұрын
Thank you Sir You are wonderfull and clever person Have a nice and safe day
@geoninja8971 3 жыл бұрын
Nice. I must admit, I just set R=1, as leaving it in the quadratic gave me the heebees.... ended up with the same answer.... happy days!
@carl13579 3 жыл бұрын
I prefer the elegance of the solution to the problem in which you replace both triangles with squares. The result is surprising. Let r be the length of the big square and q the length of the small square, and find r as a function of q. Try it!
@robertlozyniak3661 3 жыл бұрын
In your version with squares, does the smaller square meet the larger square at only one point, or along a line segment?
@carl13579 3 жыл бұрын
Along a line segment.
@harikatragadda 3 жыл бұрын
@@carl13579 @Sergio Korochinsky Here's a simple way of solving it. pasteboard.co/Ju7aALy.png
@binaryagenda 3 жыл бұрын
Nice problem and solution. I feel like if you have to explain why the angle was pi/6 you also should explain why you don't consider the negative solution from the quadratic formula though.
@hardamlimit427 3 жыл бұрын
I thought the same. Using the negative solution of the quadratic formular q would result negative: q2 = R*(-sqrt(3)-sqrt(15))/4. I‘ve called it q2 to devide it from q the positiv solution. Negative side length q2 doesn‘t make any sence, so it doesn‘t have to be considered.
@jofx4051 3 жыл бұрын
The angle of equilateral triangle is π/3 or 60° and you divide it in the middle so u get π/6 or 30°
@BroArmyCommander 3 жыл бұрын
@@jofx4051 I don't think you got the point haha
@VarunGupta3009 3 жыл бұрын
Uh... That's because lengths can't be negative? And the difference between the two terms in the numerator is obviously negative.
@lasunncty 3 жыл бұрын
@@kshuf8426 This is called radian measure. It makes some formulas a lot simpler than using degrees. Here it probably doesn't matter though. He could have used the Pythagorean theorem instead of trig to figure out the lengths.
@ramaprasadghosh717 3 жыл бұрын
sides of the larger triangle and the smaller one be a and b. Diameter of the circle is 2*(2/3)*a√3/2 = 2a/√3 Height of these triangles are a√3/2 and b√3/2 so b√3/2= 2a/√3- a√3/2 = a(4√3-3√3)/6 or b= a/3
@dippn7047 3 жыл бұрын
Me seeing you write pi/6 on the second triangle: HES FINALLY GOING TO MAKE A MISTA- Seeing you correct it immediately: ...don't know what I was expecting
@shanmugasundaram9688 3 жыл бұрын
The golden ratio appears in the denominator.
@JoseTorresMates 3 жыл бұрын
That's what I was going to add, the final result.being q=R•sqrt(3)/(2•phi)
@hybmnzz2658 3 жыл бұрын
Phi really pulls its weight for being so famous without even being transcendental. Crazy to think its just the root of the polynomial x^2-x-1. Many problems with sqrt(5) in the answer can be manipulated to bring phi into it.
@zanti4132 3 жыл бұрын
Yes, the answer looks more interesting written as q/R = sin(π/3)/Φ
@xz1891 3 жыл бұрын
???
@Mr0Redmen 3 жыл бұрын
Yes, and I've done a synthetic solution that shows where phi actually appears, with this exact problem.
@sageofsixpack226 3 жыл бұрын
Hi, I have an interesting geometry question. D is an arbitrary point on a side BC in triangle ABC. I, I1 and I2 are the centers of the inscribed circles of triangles ABC, ABD and ACD accordingly. M≠A and N≠A are the points of intersection of the circumscribed circle of triangle ABC and the circumscribed circles of triangles IAI1 and IAI2 accordingly. Prove that no matter where you put the point D, MN always passes through a fixed point
@sageofsixpack226 3 жыл бұрын
If you are wandering from which IMO this question is, it's from our school olympiad😅. And I wasn't able to solve it. Neither anyone else was. But the hint is to probably use inversion. I just don't like leaving problems without knowing how to solve them. So maybe you can help me, if you have time, of course. It's such a great idea for a video, isn't it?
