We have AD=4sin(3θ)=4(3sinθ-4sin³θ) and we have AD=9sinθ, from which 4(sinθ-3sin³θ)=9sinθ so sin²θ=3/16 so sinθ=√3/4 so cosθ=√13/4 from which cos(3θ)=4cos³θ-3cosθ=√13/16 and since x=4cos 3θ=√13/4 and y=9cosθ=9√13/4 then x/y=1/9
Applause : ) You have Such Great Vision in knowing the Exact Constructions to Draw ! I used so much Trig that I wondered how you would have done this far more easily : )
Solving using a half interval search: let AD = h, then sin(3Θ) = h/4 and sin(Θ) = h/9, or 4sin(3Θ) = h and 9sin(Θ) = h. We need to compute Θ such that 4sin(3Θ) - 9sin(Θ) = 0. We let f(Θ) = 4sin(3Θ) - 9sin(Θ) and search for a value of theta such that f(Θ) = 0. Since 3Θ must not exceed 90° for the figure to be valid, Θ must not exceed 30°. For a computer program, we replace f(Θ) by delta and Θ by theta. I wrote this program in the python language: import math # python computes trig functions for angles in radians, so convert x = 15° to radians: theta = 15*math.pi/180 dx = (7.5)*math.pi/180 # define function to compute delta def compute_delta(theta): delta = (4.0)*math.sin(3*theta) - (9.0)*math.sin(theta) return(delta) # note that delta for theta = 15° is positive and delta decreases monotonically # therefore, there is a zero crossing for theta between 15° and 30° # if a positive value is returned for delta, we must increase theta and try again # if a negative value is returned for delta, we must decrease theta and try again # start with theta = 15° and increment 7.5° (half way from middle of range to end) # once the increment dx is sufficiently small, compute and print x for a in range(0,51): # after 50 divisions by 2, dx is sufficiently small delta = compute_delta(theta) if (delta > 0): theta = theta + dx else: theta = theta - dx dx = dx/2 x = (4.0)*math.cos(3*theta) y = (9.0)*math.cos(theta) m = round(x/y , 15) # round x to 15 decimal places formatted = "{:.15f}".format(m) print(f"x/y = ",formatted) Output: x/y = 0.111111111111111 === Code Execution Successful === Note that the repeating decimal 0.11111.... = 1/9. The program can only calculate x/y to the limit of its precision. Math Booster computed an exact value. I had the program print the value of sin(Θ) as 0.433012701892. When I square it, I get 0.1875, which equals 3/16. So sin(Θ) = (√3)/4 is a good educated guess (but also confirmed in other comments). We use the identity sin(3Θ) = 3.sin(Θ) - 4.(sin(Θ))³ to compute sin(3Θ) from sin(Θ). Substituting these values for sin(Θ) and sin(3Θ), we should find 4sin(3Θ) - 9sin(Θ) = 0 exactly. Knowing sin(Θ) and sin(3Θ), we can compute cos(Θ) and cos(3Θ), compute x and y, then x/y.
*Let's note t = theta and AD = h. Then sin(3.t) = h/4 in triangle ABD and sin(t) = h/9 in triangle ACD, and then sin(3.t)/sin(t) = 9/4 We know that sin(3.t) = 3.sin(t) - 4.(sin(t))^3, so sin(3.t)/sin(t) = 3 - 4.(sin(t)^2) = 9/4 giving that (sin(t)^2 = 3/16 and (cos(t))^2 = 1 - (3/16) = 13/16. *cos(3.t) = x/4 in triangle ABD and cos(t) = y/9 in triangle ACD, so x/y = (4/9).(cos(3.t)/cos(t)). We know that cos(3.t) = 4.(cos(t))^3) -3.cos(t) so x/y = (4/9).(4.(cos(t))^2) -3) = (4/9).((13/4) - 3) = (4/9).(1/4) = 1/9.
That is nice and this is probably similar to one of the problems on your channel. That similar ratio of proportions that most likely *definitely* make uae of HL similarity with respect to the right angle. I think that that is what necessarily occurs. Also the x/y = 1/9.
