Can this actually be calculated? Unknown length of a regular hexagon! Reddit r/theydidthemath
ฝัง
- เผยแพร่เมื่อ 1 ม.ค. 2024
- Here's a fun geometry challenge from Reddit r/theydidthemath involving finding the unknown length of a regular hexagon. This question might seem impossible at first because we were only given an angle and a length, but I will show you how to use the law of sine to solve this problem.
Here's a drawn-to-scale diagram: 👉 • Post
Here's the original post on Reddit / 80mtrxnsit
Cat: 👉 • The deleted scene
Notes on Patreon: 👉 / finding-missing-95731639
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#math #algebra #mathbasics #geometry #trigonometry
![อยู่ได้แล้ว - LIPTA feat. Mirrr [OFFICIAL MV]](http://i.ytimg.com/vi/-9-hpntNym4/mqdefault.jpg)
Cat at 7:32
more here👉 th-cam.com/video/Q_47fS5AfRA/w-d-xo.html
After figuring out the angles of the blue triangle, I would have used the law of sine to solve for the long diagonal of the hexagon, finding out it's length is 2*(2+sqrt(3)). Knowing that for regular hexagons, the long diagonal is twice the side length, we conclude one side must be 2+sqrt(3) in length. From here, it’s easy to see the unknown length is sqrt(3).
Oh yea! I forgot about that 😆
That's how I did it as well 💪
How do you find the long diagonal length again? Must I use calculator to figure out value of sin 15?
Okay, even after trying to calculate it I still can't figure out how you come up with the number 2*(2+sqrt(3)), I guess I'll just give up and accept it for now.
@@Blackrobe I’m not sure what the practice in your country is, but where I’m from it’s the standard to equip students with a table full of evaluations for trigonometric functions at different angles, if they are facing a problem like this. Or a graphic calculator, depending on how far in their studies they are.
That is why I assumed the evaluations of sine (and other trigonometric functions) were a given, as figuring them out doesn’t seem to be the main focus of the exercise.
Here’s one way to work things out. Forgive the youtube formatting!
We will form an equation based on the blue triangle shown in the video as follows:
sin(15)/2 = sin(105)/D
in which D is the long diagonal of the hexagon
We can now rearrange the equation so that:
D = 2*sin(105)/sin(15) = 2*sin(105)*csc(15)
This is where we will make assumptions based on the actual meaning of this exercise. You can use lookup tables or calculators capable of giving exact values.
We will assume that:
sin(105) = 1/4(sqrt(6)+sqrt(2))
csc(15) = (sqrt(3)+1)*sqrt(2) = sqrt(6)+sqrt(2)
When we plug everything back to the equation, we get:
D = 1/2*(sqrt(6)+sqrt(2))^2 = 1/2*(8+2*sqrt(12)) = 2(2+sqrt(3))
Shout-out to the cat who can just barely be heard at 7:30 😆 This reminder of geometry was wonderful but the cat's meow really brought it up to a ten
The cat doesn’t like the right triangle trigonometry part 😂
7:30 9:05 😺
😆
You can do it fairly easily without using trigonometry. If you draw a line straight down from the top right corner of the hexagon, you can create a large 45,45,90 triangle. You can find that the bottom leg of the triangle is of length 1+sqrt(3)+x (where x is the side length of the hexagon) and that the right leg is of length xsqrt(3). Equating these allows you to solve for x as 2+sqrt(3). Subtracting 2 gives the desired result.
How did you get 1+sqrt(3)+x?
@@BryanLu0 Draw a line down from the intersection to the base. The right triangle is a 30/60/90 with the long length being 2, that makes our new line sqrt(3). Its base is 1 and the triangle to its left (45/45/90) has a base of sqrt(3): 1 + sqrt(3).
How are you getting the sqrt(3) without using trig?
@@bectionary pythagoras - which gives you the height of an equilateral triangle by breaking it into two congruent right triangels
Yes, after staring at it for a while and doing it in my head I was quite shocked to see the convoluted solution in the video. You've described it very concisely, but I think our methods were nearly identical. I think you meant (1 + sqrt(3))*x not 1+sqrt(3)+x .
Solved it using proportions. Extended the side of the hexagon opposite of the side labeled 2 down and extended the bottom side of the hexagon. labelled unknown side a and side length of hexagon 2+a.
Now you have a triangle and several lengths which are relatively easy to solve for.
The total of the bottom side of the triangle is (2+a)(radical(3)+1) and the smaller triangle with side length 2 has a bottom side length of (2+1)(radical(3)-1)
The latter length divided by the former length is equal to 2 divided by the length of the side opposite 2 on the hexagon that I extended first, which is 2(2+a). Solving for this yields that a = radical(3), which is our answer.
Nice solution!
That's a cool question. I solved it a different way (but also got the same answer). My method was to call the question mark x, then triangulate the hexagon (ie join all the vertices to the centre point of the hexagon). Since the angles of the triangles you get that way are 60 they are equilateral. The outside edge is therefore 2+x but so are the other two edges. Then when you look at the triangle you called the blue triangle, one of its sides is 2x+4 and one is 2. Then since I had all the interior angles of that triangle (the same way you did it) I used the law of sines to solve for x which is the question mark.
Brilliant solution! I solved it using a bit less angle chasing, a bit of trigonometry, and the proportionality theorem, but this method was fun to see too!
