Be warned: this problem is harder than it initially looks!

As usual, watch the video for a solution.

How To Solve For The Angle – Outside The Box Thinking!

Or keep reading.

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Answer To How To Solve For The Angle – Outside The Box Thinking!

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Method 1: Geometry

In triangle ABC, angle A is equal to 180 – 20 – 80 = 80°. Since angles A and B are equal, AC = BC. Construct an equilateral triangle AEC with AE splitting angle A into 60 and 20 degree angles, and BCE = 40°.

Then BC = CE, so BCE is an isosceles triangle with a vertex angle of 40°, meaning the other angles CBE = BEC = (180 – 40)/2 = 70°.

Now triangles BAE and DCA are congruent by side-angle-side, so angles CDA and ABE are equal, meaning CDA = 80 + 70 = 150°. Finally angle ADB is supplementary to ADC, so ADB = 180 – 150 = 30°.

Method 2: Trigonometry

Let x be the measure in degrees of angle ADB. This angle is an exterior angle of the triangle ADC, so we will have DAC + DCA = ADC, so then DAC = x – 20°.

Applying the law of sines to triangles ABD and ADC gives:

AD/AB = (sin 80°)/(sin x)
AD/CD = (sin 20°)/(sin (x – 20°))

Since AB = CD, both equations are equal and we have:

(sin 80°)/(sin x) = (sin 20°)/(sin (x – 20°))

Now we have:

(sin 80°)/(sin 20°) = (sin x)/(sin (x – 20°))
(cos 10°)/(sin 20°) = (sin x)/(sin (x – 20°))
(cos 10°)/(2 cos 10° sin 10°) = (sin x)/(sin (x – 20°))
1/(2 sin 10°) = (sin x)/(sin (x – 20°))
(1/2)/(sin 10°) = (sin x)/(sin (x – 20°))
(sin 30°)/(sin 10°) = (sin x)/(sin (x – 20°))

By inspection x = 30° is a solution, and like magic we have found an answer!

To verify there are no other solutions, we note:

–for an angle between 0° and 20° the left-hand side is positive but the right-hand side is negative
–for an angle between 20° and 100° the right-hand side is strictly decreasing, as its derivative is sin(π/9)(-csc² ((π – 9 x)/9)) which is negative.

References

previous video
https://www.youtube.com/watch?v=5vhklRWogzo
credit to YT community
https://www.youtube.com/post/UgkxCzCDIPprA97xtVXG6cEzg7gEHwfGZ_HV?lc=UgxfBeZUl5wattKxeNh4AaABAg

Cut The Knot (9 solutions)
https://www.cut-the-knot.org/triangle/80-80-20/IndexToLong.shtml
C. Knop, A Story with Geometry, or Nine Solutions to One Problem, Kvant, 1993, no 6
https://kvant.mccme.ru/1993/06/istoriya_s_geometriej.htm

Solve for x:

8^x + 2^x = 30

As usual, watch the video for a solution.

Trick You Should Know For Harvard Entrance Exam Problem

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Answer To Trick You Should Know For Harvard Entrance Exam Problem

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

Using exponent rules we have:

8^x + 2^x = 30
(2³)^x + 2^x = 30
(2^3x) + 2^x = 30
(2^x)³ + 2^x = 30

Now let’s substitute y = 2^x. Thus we have a cubic equation:

y³ + y = 30
y³ + y – 30 = 0

If we can figure out a root of this equation, we can reduce it to the product of a linear equation and a quadratic equation, both of which we can readily solve. By the rational root theorem, we have candidate rational roots are plus or minus factors of the constant term 30 divided by the factors of the leading coefficient of 1. This gives candidates of ± 1, 2, 3, 5, 6, 10, 15, 30. Testing y = -1 we get:

-1 – 1 – 30 = -32

Any larger negative value will just be more negative. Testing y = 1, 2, 3 we get:

1 + 1 – 30 = -28
8 + 2 – 30 = -20
27 + 3 – 30 = 0

So y = 3 is a root, and it is the only rational root of this equation.