Reflect ∆ADB about AD to make ∆ADM. Now, ∠MAC = 2θ Draw MN, with N on AC such that ∠MNA =2θ Both ∆AMN and ∆MNC are Isosceles, with AM= MN = NC=4 Drop a perpendicular NK on MC. KC = ½(Y-X) ∆ADC is Similar to ∆NKC. KC/DC = NC/AC ½(X-Y)/Y=4/9 X/Y=1/9
Thank you for this puzzle to which I have returned because I paused when I tried it yesterday. l am letting theta = w . The sine of 4w = sine (3w +w) = sin (3w)cos(w)+ sin(x)cos(3w) From rightangled triangles ADB and ADC: sin (3w) = AD/4 sin(w) = AD/9 cos(3w)= X/4 cos (w) = Y/9 sin(4w) will be (AD/4)(X/9) +(AD/9) (Y/4) sin(4w) = AD × (X + Y) /36 From triangle ABC there are three angles = 180º- 4w, 3w and w. sin (4w) = sin(180º-4w) The sine rule gives: sine (4w)/ (X+Y) = sine(3w)/9 = sine (w) /4 AD/36 × (X +Y)/(X+Y) = AD/4/9 = AD/9/4 Everything cancelled out ! Trigonmetry was a lot of work and no reward on this attempt. It is quite a difficult puzzle . So it is time to watch your solution. I
Using law of sines in △ABC, we get 9/4 = sin 3θ / sin θ = (3 sin θ − 4 sin³θ) / sin θ = 3 − 4 sin²θ sin²θ = 3/16 cos²θ = 1 − sin²θ = 13/16 cos θ = √13/4 cos 3θ = 4 cos³θ − 3 cos θ = 4(13√13/64) − 3(√13/4) = √13/16 From △ABD x = 4 cos 3θ = 4(√13//16) = √13//4 From △ABD y = 9 cos θ = 9(√13//4) = 9x Therefore x/y = x/(9x) = 1/9
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We have AD=4sin(3θ)=4(3sinθ-4sin³θ) and we have AD=9sinθ, from which 4(sinθ-3sin³θ)=9sinθ so sin²θ=3/16 so sinθ=√3/4 so cosθ=√13/4 from which cos(3θ)=4cos³θ-3cosθ=√13/16 and since x=4cos 3θ=√13/4 and y=9cosθ=9√13/4 then x/y=1/9
Applause : )
You have Such Great Vision in knowing the Exact Constructions to Draw !
I used so much Trig that I wondered how you would have done this far more easily : )
*Solução:*
Lei dos senos no ∆ABC:
sen 3θ/9 = sen θ/4
Ora, sen 3θ = 3 sen θ - 4sen³ θ, assim
sen θ(3 - 4sen² θ)/9 = sen θ/4
(3 - 4 sen² θ)/9 = 1/4
3 - 4 sen² θ = 9/4
4 sen² θ = 3 - 9/4 = 3/4
sen² θ = 3/16. Consequentemente,
cos² θ = 1 - sen² θ = 1 - 3/16
*cos² θ = 13/16.*
Nos ∆ABD e ∆ADC, temos, respectivamente:
cos 3θ = x/4 e cos θ = y/9
Assim,
cos 3θ/cos θ = (x/y) × 9/4. Sendo que:
cos 3θ = cos θ (4 cos² θ - 3)
Temos:
cos θ (4 cos² θ - 3)/cos θ = (x/y) × 9/4.
4 cos² θ - 3 = (x/y) × 9/4
4 × 13/16 - 3 = (x/y) × 9/4
(x/y) × 9/4 = 13/4 - 3 = (13 - 12)/4
(x/y) × 9/4 = 1/4
x/y = 1/4 × 4/9 → *x/y = 1/9.*
sin 3 t/sint = 9/4
3 - 4 (sint)^2 = 9/4
cos3 t/cost = 9 x/4 y
4 (cost)^2 - 3 = 9 x/4 y
subtracting these equations
2 = (9/4)(1-x/y)
8/9 = 1 - x/y
x/y = 1/9
Sine rule : sin3θ /9 = sinθ /4
sin3θ = 9/4 sinθ
Trigonometric identity :
sin3θ = 3sinθ - 4sin³θ
Equalling :
3 - 4sin²θ = 9/4
4sin²θ = 3 - 9/4 = 3/4
sin²θ = 3/16 ---> θ=25,6589°
x = 4 cos3θ = 0,9014 cm
y = 9 cosθ = 8.1125 cm
x/y = 1/9 (Solved √)
👎 NO CALCULATOR !