I love this channel concept, i don't remember the advanced calculus from school but i still really like solving math problems, so i often have to repeat many sections of videos and sometimes i just don't know what is going on in videos from 3blue1brown and from you, i love that you go into detail about every move on this.
I havent finished this video yet, but you explain things in such a good way! I was always pretty ok at algebra, but never *really* got trigonometry. This video made me understand everything so well!
7:30 the kitty in the background!🥺
Thank you!
Can you find the area of this square? 👉 th-cam.com/video/urmu3o05KqQ/w-d-xo.htmlsi=TexeHNYtrc-U2wl_
No...
I really liked your explanation for the sum of the internal angles of a polygon. A nice addition to a great explanation.
another thing you could do is split the blue triangle into 2 so that you have two right triangles. one of these would be a 30-60-90, the other would be 75-15-90. you can find that the shared side of these triangles is equal to sqrt(3), and then since it shares the hypotenuse with the red triangle that means that the sides are also congruent, so the side we're looking for is also sqrt(3)
Yooo!! This is super fast!!! Thanks!!!
This just blew my mind
My brain not good , i don't understand this , help m
@@TOXIC-MAIN I reread it and didn't understand it, but I figured it out again and realized i forgot to mention - the 75-15-90 triangle that's inside the blue triangle has the same angles as the red triangle. that means that they're similar triangles. then, because the hypotenuse is shared, we can conclude that they have congruent hypotenuses, therefore the legs are congruent as well.
@@spidernh ohh , you which country, my country army level we don't study this , help me
Ur math basics videos are so interesting :) Pls keep on
I just dropped a perpendicular from the upper right corner of the hexagon to make a right-angle, isosceles triangle which will have 45deg angles.
Another isosceles triangle appears which has a 120degree tip angle from the hexagon.
The sine rule can be used to find the lowest side of the small triangle (it has a top angle of 75deg since angles in a triangle add to 180).
The length plus the side length of the hex is the length of the larger triangle, find the side length of hex via subtraction and subtract 2 to get the final answer of root(3).
I did it the same way! 😊
I first changed the scale and chose the origin such that the hexagon has side lenght 1 and the two bottom vertices have coordinates (0,0) and (1,0).
Next look for the intersection point of the lines y=- √3 x and y=x+√3-1, which solves as
x = (1- √3)/(1+ √3) = √3-2; y= 2√3-3
So now our point has distance r=sqrt(x^2+y^2) = √(28-16 √3) to the origin in our scale, but 2 in reality.
In our scaled picture the wanted length would be 1-r, but our scale was too big by a factor r/2, so scaling back we get
(1-r)•2/r = 2/r -2 = 1/√(7-4 √3) - 2 = √3
For the smaller triangle outside the hexagon:
You know the angle below 2 is 60. You also know the angle above 2 is 120-45 = 75.
The angle opposite 2 is 180-60-75= 45 (you could also get that 45 from alternate exterior angles to the 45 in the top side of the hexagon.
You have 3 angles one side, you can calculate the other 2 sides with law of sines.
You can create a congruent triangle where the side congruent to 2 would be double the length of a side of the hexagon.
By congruency you calculate the length of twice the sides of the hexagon, and from there calculate the required length.
Just another way to use support different triangles to get to the same result.
If you are willing to use the angle bisector theorem, the x segment in 8:11 is an angle bisector, splitting a right triangle of 30-60-90. Therefore, we have 2:? = blue hypotenuse:red side = 2:sqrt(3) from the right triangle of 30-60-90, and therefore ? = sqrt(3)
Para os brs do canal eu pensei em uma outra solução um pouco mais simples:
Encontre os ângulos do triângulo pequeno (45, 75 e 60)
Encontre suas medidas a partir da altura relativa à base
Note que o triângulo grande (traço vertical abaixo dos 45 iniciais) é isoceles, então dali temos a base para o triângulo médio (lado direito com base no traço acima)
Com o cos(30) encontramos a aresta do hexágono
Aresta - 2 = v3
I just love your teaching style!
Thank you!
I am highly impressed by the method used to solve the question.
I chose a different approach that doesn't require you to know the law of sines. Let's call the question mark "x", then hexagon side is (x+2) and hexagon height is (x+2)*sin(60) = (x+2)*sqrt(3). Let's first consider the large 45-90-45 triangle that has hexagon height as one of its legs. This is an isosceles triangle, hence the whole line segment on the bottom has length of (x+2)*sqrt(3), and it's left part has length (x + 2) * (sqrt(3) - 1). Let's now consider the bottom-left triangle that has 2 as one of its sides. If we draw and altitude (h) to it's bottom side, the triangle will be split into two triangles: 45-90-45 and 60-90-30. This will allow us to calculate triangle's bottom side as (h + 2*cos(60)), or (2*sin(60) + 2*cos(60)), or (1 + sqrt(3)). But we already know that this side has the length of (x + 2) * (sqrt(3) - 1), which leads us to the equation (x + 2) * (sqrt(3) - 1) = 1 + sqrt(3). After solving for x you'll get x = sqrt(3).
Wow! So unexpectedly simple yet hard to solve!
Very clear presentation. Excellent.
I was going to use the external extra triangle - which also would have worked. You can find the angles and use law of sines to find the shared side outside the hexagon, and it works similarly.
I solved this question in an other way and answer is different.