@minh9545 3 жыл бұрын
Can D reach outside B and C?
@sageofsixpack226 3 жыл бұрын
@@minh9545 no, D is a point on a side BC
@TechToppers 3 жыл бұрын
@@sageofsixpack226 In which class, they teach you inversion?
@sageofsixpack226 3 жыл бұрын
@@TechToppers No, we don't learn anything even close to such advanced topics as inversion, but for olympiads we are supposed to know way more than what they teach us
@digvijaygadhavi7418 3 жыл бұрын
Did Anyone Noticed final answer can be further Simplified as R×tan(60⁰)×sin(18⁰)
@pandas896 3 жыл бұрын
Yes
@alpercancakr6719 3 жыл бұрын
Ah yes the sin(18°)
@jofx4051 3 жыл бұрын
Wait how u define sin 18 😂
@alpercancakr6719 3 жыл бұрын
@@jofx4051 by memory
@jofx4051 3 жыл бұрын
@@alpercancakr6719 🤣🤣🤣
@ahmadhabil7933 3 жыл бұрын
What about another smaller equilateral triangle under the small triangle?
@meirkarlinsky7497 3 жыл бұрын
There is a much simpler way - with no trig, and drawing only one additional line to the given draw: In equilateral triangle, each altitude is also a perpendicular bisector, and also an angle bisector. Angles bisectors trisect each other at the center of the given circle - using all these + pythagorean theorem and proportionality in similar triangles enables a shorter solution.
@nombreusering7979 3 жыл бұрын
You can factor sqrt(15) to sqrt5 times sqrt3 and get R*sqrt(3)/2 divided by the golder ratio Which is a nicer value
@simranjeetsinghmatharoo3877 3 жыл бұрын
There still be aa sqrt5-1 left in equation
@mohaben82 3 жыл бұрын
please can you solve this problem for me a,b and c are three chosen points from circle (o,R) where a b and c are not identical. when ab+bc+ac is the biggest value and when is also the smallest . thanks alot
@andreischor2574 3 жыл бұрын
I solved it without trigonometry, just by using Pythagorean theorem for the various triangles involved. Nice little problem.
@einzeller85 3 жыл бұрын
Why can we conclude that all the triangles we devide the yellow triangle in are equal? can we assume that the center of the triangle is "the same" as center of the circle cause that wasn't explicitly mentioned.?
@HagenvonEitzen 3 жыл бұрын
3:58 or use that edge-bisectors intersect in ratio 2:1
@jeanf6295 2 жыл бұрын
Did it using complex numbers : the small triangle is the image of the big triangle by a rescaling and translation operation f:z->a*(z-1)+b such that : f(1) = -1/2 so b = -1/2 and |f(j)|² = 1 so a²|j-1|²+1/4-2aRe[(j-1)/2)] = 1 and a²+a/2-1/4 = 0 Thus a = (sqrt(5)-1)/4 and the side of the small triangle is a*sqrt(3).
@Ensivion 3 жыл бұрын
a fun follow up question: iterate this equilateral triangle n times and see how the length of the smaller equilateral triangles changes with respect to R, or you can just take R =1. What I mean is you draw another equilateral triangle like the smaller one that is tangent to the circle and touches the second triangle at its midpoint, and keep doing that. You might find the ratio of the n-th and the n+1-th triangles' sides doesn't depend on R at all.
@sergiokorochinsky49 3 жыл бұрын
@Pedro Abreu Fractal?... Is it?...
@Ensivion 3 жыл бұрын
@@sergiokorochinsky49 it's a fractal in the sense that it's a self similar iterated geometric shape, but the rigorous definition is that it's a subset of euclidian space that has a fractal dimension that exceeds its topological dimension. the word fractal is used loosely as a subjective thing and even in more formal discussions it's used as a description rather than a definition. Like the coastline of Britain has a fractal dimension that can be studied. Maybe you should calculate out the fractal dimension of this object. My guess is that it'll be related to the golden ratio.
