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Triangle

cinelli
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Triangle

#463836

Postby cinelli » December 7th, 2021, 12:51 pm

Inside a square ABDE, place a point C so that CDE is an isosceles triangle with angles 15 degrees at D and E. What are the values of the angles of triangle ABC?

Cinelli

UncleEbenezer
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Re: Triangle

#463855

Postby UncleEbenezer » December 7th, 2021, 1:28 pm

cinelli wrote:Inside a square ABDE, place a point C so that CDE is an isosceles triangle with angles 15 degrees at D and E. What are the values of the angles of triangle ABC?

Cinelli

Are we allowed trig functions? If so, surely far too obvious!

If (for convenience) we take the side of the square as 2 units and X as the midpoint of DE, then DX = XE = 1, and CX = tan(15).
So if Y is the midpoint of AB, then CY = 2 - tan(15).
Thus the angles at A and B are each arc tan(2-tan(15)), and at C is the remainder subtracting those at A and B from 180.

9873210
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Re: Triangle

#463930

Postby 9873210 » December 7th, 2021, 6:00 pm

UncleEbenezer wrote:
cinelli wrote:Inside a square ABDE, place a point C so that CDE is an isosceles triangle with angles 15 degrees at D and E. What are the values of the angles of triangle ABC?

Cinelli

Are we allowed trig functions? If so, surely far too obvious!

If (for convenience) we take the side of the square as 2 units and X as the midpoint of DE, then DX = XE = 1, and CX = tan(15).
So if Y is the midpoint of AB, then CY = 2 - tan(15).
Thus the angles at A and B are each arc tan(2-tan(15)), and at C is the remainder subtracting those at A and B from 180.


If Φ is the angle ABC we can rewrite that as
tan(Φ) + tan(15) =2

You can calculate trig values for 15 exactly using the half angle formula for tan and known values for cos(30) and sin(30)

tan(Θ/2) = (1- cos(Θ)/sin(Θ)
so
tan(15) = (1-cos(30))/sin(30) = (1-sqrt(3)/2)/(1/2) = 2- sqrt(3)

Substituting
tan(Φ) + 2-sqrt(3) =2
tan(Φ) = sqrt(3)

So by inspection Φ = 60

There's probably a trig identity that lets you get there in one step, but that requires memorizing more trig identities.
Once you have that trig identity you can probably disguise it as a calculation that does not use trig.

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Re: Triangle

#463955

Postby 9873210 » December 7th, 2021, 7:07 pm


Once we know the answer we can prove it simply.

Construct the equilateral triangle AGB.
Then ∠EAG = 30 and AE = AG.
This is an isoscelese triangle so ∠AEG = (180 -30)/2 = 85
Thus ∠EGD = 15 and points G and C are the same.

Verification simple, derivation requires clairvoyance (or trig).
Please sir may I have another Green's function?



cinelli
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Re: Triangle

#464348

Postby cinelli » December 8th, 2021, 8:40 pm

I like these ingenious methods. I have another solution which doesn't involve trig or clairvoyance which I will post in a couple of days (to give time for anyone else to have a go).

Cinelli

cinelli
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Re: Triangle

#464822

Postby cinelli » December 10th, 2021, 5:44 pm

Here is my alternative solution.

A................................................E
. . . . ..
. . . . ..
. . F . .
. . . . . .
. . . . . .
. . . G .
. . . . .
. . . . .
. . . . .
. . . . .
. .C .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . ..
. . .
. . .
B................................................D

Construct another isosceles triangle AFE with the angles at A and E 15 degrees. Produce AF to meet CE at G. Then CEF Is equilateral and since angle EAF is 15 degrees and angle AEG is 85 degrees, angle AGE is a right angle. Hence G is the mid point of CE and triangle CAE is isosceles. So AE = AC = AB = BC (by symmetry) and ABC is equilateral.

Cinelli
~
~
~

cinelli
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Re: Triangle

#464845

Postby cinelli » December 10th, 2021, 7:12 pm

Typo: angle AEG is 75 degrees, not 85.

Cinelli


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