As we have a triangle outside hexagon. Its one angle would be 45 by alternate angles one angle would be 60 (as 180-120) so third angle would be 75 (180-60 -45) and here one side of triangle is 2 . Then I produced the side of hexagon adjecent with our part of side which produced side will touch the outer corner of triangle as in regular hexagon produced sides intersect each other as same lenght as side of hexagon making an equilateral triangle. In this equilateral triangle we have two triangles one is discussed already.Other one would have one angle of 60 degree (180-120) , one angle would be 105(180-75, here 75 from adjecent triangle) and third angle would be 15 (180-105-60) now we have two triangles having one common side, all internal angles of two trianles known and one side of one triangle is known which is 2 . Now by using law of sines we would find common side of two trianlges which would be sqrt(6) and then we will find unknown side of other trianles by using law of sines again and the answer would be sqrt(3) - 1 instead of just sqrt(3).
And when we see diagram sqrt(3) -1 looks like a more probable answer.
Yes, I was looking for this comment, because that's exactly what I decided! And I can't understand why the other answers differ from sqrt(3)-1
i have won faith back in myself by pausing from a certain point and solving this myself. although i acknowledge that i wouldnt have been able to solve it from the starting position. very well explained.
This can be done without any trig, just high school level geometry. Drop a perpendicular from each of the the two top vertices, to the corresponding bottom vertex. You get several 45-45-90 triangles which allow you to show that the extension at the bottom left is equal to a side of the hexagon, and that the left perpendicular intersects the diagonal exactly at the vertical center of the hexagon. You can then draw a small horizonal line from the left vertex of the hexagon the point where the left vertical intersects the diagonal. This gives two similar triangles, that share the vertical angle just above the "2". Write the ratios of sides and done.
My apologies if this sounds too Fermat-ish.
9:07 Moaw 😂
😆
An alternative solution is to take advantage that the given length is part of the hexagon's sidelength b (which contains the unknown length x + 2) and construct a right triangle by using the secant as the hypotenuse.
The catheti are made using the extended bottom side and the short diagonal as the catheti (whose length a is sqrt(3) × b). One of the triangle's angle is 45° and since we've got a right triangle, it implies the triangle is isolesces. This is an important factor because it establishes that the other cathetus, the hexagon's extended bottom side is the same as the short diagonal so a = b + c (c being the extension of the hexagon's bottom side / the bottom side of the small triangle) and you can create the equation sqrt(3)×b = b + c. Rearrange some stuff and you get b = c/(sqrt(3) - 1).
The next step is to find c and for that, we have to find the angles of the small triangle. The angle opposite of 2 is 45° (shared by the constructed right triangle), the angle opposite of c is slightly more complex: I use the triangle angle sum, where the opposite angle of the irrelevant side is 60° (since it sums together with the inner angle of a regular hexagon - 120° - to 180°) and with the 45°, the remaining angle (c's opposite angle) must be 75° to add up to 180°. c is then solved using the law of sines: c/sin(75°) = 2/sin(45°) or rearranged to c, c = 2 × sin(75°)/sin(45°).
Thus, the hexagon's sidelength must be 2/(sqrt(3) - 1) × sin(75°)/sin(45°) or around 3.732. x = b - 2 so subtract 2 from b and you get x = 1.732.
The downside is that it isn't immediately obvious that you get square root of three as a result (you have to realise that sin(75°) = (sqrt(6) - sqrt(2))/4) so you lose some of the beauty in my solution over yours but it also shows, why the given side is 2 is fairly natural.
I noted the length is l-x if l is the side of the hexagon. Then the line creates a 45 degree triangle so the vertical and horizontal lengths are the same. Vertical is 2l sin(60), horizontal is 2 sin(60) + 2 cos(60) +l (you know angles and 1 side for external triangles find angles and using basic sin/cos). Set these expressions for vertical and horizontal equal then it’s easy to solve for l, then subtract 2 and you get it.
Another way to solve it is from figure at 5:00, create a red triangle rectangle 90 degrees on both sides using same long line blue and red are using as a common line (you will get 2 equal triangles mirroring each other). You will find the wanted side on the other triangle is just one of the sides a new triangle formed(30,90,60 degrees) which has the hypotenuse of 2:
sin(60) = ?/2
? = sqrt(3)
I extended the top line and the lower left line of the hexagon out to meet in the top left to form an equilateral triangle. Then i had a similar triangle making all the angles pretty easy to work out. Then i just needed to rearrange the resulting sine rule expression: (? + ? + 2)/sin(45)=(?+2+?+2)/sin(75)
=> 2(sin(75)-sin(45))*? = 4*sin(45) - 2*sin(75)
=> ? = (2*sin(45)-sin(75))/(sin(75)-sin(45))
Gives the same answer, your method is interesting tho!
Now all I need to do is figure out what the square root of 3 is. But I'm being irrational.
Oh wow, i knew the formula for interior angles, but somehow the "it's minus 2 because it's how many triangles you can draw inside times 180"
Makes so much sense yet somehow i never made that connection.
Even the car was interested in the math
Math car
😆
Non-trig solution:
Note that the bottom left triangle has angles 45 (by Z-angles), 60 (since hexagon internal angles are 120) and 75 (by 180-45-60 = 75).
Drop a horizontal line from the top point, splitting it into two triangles. The left one is then a 45/45/90 triangle, the right one a 30/60/90.