@sergiokorochinsky49 3 жыл бұрын
@@Ensivion, I thought about the circular segment formed between the circle and the ever shrinking triangle, which gets flatter and flatter in each iteration; but if you consider only the triangles, I guess you are right. I also guessed that the total perimeter of the triangles is bounded, but I didn't do the numbers, so I might be wrong.
@simonux1226 3 жыл бұрын
8:24, yes, nice
@factorization4845 3 жыл бұрын
After finding x, I directly used the laws of cosines where the angle is 150°. Then q is found directly
@professorpoke 3 жыл бұрын
Thats a smart move indeed.
@elangavinindrav.a.h3725 3 жыл бұрын
I came up with q=(4√3r-6)/3 is this considered as "q in the terms of r"?
@sl0wsn0w 3 жыл бұрын
What about the second q of quadratic formula with the minus inside?
@Walczyk 3 жыл бұрын
R*(sqrt(3)/2)*(sqrt(5)-1)/2=q seems to have some geometric interpretation but i dont see it
@user-mx6uf2oh1z 3 жыл бұрын
medians of an equailateral trisect one another problem done :)
@sergiokorochinsky49 3 жыл бұрын
really?...
@charlescastleman4588 3 жыл бұрын
I did it a different way, using the law of cosines. We have the obtuse triangle with side lengths R, R/2, and q. The obtuse angle is 150 degrees (360 - 90 - 90 - 30) I then used the law of cosines: a^2 + b^2 - 2abcosC = c^2. q^2 + (R/2)^2 - 2(q)(R/2)(cos150) = R^2 q^2 + R^2/4 - qR(-sqrt(3)/2) = R^2 Multiply both sides by 4 4q^2 + R^2 + 2Rsqrt(3)q = 4R^2 4q^2 + 2Rsqrt(3)q = 3R^2. Then plug into the quadratic formula using only the + out of the +/- to get q = R*(sqrt(15) - sqrt(3))/4
@anirudhgupta4296 3 жыл бұрын
pls do Q6 1988 British Mathematics Olympiad. its about number theory involving a little bit of inequalities as well.
@takyc7883 3 жыл бұрын
i like this question
@louisromao7183 3 жыл бұрын
Two modifications: 1) If there was a third equilateral triangle where one corner touched the second triangle and the remaining corners touched the circle (in the same manner as the second triangle), what would be the length of one side of the third triangle. 2) If more triangles were added in the same manner, is it possible to define a sequence made up of one side of each of these triangles?
@viliml2763 3 жыл бұрын
It's easy, just look at the new right triangle. The new side length turns out to be (sqrt(16 - 3 S^2) - 3 S)/4 where S is the sum of all side length so far if we say the "zeroth" side length is (sqrt 3)/3, the side length of the hypothetical equilateral triangle that has X from the video as its height. For the first triangle you plug in S=(sqrt 3)/3 and you get (sqrt 3)(sqrt(5)-1)/4 as expected/ Plug in S= (sqrt 3)/3+(sqrt 3)(sqrt(5)-1)/4 and you get (sqrt 3) (sqrt (70 - 2 sqrt 5) - 3 sqrt 5 - 1)/16~~0.04 Asymptotically the series approaches (4 sqrt 3) * 0.27881^(2^n) The base of the exponent is probably a transcendental number. Interestingly I don't see it in the OEIS. Maybe I should publish a paper about it.
@viliml2763 3 жыл бұрын
A simple recurrence formula is x^2/(sqrt(12 + x^2) + sqrt(12 - 3 x^2)) where x is the previous side length. The recurrence is asymptotically x^2/(4 sqrt 3) for x
@davidgillies620 3 жыл бұрын
This is straightforward using the cosine rule, as many others have noted. But I would like to know how the Japanese of the day would have approached the problem. Prior to Peary's gunboat diplomacy in the 1850s Japan was a very hermetic society, and as one of the books in the description shows, Japanese mathematicians had to develop theorems on their own. Did they have the cosine rule or even an equivalent to trigonometry in its Greco-Arabic incarnation? Maybe I'll have to order that book...