Therefore y³ + y – 30 is divisible by y – 3. Performing long division we get:

This means we have:

(y – 3)(y² + 3y + 10) = y³ + y – 30

Setting the equation equal to 0, we have the linear factor or the quadratic factor has to be equal to 0, giving:

y – 3 = 0
y = 3

y² + 3y + 10 = 0
(using quadratic formula)
y = -1.5 ± 0.5 i√31

We previously made the substituton:

y = 2^x
ln y = ln 2^x
ln y = x ln 2
x = (ln y)/(ln 2)

Thus we have 3 possible solutions:

y = 3
x = (ln 3)/(ln 2)

y² + 3y + 10 = 0
x = (ln (-1.5 ± 0.5 i√31))/(ln 2)

It is important to check for extraneous solutions, so it is good to test these values to make sure they do solve the original equation.

You can check in WolframAlpha, for example:

x = (ln 3)/(ln 2)
8^x + 2^x = 30
https://www.wolframalpha.com/input?i=8%5E%28ln+3%2Fln+2%29%2B2%5E%28ln+3%2Fln+2%29

x = (ln (-1.5 + 0.5 i√31))/(ln 2)
8^x + 2^x = 30
https://www.wolframalpha.com/input?i=8%5E%28ln+%28%28-1.5%2B0.5i*sqrt%2831%29%29%29%2Fln+2%29%2B2%5E%28ln+%28%28-1.5%2B0.5i*sqrt%2831%29%29%29%2Fln+2%29

x = (ln (-1.5 – 0.5 i√31))/(ln 2)
8^x + 2^x = 30
https://www.wolframalpha.com/input?i=8%5E%28ln+%28%28-1.5%2B0.5i*sqrt%2831%29%29%29%2Fln+2%29%2B2%5E%28ln+%28%28-1.5%2B0.5i*sqrt%2831%29%29%29%2Fln+2%29

Thus we have found 3 solutions to the original equation. What a fun problem!

Anna says the number 2021 is fantabulous. All such numbers obey a property.

If any element of the set {m, 2m + 1, 3m} is fantabulous for a positive integer m, then all elements of the set are fantabulous.

Is the number 2021²⁰²¹ fantabulous?

As usual, watch the video for a solution.

Fantabulous Numbers

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Answer To Fantabulous Numbers

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

A mathematician would likely think all positive integers are fantastic and fabulous, so it would stand to reason that all numbers are fantabulous. Can we prove it?

Since m = 2021 is fantabulous, we can readily calculate other numbers

2m + 1 = 4043
3m = 6063

But now let’s make a set {n, 2n + 1, 3n} with 2n + 1 = 6063. We then have:

n = 3031

Iterating with 2p + 1 = 3031, we have:

p = 1515

Taking 1515 = 3q, we have

q = 505

Take this as the form 2r + 1 to get

r = 252

We can iterate in this fashion to find fantabulous numbers

252/3 = 84
84/3 = 28
2(28) + 1 = 57
57/3 = 19
(19 – 1)/2 = 9
9/3 = 3
3/3 = 1

So 2021 being fantabulous implies 1 is.

We already know m being fantabulous implies 2m + 1 is. If we could also show it implies 2m is fantabulous, then we would have 1 being fantabulous implies 2, 3 are, and then iterating we would have 4, 5, 6, 7 are, and so on, so that all positive integers are fantabulous, including 2021²⁰²¹.

We can proceed like in the last calculation.

Let m be fantabulous. Then we will have fantabulous numbers from the operations 2m + 1, 3m. If m is odd we can perform (m – 1)/2, and if m is a multiple of 3 we can do m/3. So the fantabulous numbers are:

m
3m
2(3m) + 1 = 6m + 1
2(6m + 1) + 1 = 12m + 3
(12m + 3)/3 = 4m + 1
((4m + 1) – 1)/2 = 2m

Thus m being fantabulous implies 2m is. Along with m implies 2m + 1 and 1 being fantabulous, we have that all positive integers are fantabulous, including 2021²⁰²¹.

Reference

2021 European Girls’ Mathematical Olympiad (EGMO)
https://www.egmo.org/egmos/egmo10/solutions.pdf
https://www.egmo.org/egmos/egmo10/