Remembering that a 30/60/90 has 1, sqrt(3), 2 sides and a 45/45/90 has 1, 1, sqrt(2), it is easy to compute all the sides. (This implicitly uses trig, but most highschoolers have to memorise 30/60/90 triangles specifically anyways.)
We find that the triangle had sides 2, sqrt(3)+1 and sqrt(6), and the internal wall has sqrt(3).
Now drop an orthogonal line from the top-right point of the hexagon to the bottom-right. We first call the sides of the hexagon 'x', then we can find with another 30/60/90 hexagon that the side we dropped has length sqrt(3)*x. Then note that this line, together with the bottom left triangle, forms another 45/45/90 triangle. This means that:
1+sqrt(3)+x = sqrt(3)*x.
Solving for x gives x = (sqrt(3)+1)/(sqrt(3)-1) = 2 + sqrt(3).
Subtracting the 2 gives ? = sqrt(3).
EDIT: Ah, I see other people already mentioned this (and similar solutions) before. The trick is always to split your weird triangle into triangles you understand with orthogonal lines.
Mom to 9th grader enjoying this immensely in prep for being default/free geometry tutor in the upcoming months.... TY for this refresher!!!!
I was able to solve this one as well, but instead of the red triangle, I used the big triangle consisting of your blue triangle + red triangle. After I got the side of the hexagon that way, I just subtracted 2.
I did the problem by using co ordinate geometry, i considered the the bottommost left vertex to be (0,0) and respectively wrote the equation of the 45 degree line and the line with the unmarked length, now just find the intersection and its distance from origin and you get the value of s(side length) and then just subtract it from two to get √3.
Here's my solution, no trig involved:
Label the hexagon's side as A, that means we're looking for A-2. The hexagon's height is Asqrt3, and since the line is at a 45deg angle, then the bottom line (hexagon's bottom side + triangle's bottom side) is also Asqrt3, which means that the triangle's bottom side is A(sqrt3-1).
The triangle's altitude from its top vertex divides that side into two segments. The right one forms a 30-60-90 triangle with the altitude and the length-2 line, so its length is 1 and the altitude's length is sqrt3.
The length of the left segment is thus A(sqrt3-1)-1, but it's also a leg of a 45-45-90 triangle along with the altitude, so it has the same length sqrt3.
This gives us an equation A(sqrt3-1)-1 = sqrt3, which solves nicely to A = 2+sqrt3, so the value we're looking for is A-2 = sqrt3.
Another way to do it is finding the cross-section's length instead of "x" when doing sine law. Then, split one of the two halves into three triangles. You know they're equilateral because it's a regular hexagon. Now, you can find the length of one of the hexagon's sides by halving the cross-section. Finally, subtract two from that and you end up with 1.73.
9:06 i swear i head a cat meowing 😂😂😂😂
Edit: I think I realized where I went wrong. I assumed that when I extended the line to form the triangle that the left most angle of the diagram was part of the left most angle of the equilateral triangle. This is not the case.
Original: Can someone explain where I went wrong in my logic? I thought of creating a line from the left most angle in the hexagon to the left most angle in the whole diagram. By doing this, I think you create an equilateral triangle where ? + 2 is one of the sides. From here, you consider the triangle inside the triangle that has 2 as one of the sides. The angles of this triangle are 60, 75, and 45. The angle opposite to the 2 side is 45 so using the law of sines this means that the bottom side is 2.73. Since the bottom side is also one of the sides of the equilateral triangle this means that ? + 2 = 2.73 and ? = .73. However, as others have said the actual answer is 1.73 or sqrt(3). I am not sure where I messed up
I did the same mistake and got -1+root3 = 0.73 as my answer aswell after using the sine rule. I was wondering what I did wrong until I read this comment and I realised that I also made the incorrect assumption that the continuation of the straight line on the top left side of the hexagon would meet the point of the little triangle on the left and form a larger equilateral triangle when that is not necessarily the case 🤦. Thank you for pointing this out.
@@tobiascorrintomas8336 np, diagrams be lying
Drop an altitude at the angle labeled 45⁰. This is length sqrt(3)×s. This is the same length and the horizontal line at the bottom, giving (sqrt(3)-1)×s for the fraction on the triangle.
Draw a horizontal line across the middle of the hexagon. The right most section is s/2, the middle is sqrt(3)/2 (half the bottom line), leaving (3-sqrt(3))/2×s for the left.
Now you're left with two similar triangles. (3-sqrt(3))/2/(sqrt(3)-1) = ?/2 simplifies to ?=sqrt(3)
I solved it with different method
First draw a horizontal line from the left angle of hexagon, which forms a pair of similar triangles, find the angles of the triangles sqrt6/sin60 = 2/sin45 = (1+sqrt3)/sin75 and find the bottom side of the triangle with sine law which is 1+sqrt3
Then, draw a right equilateral triangles from 45 degree, 1 + sqrt3 + S = V vertical line of hexagon, with cosine law, V^2 = S^2 +S^2 -2(S^2)(cos120) and V=(sqrt3)S
After that, solve 1 + sqrt3 + S = (sqrt3)S, and gets S = 2+sqrt3, hence the unknown is sqrt3
I started with the red triangle. Which forms an isoceles triangle where the non-congruent angle is 2pi/3, leaving the other two angles as pi/6. pi/4-pi/6=pi/12 for the angle opposite the ?. Dividing the regular hexagon into 6 equilateral triangles you can show the leg adjacent to the pi/12 angle is sqrt(3)*the side length of the hexagon, which we know to be 2+?.