@vanditseksaria5897 Жыл бұрын
they would have subconsciously proved the cosine formula for angles like 30 degrees and 60 degrees due to their specific property
@think_logically_ 3 жыл бұрын
Need to prove that sides of triangles are parallel (in other words that a line perpendicular to one side is also perpendicular to another side). It's really easy to prove, but worth mentioning.
@ittaloceara 3 жыл бұрын
how?
@think_logically_ 3 жыл бұрын
Voltaire Connecting base vertices of the smaller triangle to the centre gives congruent triangles so we can prove that bisector of the bigger triangle is also the bisector of the smaller one.
@Amipotsophspond 3 жыл бұрын
well that was fun.
@kenjomikagami3384 3 жыл бұрын
All I know about Sangaku is use kamehame wave to blast an equilateral mountain and find q in the rubble. Easy!
@SilplanConsulting 3 жыл бұрын
q=πd((9√30-9√6)/320), where π=√2/0.45
@jumafasi 3 жыл бұрын
Wow, fantástico! Lindo problema. Por medios menos elegantes obtuve: R(206/385) y posteriormente con la razón aurea (y mucho tiempo): q=R(√3+√15)/(6+2√15)=R√3(√5-1)/4
@JSSTyger 3 жыл бұрын
We have BlackpenRedpen already. Now we have WhitechalkYellowchalk.
@jofx4051 3 жыл бұрын
Haha 😂
@ElchiKing 3 жыл бұрын
There is one piece missing: Why do we know that the base lines are parallel (i.e. why is the extension perpendicular to the smaller base and in the center?) In fact, there is exactly one equilateral triangle with different side lengths: Just turn the smaller triangle upside down and extend until it reaches the circle.
@ElchiKing 3 жыл бұрын
Oh, ok, this is easy: Since the triangle is isosceles, its two base nodes are the two (unique) intersection points of a circle around the upper point. But since both circles are symmetric wrt to the same axis, their intersections must be as well (and they can only lie on the axis of symmetry if both circles are equal), hence the orthogonality.
@kaloan999 3 жыл бұрын
Another proof: Let Ha be the height of the large equilateral and Hq the one of the small one. Then Ha=sqrt(3)/2*a, and Hq=sqrt(3)/2*q. However 2R=a/(sqrt(3)/2),i.e. R=sqrt(3)/3*a, and so Ha=3/2*R. Then we draw a diametre down from the top point of the large triangle. Since it is also perpendicular to the opposing side in the small triangle, and bisects it we get that: q/2*q/2=(Ha+Hq)*(2R-Ha-Hq) and after we use the aformentioned transition from Ha to R we get the same quadratic equation.
@Acriimony 3 жыл бұрын
damn rommel. nice math bro
@Scrogan 3 жыл бұрын
completing the square > quadratic formula
@RiotGearEpsilon 3 жыл бұрын
Dang! Supercool!
@darreljones8645 3 жыл бұрын
Rounded to four decimal places, q is about 0.5352 times R.
@larrybud 3 жыл бұрын
The real problem solving isn't the math, it's the vision to see the steps in order to come up with the required math.
@sthinvc 3 жыл бұрын
Well, what is the problem if I calculate x in this way x^2+(R/2)^2=R^2, by Pyth, theorem, x will become sqrt(3)*R/2 instead of R/2, what's wrong with that actually?
@markwalker8846 3 жыл бұрын
The sides of the equilateral triangles are not radii, so you cant use (R/2) as one of the sides to find X.
@zoranocokoljic8927 3 жыл бұрын
Knowing basic geometry of equilateral triangle ( h=a*sqwr(3)/2 and R=2*h/3) no trigonometry needed.
