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Triangle
Posted: December 7th, 2021, 12:51 pm
by cinelli
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
Re: Triangle
Posted: December 7th, 2021, 1:28 pm
by UncleEbenezer
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.
Re: Triangle
Posted: December 7th, 2021, 6:00 pm
by 9873210
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.
Re: Triangle
Posted: December 7th, 2021, 7:07 pm
by 9873210
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?
Re: Triangle
Posted: December 8th, 2021, 8:40 pm
by cinelli
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
Re: Triangle
Posted: December 10th, 2021, 5:44 pm
by cinelli
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
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Re: Triangle
Posted: December 10th, 2021, 7:12 pm
by cinelli
Typo: angle AEG is 75 degrees, not 85.
Cinelli