so the tan(pi/12)=?/sqrt(3)(2+?)
using the double angle formula for the tan, tan(pi/6)=(2?/sqrt(3)(2+?))/(1-(?/sqrt(3)(2+?))^2)=2sqrt(3)?(2+?)/(3(2+?)^2-?^2)=2sqrt(3)(2?+?^2)/(12+12?+2?^2)=sqrt(3)(2?+?^2)/(6+6?+?^2)=1/sqrt(3)
-> 3(2?+?^2)=6+6?+?^2
-> 2?^2=6
-> ?=sqrt(3)
I think my method was kind of interesting. I used the alternate interior angles theorem on the 45⁰ angle that was given to find the first angle of the triangle with the 2 in it. I then found the other 2 angles using the fact that a hexagon has 120⁰ angles. Then I drew the line to complete a triangle where the question mark is outside of the hexagon. We essentially create a large equilateral triangle in the process, meaning the tiny triangle's bottom left angle is 60⁰-45⁰=15⁰. Now we have enough info of each of these outside triangles to find the length of the side they have in common, then I used the law of sine to find √3. I still have to think of a trig-free solution though
Ameowzing explanation as always.
If you consider all the diagonal drawn, there are 6 equilateral triangle (hence 60 identified). Nice question. Explanation is clear. Thank you
3:28 A simpler way to find this is to construct triangles at the center of the polygon connecting to the vertices. These center angles are simply 360 / n where n is the number of sides of the polygon.
Because it's a regular polygon, these triangles will at least be isosceles. Due to the symmetry of being regular, the sum of these 2 base angles will equal the polygon's internal angle by adjacent triangles. So, [180 - (360 / 6)] / 2 = 90 - (180 / 6) = 60.
i found the point of the intersection using basic trig to be
(-2cos60, 2sin60)
assuming the bottom left corner is at (0,0), the equation for the line is:
y = x + 2sin60 + 2cos60
let L be the length of a side of the hexagon, such that the top right corner is at
(L, 2Lsin60)
plug this point into the line equation to solve for L:
2Lsin60 = L + 2sin60 + 2cos60
L(2sin60 - 1) = 2sin60 + 2cos60
L = (2sin60 + 2cos60)/(2sin60 - 1)
plug in values for sin and cos:
L = (2/2sqrt3 +2/2)/(2/2sqrt3-1)
L = (sqrt3 + 1)/(sqrt3 - 1)
multiply by (sqrt3 - 1)/(sqrt3 - 1)
L = (4 + 2sqrt3)/2
L = 2 + sqrt3
final answer is L - 2, equal to sqrt3
"doesn't have to be regular" proceeds to draw the most oblong hexagon
I got to the answer in a slightly different way that i find quite interesting:
- I saw the 45 degrees angle and thought that i should draw a square with that line as the diagonal (so the height of the square is the same as the height of the hexagon)
- Then, I calculated the angles for the bottom triangle in which the 2 is.
- Seeing that the angle below where the 2 is written was 60 degrees, that meant that if I prolonged the line of "? + 2", it would make a right-angled triangle (with the square) with an angle of 60 degrees (so half an equilateral triangle)
- I then used the law of sin to find the value of the base of that triangle (with the 2) to find the hypotenuse of that right-angled triangle
The answer was then just (hypotenuse - 2)/2
I figured it out by dropping the perpendicular from the point where the line intersects the hexagon, use it to figure out how far the 45 degree line is from the hexagon and then use a bit of algebra to get the side length of the hexagon... in my head.
I loved this video. Anybody know what software or hardware was used to make it?
8:12 I didn't realize what you meant by really cool until around 40s later. And I agree. That was cool
A simple approach is to construct a 45 degree right trianle by dropping a perpendicular from the right vertex of the hexagon to the vertex at the base of te hexagon. If we assume the side of the hexagon is 2x, the side of this right triangle will be 2x*sqrt(3). Now the base of the small triangle inside this large right triangle is 1+sqrt(3) to which if we add the side of the hexagon , 2x, will give an equation 2x+1+sqrt(3)=2x*sqrt(3). Solving for 2x gives 2x=2+sqrt(3)
I drew different triangles (right triangles) and used symmetry of the regular hexagon, but also got the sqrt3 answer.
Denote the hexagon as ABCDEF and O is the point outside the hexagon where angle ABO = 45 deg.
Let the length of the hexagon be k.
BD = 2 k sin(60 deg) = sqrt(3) k
Note that triangle BDO is a right-angle isos. triangle, so OD = BD = sqrt(3) k
Therefore, EO = sqrt(3) k - k = (sqrt(3) - 1) k
Join CF. Denote the intersetion of CF and BO as G and the intersetion of EF and OB as H.
Note that CF = 2k. Let FG = x.
Also, triangle BCG ~ triangle HFG
BC / HF = CG / FG
k / (k-2) = (2k - x) / x
kx = 2k^2 - kx - 4k + 2x
x = (k^2 - 2k)/(k-1) = (k(k-2))/(k-1)
Next, triangle HFG ~ triangle HEO
FG/FH = EO/HE
k(k- 2)/(k-1) * 1/(k-2) = (sqrt(3) - 1) k / 2
1/ (k - 1) = (sqrt(3) - 1) / 2
k - 1 = 2(sqrt(3) + 1)/ 2 = sqrt(3) + 1
k - 2 = sqrt (3), which is the length of FH.