@suou7938 3 жыл бұрын
i wonder how they did it back then. i don’t think trigonometry and algebra was so sophisticated in japan at that time, although relations of certain triangles were probably known.
@alonamaloh 3 жыл бұрын
I think my solution is much easier. Call the side of the triangle a. Applying the notion of the power of a point with respect to a circle to the obvious place in the diagram, you get q*(a/2+q)=(a/2)^2. This quadratic equation in q has a single positive root, which is a*((sqrt(5)-1)/4). Now write a as R*sqrt(3) and we are done.
@ccm_priv 3 жыл бұрын
but why when you use the quadratic formula you omit the 2nd possible result for q? I mean it not only can be plus the square root of b squared minus four a times c, but it also can me minus that same square root.
@Zolhungaj 3 жыл бұрын
Because we know the length q is a postive number, and thus the answer q = -R[(sqrt(15)+sqrt(3))/4] is nonsensical and can be ignored.
@zdrastvutye 3 жыл бұрын
i would apply the cosinus formula with rg=the unknown radius (der gesuchte r) and r= the given exterior circle radius: ld=see the following source code (runs in bbc basic sdl) dr=sqr(r*r-ld^2/4) so that dr^2+rg^2-2dr*rg*cos(150)=r^2 10 r=1:ld=r*sqr(2*(1-cos(rad(120)))) 20 dr=sqr(r*r-ld^2/4):disr=(dr*cos(rad(150)))^2+r^2-dr^2 30 if disr
@bobmarley9905 3 жыл бұрын
law of cosines works for this one too
@seebe2084 3 жыл бұрын
All I know is that 9 little triangles fit in that big one. So if I take 1/3 the perimeter of the big one and divide it by 3, I get q. Should I prove it?
@wayneblackburn9645 3 жыл бұрын
An equilateral triangle can easily be split into 4 or 9 smaller triangles but the small triangle here is not the same size as one that is one-ninth of the larger one. That's a tempting route but easily shown to be false - it would be true if the small triangle was pointing downwards but since it is pointing upwards it has to be smaller.
@georgecaplin9075 3 жыл бұрын
I know this isn’t the point and I enjoyed watching him solve this in quite a cool way, but it was only when it got about half way through I realised he was writing on an actual blackboard. With chalk and everything.
@JLvatron 3 жыл бұрын
What-chew think 'twas?
@georgecaplin9075 3 жыл бұрын
@@JLvatron I didn’t think it was anything. I paid it no mind. Then it occurred to me that it wasn’t an electronic whiteboard. That’s all.
@JLvatron 3 жыл бұрын
@@georgecaplin9075 That circle was drawn very well, so I can see how that gave an electronic vibe.
@AlejandroTorres-jw4vc 3 жыл бұрын
These are like the best videos to see when u go to the bathroom xd
@LorxusIsAFox 3 жыл бұрын
I beat the crap out of this one using coordinate geometry. Inelegant, I know, but I find an overpowered vector-based approach satisfying.
@ROCCOANDROXY 3 жыл бұрын
Letting phi be the golden ratio we have q = (sqrt(3)/2) * (phi - 1)R.
@o3235 3 жыл бұрын
And here I was thinking in 1800s ppl used stones to carve things on metal or wood
@rudychan2003 3 жыл бұрын
Strange writing your 'q' ^ q = (1/4).√3.(√5 - 1).R = 0,5352.R Good Sir. 👍 Thanks^
@brettgbarnes 3 жыл бұрын
Could have just skipped the trig for calculating x and y by using the 30-60-90 triangle side ratios of { 1 : sqrt(3) : 2 }.