@rabotaakk-nw9nm
Why you say that?
It doesn't have sense !!
@@marioalb9726 😡
θ≈25.6589; x≈0.9014; y≈8.1125
x/y=0.9014/8.1125≈0.111112481
1/9=0.111111111… => x/y≠1/9 😬
@@rabotaakk-nw9nm
And ???
It's perfect my solution !!!
Do you need help to download a calculator ??
@@marioalb9726 😆😅😂🤣
You need to download the geometry and trigonometry! => x/y=1/9
Solving using a half interval search: let AD = h, then sin(3Θ) = h/4 and sin(Θ) = h/9, or 4sin(3Θ) = h and 9sin(Θ) = h. We need to compute Θ such that 4sin(3Θ) - 9sin(Θ) = 0. We let f(Θ) = 4sin(3Θ) - 9sin(Θ) and search for a value of theta such that f(Θ) = 0. Since 3Θ must not exceed 90° for the figure to be valid, Θ must not exceed 30°. For a computer program, we replace f(Θ) by delta and Θ by theta. I wrote this program in the python language:
import math
# python computes trig functions for angles in radians, so convert x = 15° to radians:
theta = 15*math.pi/180
dx = (7.5)*math.pi/180
# define function to compute delta
def compute_delta(theta):
delta = (4.0)*math.sin(3*theta) - (9.0)*math.sin(theta)
return(delta)
# note that delta for theta = 15° is positive and delta decreases monotonically
# therefore, there is a zero crossing for theta between 15° and 30°
# if a positive value is returned for delta, we must increase theta and try again
# if a negative value is returned for delta, we must decrease theta and try again
# start with theta = 15° and increment 7.5° (half way from middle of range to end)
# once the increment dx is sufficiently small, compute and print x
for a in range(0,51): # after 50 divisions by 2, dx is sufficiently small
delta = compute_delta(theta)
if (delta > 0):
theta = theta + dx
else:
theta = theta - dx
dx = dx/2
x = (4.0)*math.cos(3*theta)
y = (9.0)*math.cos(theta)
m = round(x/y , 15) # round x to 15 decimal places
formatted = "{:.15f}".format(m)
print(f"x/y = ",formatted)
Output:
x/y = 0.111111111111111
=== Code Execution Successful ===
Note that the repeating decimal 0.11111.... = 1/9.
The program can only calculate x/y to the limit of its precision. Math Booster computed an exact value. I had the program print the value of sin(Θ) as 0.433012701892. When I square it, I get 0.1875, which equals 3/16. So sin(Θ) = (√3)/4 is a good educated guess (but also confirmed in other comments). We use the identity sin(3Θ) = 3.sin(Θ) - 4.(sin(Θ))³ to compute sin(3Θ) from sin(Θ). Substituting these values for sin(Θ) and sin(3Θ), we should find 4sin(3Θ) - 9sin(Θ) = 0 exactly. Knowing sin(Θ) and sin(3Θ), we can compute cos(Θ) and cos(3Θ), compute x and y, then x/y.
Is there an AI app which can make all the MathBooster problems a cakewalk?
*Let's note t = theta and AD = h. Then sin(3.t) = h/4 in triangle ABD and sin(t) = h/9 in triangle ACD, and then sin(3.t)/sin(t) = 9/4
We know that sin(3.t) = 3.sin(t) - 4.(sin(t))^3, so sin(3.t)/sin(t) = 3 - 4.(sin(t)^2) = 9/4 giving that (sin(t)^2 = 3/16
and (cos(t))^2 = 1 - (3/16) = 13/16.
*cos(3.t) = x/4 in triangle ABD and cos(t) = y/9 in triangle ACD, so x/y = (4/9).(cos(3.t)/cos(t)). We know that cos(3.t) = 4.(cos(t))^3) -3.cos(t)
so x/y = (4/9).(4.(cos(t))^2) -3) = (4/9).((13/4) - 3) = (4/9).(1/4) = 1/9.