I miss doing math like this in school
What software you use?
If "b" is the length of the diagonal that divides the hexagon into two equal areas, and "a" is the side length of the hexagon, then b=2a. i.e. b=a+2a sin(30)
The diagonal of a hexagon also divides the 120 internal angle in half and the opposing angle to the side length of 2 can be calculated. i.e. 60 - 45 = 15
From the figure, the remaining angle is 105 and the sine rule gives, 2/sin(15) =b/sin(105) = 2a/sin(105) or a= sin(105)/sin(15) ~ 3.73205
Bro wrote down the entire biodata by just one angle
7:00 - You almost causing me the pain there :D
you can find the length of the longest side instead, its length is 2*hexagon side's length
Beautiful!
i did it with similar triangles. no trig required. if L is hte side length of the hexagon, (2L - 2) / 2L = 2 / (sqrt(3) - 1)L. Invert, cancel Ls etc. The ratio on the right is from the triangle in hte lower left, the length of the base is the denominator. The ratio on the left is from the triangle obtained by extending the length 2 side, and the uppermost side of teh hexagon.
I seen this on reddit the other day! This is how I solved it algebraically on the post:
Let's set a few assumptions first:
If the shortest limb (opposite 30) of a 30-60-90 triangle is a, the hypotenuse is 2a and the longer limb is √3a
The limbs of a 45-45-90 triangle are equal (represendted by a) and the hypotenuse is √2a
The angle between two sides of a regular hexagon is 120°
Z angles rule dictates that the internal angle of the 2 bends in a letter Z are equal
The length of a side of the hexagon is x.
Shorten bottom left triangles name to BLT (yum)
Math time:
Given angle (45°) can be copied to bottom left of BLT given the z angle rule. Bottom right angle of BLT is 60° because the straight line is 180° minus internal angle of regular hexagon. 180° - 120° = 60°
Drawing a vertical line starting at the top corner of BLT that is perpendicular to the bottom side we split BLT into a 45-45-90 triangle on the left and a 30-60-90 triangle on the right.
Right hand triangle of BLT (BLTR) has a hypotenuse of 2 from the given length. Due to rules of 30-60-90 triangles, the base of BLTR is 1, and the height is √3.
Left hand triangle of BLT (BLTL) has a height of √3, calculated from BLTR. Due to rules of 45-45-90 triangles, the base is also √3.
The base of BLT is base of BLTL + base of BLTR which = √3 + 1
If we draw a vertical line from the given angle in the top right down to the bottom right angle of the hexagon such that it is perpendicular to the top and the bottom of the hexagon, we create a large 45-45-90 triangle which contains BLT.
The base of this triangle is base of BLT + side length of hexagon or √3 + 1 + x. This also means the height of the hexagon is √3 + 1 + x.
If we draw a line from the rightmost corner of the hexagon to be perpendicular to our height line we drew, we create a 30-60-90 triangle in the top right of the hexagon (TRT). Given the line bisects the height, the height of TRT is (√3 + 1 + x)/2
Given it's a 30-60-90 triangle, and we know the side length of the hexagon is x, our hypotenuse for TRT is x, meaning the base is x/2, and the height is √3x/2
We now have 2 equations for the height of TRT using x, and can therefore solve for x:
√3x/2 = (√3 + 1 + x)/2
Multiply both sides by 2
√3x = √3 + 1 + x
Subtract x from both sides
√3x - x = √3 + 1
Take out x as common factor on left
x(√3 - 1) = √3 + 1
Multiply both sides by √3 + 1
x(√3 - 1)(√3 + 1) = (√3 + 1)^2
Simplify
2x = 4 + 2√3
Divide both sides by 2
x = 2 + √3
Now we have x we can solve for ?:
? = x - 2
? = 2 + √3 - 2
? = √3
Imagine if the title was: “Yes this can be calculated, use calculus” i'd actually lose my mind. 😭
Can someone explain to me how the red and blue 15° angles (8:14) work? They don't look identical at all to me... is it just a drawing error?
What app did you use to make this video?
How do you make these videos? Please reply
one of the triangles has the angles 45, 30 and 120 but has 2 sides equal. so, should it not be a isoceleus triangle which would have 2 angles equal??
I feel like in the case of an even number side shape, you can skip knowing the trapezoid and just know that when you split it in two, you have split on angle in half. Which is how the trapezoid formula comes about anyway. So 120/2 instead of all the other math. So one angle is 60 and the other is 60-45.
i found a easier method
use blue triangle then law of sine
if we think abt the triangle outside the hexagon :-
angles are 45, 60 and 75
2/sin45=m/sin 75
then m= root 3 +1
Equation:-
m/sin 15 =(4+2x)/sin 45
then solve
x= root 3
Haven't read any comments yet so here's a non-trig solution.
Setup: Let the side with the labels 2 and ? be AB, and let C be on AB where the line intersects it. Let that line be DE (so that C is at the intersection of AB and DE and A, B, and D are vertices on the hexagon, and A is the leftmost vertex. Also let F be the rightmost vertex of the hexagon. Finally, let the side length of the hexagon be x.