@santoriomaker69 3 жыл бұрын
I solved this with mostly just law of sines and I'm pretty pumped that I get the same answer from that. After looking at the comments, it really shows how there are just sooo many ways to solve this problem with trigonometry and geometry. And my dumb brain forgot that cosine law works for any one angle and two sides, which would've made my work easier, but I guess it does make my solution a bit more unique. It was a fun problem, and I really enjoyed solving it. Here's my solution: (warning: very long) The setup is pretty much the same from 3:00, but it's obvious that x = R/2 (from solving way too many problems involving equilateral triangles >.>) Let's stick with that for now, and say that we have to find the two angles adjacent to the purple line of length R. Once we find the upper adjacent angle, we can perform the sine law to get the side opposite to it, which is q. The angle opposite to R should be 150°, as you can form an isosceles triangle with two radii connecting to the base vertices of the small equilateral triangle, making the two angles opposite to those two radii equal. The two angles (let's call them x) and the 60° angle from the small equilateral triangle form a 360° angle, thus x = 150°. Let's call that lower angle α. With the 150° angle opposite to R, and angle α opposite to R/2, we can now do some sine law: R/(sin 150) = (R/2)/(sin α)
@panacea11 3 жыл бұрын
This is exactly how I solved it too.
@santoriomaker69 3 жыл бұрын
@@panacea11 nice :D
@manucitomx 3 жыл бұрын
I loved this problem, but I’m a trigonometry fein.
@procerpat9223 3 жыл бұрын
DELIBERATELY SHARP
@wesleydeng71 3 жыл бұрын
To have some fun, can we make it n stacked equilateral triangles?😁
@SuscribersWithoutvideos-nn1xi 3 жыл бұрын
I just came here to read the comments. We all know that comment section has more meaningful content than that of the video. Agree ?? If not the scroll down for another comment.
@thethug1946 3 жыл бұрын
Why do we take it for granted that the smaller triangle is equilateral? (I am not suggesting that it is not, but ..)
@erosravera3721 3 жыл бұрын
And the other solution of the quadratic equation?
@wayneblackburn9645 3 жыл бұрын
As indicated above that gives a negative value for the length so can be ignored.
@hxhdfjifzirstc894 3 жыл бұрын
Somebody has this symbol on their mailbox. I was wondering what it was. I guess the algorithm is getting better, because I didn't even search for this.
@johnfonseka6728 3 жыл бұрын
That is weird. What, they are reading minds now?
@jonathanhanon9372 3 жыл бұрын
Why didn't we go one step further and say R sqrt(3) / 2 * (sqrt(5) - 1) / 2 = R sin(pi/3) / phi, we find the golden ratio and sin(pi/3).
@andolinismic 3 жыл бұрын
1:59 oh no you didnt reach the corner properly
@indusand3720 3 жыл бұрын
It's amazing how many 200 years it is.I solved such tasks in primary school.
@sjdpfisvrj 3 жыл бұрын
I'm gonna have to throw this out there, but this math video is a lot like math lectures where the lecturers just explain "backwards" the solution that was found. I feel the trick about mathematics is showing the struggle of finding the solution, not just showing the final shortcut.
@jacemandt 3 жыл бұрын
The key is to figure out which other lines to draw. In this problem, the points of contact between the small triangle and the circle have to be crucial. And the primary advantage to working inside a circle is that all the radii are congruent. So the natural first attempt would be to draw a radius to touch one of those points. After that, the ratios follow pretty naturally.
@JCrashB 3 жыл бұрын
Good.
@Mynthio 3 жыл бұрын
You should acknowledge the second solution to the quadratic equation and argue why it isn't also a solution to the initial problem otherwise you might have missed another solution.
@ciderfan823 3 жыл бұрын
Where does the π come from in π/6?
@Victor_Marius 3 жыл бұрын
And it is about half of R, ~0.5352
@marvinkitfox3386 3 жыл бұрын
I wish I could skip steps like his in my math exams. For example: at 3:45 he states that "we have right angles here". Yes, they are right angles. But where is the rule that states that the equilateral triangle at the bottom's top point is exactly on and in the middle of the bottom of the bigger triangle? There is no such a direct rule, one has to derive it from other info! If I were to skip so many steps, then I would get a failing grade on a test.
@DeeFeeCee 3 жыл бұрын
That was stated by construction.
@ArthurvanH0udt Жыл бұрын
These problems from Japanese temples are called san goku!