That is nice and this is probably similar to one of the problems on your channel. That similar ratio of proportions that most likely *definitely* make uae of HL similarity with respect to the right angle. I think that that is what necessarily occurs. Also the x/y = 1/9.
Reflect ∆ADB about AD to make ∆ADM. Now, ∠MAC = 2θ
Draw MN, with N on AC such that ∠MNA =2θ
Both ∆AMN and ∆MNC are Isosceles, with AM= MN = NC=4
Drop a perpendicular NK on MC.
KC = ½(Y-X)
∆ADC is Similar to ∆NKC.
KC/DC = NC/AC
½(X-Y)/Y=4/9
X/Y=1/9
th-cam.com/video/JF_cp-izoTQ/w-d-xo.htmlsi=m4FFz1AUIsf5b_Tx
Auto translation⁉️
Uau! Que questão bonita e elegante, muito elegante. Eu a resolvi de duas maneiras distintas. Parabéns pela escolha. Brasil - Novembro de 2024.
th-cam.com/video/JF_cp-izoTQ/w-d-xo.htmlsi=m4FFz1AUIsf5b_Tx
Thank you for this puzzle to which I have returned because I paused when I tried it yesterday.
l am letting theta = w . The sine of 4w = sine (3w +w) = sin (3w)cos(w)+ sin(x)cos(3w)
From rightangled triangles ADB and ADC: sin (3w) = AD/4 sin(w) = AD/9 cos(3w)= X/4 cos (w) = Y/9
sin(4w) will be (AD/4)(X/9) +(AD/9) (Y/4) sin(4w) = AD × (X + Y) /36
From triangle ABC there are three angles = 180º- 4w, 3w and w. sin (4w) = sin(180º-4w)
The sine rule gives: sine (4w)/ (X+Y) = sine(3w)/9 = sine (w) /4
AD/36 × (X +Y)/(X+Y) = AD/4/9 = AD/9/4
Everything cancelled out ! Trigonmetry was a lot of work and no reward on this attempt.
It is quite a difficult puzzle . So it is time to watch your solution.
I
th-cam.com/video/JF_cp-izoTQ/w-d-xo.htmlsi=m4FFz1AUIsf5b_Tx
Can someone tell that what is wrong in this solution:
Cos(3∅) = X/4 and Cos(∅) = Y/9
As in triangle ADC angle ∅ would be less than 90⁰ so ∅
Using law of sines in △ABC, we get
9/4 = sin 3θ / sin θ = (3 sin θ − 4 sin³θ) / sin θ = 3 − 4 sin²θ
sin²θ = 3/16
cos²θ = 1 − sin²θ = 13/16
cos θ = √13/4
cos 3θ = 4 cos³θ − 3 cos θ = 4(13√13/64) − 3(√13/4) = √13/16
From △ABD
x = 4 cos 3θ = 4(√13//16) = √13//4
From △ABD
y = 9 cos θ = 9(√13//4) = 9x
Therefore
x/y = x/(9x) = 1/9
th-cam.com/video/JF_cp-izoTQ/w-d-xo.htmlsi=m4FFz1AUIsf5b_Tx
How to draw triangle ABC correctly ?
Well, θ = (180° - arccos(7/32))/4 ≈ 25.66°
(4)^2 (9)^2={16+81}=97;{30°A+30°B+90°C}={150°ABC+30°D}=180°ABCD/97=1.83 1.83^1 1.1^1 1^1(ABCD ➖ 1ABCD+1).
th-cam.com/video/JF_cp-izoTQ/w-d-xo.htmlsi=m4FFz1AUIsf5b_Tx
just check this.
I am doubtful.
let perpendicular = k and theta = m
sin(3m)/k = cos(3m)/x = 1/4
sin(m)/k = cosm/y = 1/9
sin(m)/4 = sin(3m)/9 = sin(4m)/(x+y)
sin(m)/4 = (3 Sin(m) - 4 (sin(m))^3)/9
solving
sin(m)^2 = 3/16
sin(m) = sqrt(3)/4
cos(m) = sqrt(13)/4
k/9 = sqrt(3)/4
k = 9sqrt(3)/4
y/9 = sqrt(13)/4
y = 9sqrt(13)/4
x^2 + k^2 = 16
gives x = 13/16
x/y = sqrt(13)/36
asnwer=1/9 hmm gmm