Solution: First, extend DF and EB until they intersect at G. Next, drop an altitude from D, which will hit the line EB at H, another vertex of the hexagon. Then the angles DHE and DHG are right angles, and HDE and DEH are both 45 degrees. Because the interior angles of the hexagon are 120 degrees, this means HDG is 30 degrees. This means that HDG is a 30-60-90 triangle and HFG is an equilateral triangle with side length x. Therefore, DG has length 2x and DH has length xsqrt(3) (and so does EH because it is part of a 45-45-90 triangle).
Clearly, DG and AB are parallel so the triangles ECB and EDG are similar, so we can set up the ratio as follows:
DG/CB = EG/EB ==> 2x/2 = (EH+HG)/EB ==> x = (xsqrt(3)+x)/(xsqrt(3)+x) ==> x = 2+sqrt(3).
Therefore, the length labelled ? is x-2 = sqrt(3).
You don't need the trapezoid to get 60˚. The blue line bisects the 120˚ angle.
Also the right triangle is a given. In a regular hexagon opposite sides are parallel. A triangle from 3 of the points of opposite sides is a 90˚60⁰30⁰ right triangle.
Homie completely forgot about the whole concept of bisection and angle summation
Solved it with the same exact approach
Nice question.
I feel like Ive done something interesting. So I drew a line that is straight on the bottom left triangle. With this I found the base of it as √3+1. Then I made a huge 45 45 90 triangle by drawing a line from top right corner to the bottom of the hexagon. We know that one side of hexagon is x+2. So that makes one side of the 45 45 90 triangle equal to (x+2).√3 which is also equal to x+3+√3. From there I solved it by leaving x alone and getting (3-√3)/(√3-1)=x. I divided and multiplied the left side of the equation with (√3+1) and got √3=x
You actually don't need the sine rule OR the angle bisector theorem. you only need to know that 60-30-90 triangles have sides proportional to (√3, 1, 2), and some basic geometry and algebra.
lets say x is the unknown length. we know the length of the segment across the bottom is the same as the height of the hexagon since the diagonal segment makes a 45 degree angle with the top of the hexagon. we can calculate the amount of the bottom segment to the left of the vertex by using the 60-90-30 triangle with hypotenuse 2 and a vertical segment to the base. then by 60-30-90 proportionality, we know the height of that triangle is √3 and the base is 1. then since the diagonal is 45 degrees, we know the length of the segment to the left of the hexagon's base is 1 + √3.
so the length of the segment across the bottom (and thus the height of the hexagon) is the length of one side of the hexagon (x + 2) plus the remaining length (1 + √3). from there we can construct another 30-60-90 triangle whose hypotenuse is (x + 2), and whose height is (x + 2 + 1 + √3) / 2 (half the height). lets call the height y and the base z, then by the 30-60-90 proportionality we know that the height y = (x + 2 + 1 + √3) / 2 = √3z, and (x + 2) = 2z. a little algebra autopilot gives you that z = (√3 + x + 3)/(2√3). you can substitute (x + 2)/2 for z on the left hand, which gives you an equation in terms of only x: (x + 2)/2 = (√3 + x + 3)/(2√3). more autopilot to rearrange and get x on just one side and simplifying gives you x = √3 💥
Nice 👍!!!
Without using the law of sine (but using the properties of a 30-60-90 triangle):
Set the side length to (2+x), where x is the "?". Then, by the properties of a 30-60-90 triangle (and of course knowing that a regular hexagon is made up of a bunch of equilateral triangles), we draw a vertical line between the bottom right corner and the upper right corner, and calculate the height of the hexagon to sqrt(3)*(2+x). Since we now have a 45-90-45 triangle with the sides sqrt(3)*(2+x), we now know that the whole length of the bottom line is also sqrt(3)*(2+x). This means that the part of the bottom line that is outside of the hexagon is (sqrt(3) - 1)*(2+x). The last step is to draw the vertical height of the small triangle outside of the hexagon. Using 30-60-90 triangle, we can see that the height is sqrt(3), which is also the length of the base of the small 45-90-45 triangle at the bottom left of the figure. Subtract the base of the small 30-60-90 triangle, which we know is 1, from the total length (sqrt(3) - 1)*(2+x) of the small triangle. We now have the simple equation:
(sqrt(3) - 1)(2+x) - 1 = sqrt(3)
2 + x = (sqrt(3) + 1)/(sqrt(3) - 1) = (sqrt(3) + 1)^2/2 = 2 + sqrt(3) (by multiplying both numerator and denominator by (sqrt(3) + 1) )
x = 2 + sqrt(3) - 2 = sqrt(3)
3:23 absurdly long when you can infer that because it is a perfect split on line C, you can just do half of the inner radius to find the bottom right angle
The picture must be out of scale because the height of the point where the 45° line touches the hexagone side is sqrt(3) from the bottom side, yet the unknown part of the side is much shorter.
I agree with you.
Knowing me, I’d instantly turn to coordinate geometry and not regret anything
My method was to find the diagonal of the hexagon and determine what the side length should be from that. 2/sin(15) = y/sin(105). Then y is the hypotenuse of the 30/60/90 triangle and the ? is the short side - 2. That definitely required a powerful calculator. :)
This is a really interesting one in that it doesn't seem at first you could solve this with no trig. But if you know the 45-45-90 and 30-60-90 triangles you're all set.
The key for me was looking at the small triangle in the lower left. By parallel lines the lower left angle is 45. By properties of a hexagon the lower right angle is 60. This leaves 75 on top.
