Find the angle X | Norway Math Olympiad Problem

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[MarieAnne.]

I'm not sure why you bothered calculating extra angles when we can see that
tan x = b/a, tan 16 = d/a, tan 44 = b/c, tan 48 = d/c
Therefore
tan x tan 48 = tan 16 tan 44
*tan x tan(3*16) = tan 16 tan(60-16)* **
And instead of using a formula that few have seen, we can stick to those that are well known.
Let t = tan 16 = tan A
tan(3A) = tan(A+2A) = (tan A + tan 2A) / ( 1 − tan A * tan 2A)
= (t + 2t/(1−t²)) / (1 − t*2t/(1−t²))
= t(3−t²)/(1−3t²)
tan(60−A) = (tan 60 − tan A) / (1 + tan 60 * tan A)
= (√3−t)/(1+t√3)
Plugging into equation ** above, we get
tan x * t(3−t²)/(1−3t²) = t * (√3−t)/(1+t√3)
tan x = (1−3t²)/(3−t²) * (√3−t)/(1+t√3)
tan x = (1−t√3)(1+t√3)/((√3−t)(√3+t)) * (√3−t)/(1+t√3)
tan x = (1−t√3)/(√3+t)
tan x = (1/√3−t)/(1+(1/√3)t)
tan x = (tan 30 − tan 16) / (1 + tan 30 tan 16)
tan x = tan(30−16) = tan 14
*x = 14*

[jimlocke]

Math Booster overlooked a second solution: tan(Θ) = tan(Θ + 180°). So, x = 194° is also a solution to tan(14°) = tan(x). However, x is acute in the problem. Therefore x = 194° is an invalid solution to this problem and must be discarded, just as a second root to a quadratic equation must often be discarded. To be complete, this solution should also be presented and then discarded as invalid.

[oscarcastaneda]

I drew so many extra lines and semicircles that my whole body became dizzy !

Excellent Solution Amigo : )

[pwmiles]

By ratios of the known angles
tan x = tan 44 tan 16 / tan 48
tan 16 = t
By triple angle formula for tan
tan 48 = 3t - t^3 / (1 - 3t^2)
tan x = tan 44 (1-3t^2) / (3 - t^2)
= tan(60 - 16) (1-3t^2) / (3 - t^2)

= [(sqrt(3) - t)/(1 + t sqrt(3))] (1 - t sqrt(3)) (1 + t sqrt(3) /[(sqrt(3) - t)(sqrt(3) + t)]
= (1 - t sqrt(3)) / (sqrt(3) + t)
= (1/sqrt(3) - t) / (1 + t/sqrt(3))
= tan(30 - 16)
= tan(14)
x = 14 deg

[Pryszczyk]

tan(x)=tan(44)×tan(16)÷tan(48)≈0, 249
x=arctan(0, 249)=14°

[kateknowles]

tan (X) = BO/AO tan(44º) = BO/CO tan(48º) = DO/CO tan (16º) = DO/AO
Tan (48º)/ tan(16º) = AO/ CO because DO/DO =1
Tan (X) /tan (44º) = CO / AO because BO/BO =1
so tan(16º) /tan (48º) = tan (X) / tan(44º) tan(X) = tan (44º) × tan (16º) / tan (48º)
reaching for calculator... approx arctan (0.249) X = 14 º
Now time for the video and the other comments. Thank you.

[shaozheang]

I found a video on a polish math olympiad question by Mind Your Decisions, he uses a similar method too.

[huss]

Such a long complicated answer. It's 14, 96°C or approximately 15°C.

[vcvartak]

How can we presume the formula for tan3A . it's not very popular ( in terms of (60-A))

[oguzhanbenli]

The angle values for this figure are not correct. The value 14 does not verify that the sum of the inner angles is equal to 360 degrees. The angles at the center cannot be 90 degrees each

[RealQinnMalloryu]

{16°A+44°B+C48°}=108°ABC {108°ABC+72°D}=180°ABCD 72^108 72^16^44^48 8^9^16^2^2^2^2^2^2^2^3 8^3^2^4^4^1^1^1^1^1^1^1^1^1 2^3^3^1^2^2^2^2 1^1^13^1^1^1^1^2 32(ABCD ➖ 3ABCD+2).

[giuseppemalaguti]

Utilizzando i teoremi del seno risulta tgx=tg44/tg48tg74...x=14

[paulfernandez]

Very Beautiful Way Of Approach And Execution Of Very Elementary Fact And Basic Trigonometry Identity. Thank You For Sharing And Keep Up The Good Work.

[ДмитрийИвашкевич-ят]

Есть секретная формула:
tg3α=tgα*tg(60°-α)*tg(60°+α). Проверял лично.🙂
α=16°


α=14°

[kateknowles]

At 6.05 I am learning something new:- tan (60º-A) × tan (60º+A) is equal to tan (3A)/ tan (A) This is something nice.

[antoniopedrofalcaolopesmor]

Wondering how this problem could be solved if we were not using trigonometry, I mean just using elementary geometry.