If you drop an altitude and decompose this triangle, you get our old friends, 45 45 90 on the left, 30 60 90 on the right. A little figuring tells you the base of this small triangle is sqrt3+1.
Drop a vertical line from the top right vertex of the hexagon. also connect this vertex with the lower right vertex of the small triangle. This forms a 30-60-90.
For clarity let ? = x. The base is 2 + x and so the height is 2sqrt3 + sqrt3x..
But this vertical line is also a leg of a 45 45 90 and we know the other leg is (sqrt3 + 1) + (2 + x) or sqrt3 + 3 + x. So 2sqrt3 + sqrt3x = sqrt3 + 3 + x.
Move all the x terms to one side, factor out x, and divide, and we get x = (3 - sqrt3)/(sqrt3 - 1). After multiplying top and bottom by sqrt3 + 1, this reduces to sqrt3.
I've arrived at the answer in exactly the same way! I like it more because it uses very simple reasoning. But since I'm weak at math and didn't know the law of sines, Steve's solution was very enlightening for me.
@@tonygluk1 Well ya, you gotta know law of sines ;) It's an awesome property of triangles you can leverage all over the place. And I considered that approach but when there's a geometric approach I sure prefer it to a trigonometric approach. I guess I find it more intuitive and how often can you be intuitive, right?
@@Qermaq Agree on both counts )
I made an equilateral triangle in the bottom left and tried using the law of sines in the upper triangle that was created. I got (sin(15)/x) = (sin(105)/2+x)..where did I go wrong?
same, I'm getting sqrt( 3 )- 1 and I also didn't understand.
I think the solution shown here is way more difficult than it would need to be. After connecting the opposite corners of the hexagon I would've calculated the length of the diagonal. Then I would've divided half the hexagon in three regular triangles, showing that the diameter equals 2*(2+?) and therefore: 2:sin(15°)=d:sin(105°) d=2*sin(105°):sin(15°). This means that (since d is twice the sidelength) 2*(2+?)=2*sin(105°):sin(15°) und therefore ?=sin(105°):sin(15°)-2. This means ?=√3≈1,732
I haven’t finished the video, but my instinct is to construct two right triangles, one 15 75 90 and another 30 60 90. Tan 15 would be x/L, where L is the distance that skips one corner of the hexagon. Tan 30 is (2+x)/L. Now we just need to find out how many times larger than tan 15 tan 30 is. 2 + x is that many times larger than x. Then we simply solve the resulting system of equations.
Don’t bother solving for L. It cancels out.
Side note: To visualize the two triangles, label points A-G starting at the top left vertex and going clockwise, plus G as the point connected to B across the hexagon. The triangles I had in mind were FBG and EFC.
I did it by a MUCH more complicated way, even unecessarily. I will not go into the details, but I found out this linear equation:
(3+x)/(1+x) = sqrt(3), being x the unknown length.
Doing some manipulations, I ended with:
x = (sqrt(3)-3)/(1-sqrt(3)).
I don't know exactly why, but it is exactly (or almost, I just got up to the third decimal) sqrt(3). I find it quite curious. Maybe it can be useful in some substitution, idk.
Edit: figured out why it ended as sqrt(3). At first I did it with a calculator, but since it is the same as sqrt(3)(1-sqrt(3))/(1-sqrt(3)), the division goes smooth.
(Sorry for any misspellings or grammatical errors. Practicing English yet)
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How is it that the height of the bottom left triange is also sqrt(3) yet it looks so much longer than the "?" bit?
7:35
couldn´t we just calculate the angle between "?" and the 105°? I think this would save us a bit of work
I think there's an easier way of looking at this once you realise that both the red and blue triangles have a 15 degree angle and one length in common. That means that if you mirror the red triangle around the common side, the two other long sides will line up and the question mark side will become a line perpendicular to the line bisecting the hexagon. This gives you a 30-60-90 right triangle with hypotenuse of length 2 and the question mark on the side adjacent to the 30 degree corner (opposite to the 60 degree corner). We know the proportions of such a triangle's sides (it's half of an equilateral triangle) so plugging in the known hypotenuse length will give us our answer.
Edit: Aha, I think @spidernh has the same idea, expressed even more simply.
And I am here waiting for another speed run with no trig😅
That's a long way around, the bisector can be found from the blue triangle. Half of it will be the same as the ?+2 as the hexagon can be decomposed into 6 equallateral triangles.
label the hexagon from top left as ABCDEF and the intersection P
2+? = L [ the side is made of the parts ]
diag = 2L [ the diameter of a hexagon is twice the side length ]
(sin 15)/2 = (sin BPE)/diag [ law of sines]
BPE + 60 + 15 = 180 [ triangle sums to 180 ]
4 equations 4 unknowns rearrange to
2+? = sin 105 / sin 15
but
sin 105 = sin 90+15 = sin 90-15 = cos (90 - (90-15)) = cos 15
so
2+? = cot 15 = cot 30/2
but by half angle formula
=cot 30/2 = csc 30 + cot 30 = 2 + sqrt(3) = 2+? [ sin 30 = sqrt(1/4), cos 30 = sqrt(3/4)]
? = sqrt 3
My way was muuuuuuch harder, and super prone to error.
haven't solved anything yet but you should be able to easilly extract all the angles in the shape and use the sin funktion to get what ever lenght you need.