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Find the value of the angle x.

enter image description here

Plus : Someone could recommend me some good book about this subject ?

Joel
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3 Answers3

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Purely geometric proof

enter image description here

Let the triangle be $\triangle PQR$ with $\angle QPR = 20^\circ$ and $\angle PRQ = 90^\circ$

Let $O$ be on the side $PQ$ such that $O$ is the centre of a regular 18-gon with radius $OQ$ and one side $AB$ lying on $PR$, where $B$ is between $P$ and $A$, which is possible because that side $AB$ and $OQ$ meet at an angle of $20^\circ$

Then $QR$ clearly passes through another vertex $C$ of the 18-gon

Let $D$ be another vertex of the 18-gon such that $CD \perp OA$

Then $CD$ bisects $OA$ because $\triangle OAC$ is equilateral

Thus $CD$ passes through $P$ because $\triangle OPA$ is isosceles since $\angle AOP = \angle PAO$

Thus $\angle QPC = 10^\circ$ and $\angle CQA = 30^\circ$ and so they are the desired points

Therefore $\angle QAC = 20^\circ$

user21820
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  • Woww !! +1) Nice solution. – Joel Dec 31 '13 at 13:38
  • Thanks! It was fun finding it too! =D – user21820 Dec 31 '13 at 13:38
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    By the way, a lot of these triangle problems can be solved using some similar regular polygon somewhere. It took me a while to find where I could put the polygon, but after that it was not too hard. And if you like this solution best, you can accept it as the answer. =) – user21820 Dec 31 '13 at 13:41
  • And you might be interested to take a look at http://www.cut-the-knot.org/triangle/80-80-20/index.shtml, which give other such triangle problems. I've never come across yours though. – user21820 Dec 31 '13 at 13:44
  • I've already seen this problem in your link but not the variances. Thanks and do you know some good book about this subjects ? – Joel Dec 31 '13 at 13:53
  • Nice! Do you want to try your skill with this question as well? Looks kind of similar. – MvG Dec 31 '13 at 14:25
  • I don't know of any book on this subject, and I suspect it is too small a topic? I can try the other question but it looks like a different sort of trick is necessary if an elementary geometry proof is possible. – user21820 Dec 31 '13 at 15:39
  • Hello @user21820 ! That is an awesome answer. I wish I could transfer up votes from here to EL&U :) You are deserving. – Ellie Kesselman May 04 '15 at 05:20
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    @EllieKesselman: Hello! Nice to see you here and thanks! I didn't know you were on Math SE too. =) – user21820 May 04 '15 at 05:24
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Claim

The angle is $20°$, as found in a construction performed with Cinderella.

Cinderella snapshot

Proof

If you want to proove this, you could use the following computation using sage. All arithmetic is done using exact algebraic numbers, to avoid floating point rounding issues.

def sin_cos(degrees):
    degrees = QQ(degrees)/360
    z = QQbar.zeta(degrees.denominator())^(degrees.numerator())
    return z.imag(), z.real()
sin10, cos10 = sin_cos(10)
sin20, cos20 = sin_cos(20)
sin30, cos30 = sin_cos(30)
sin40, cos40 = sin_cos(40)
lenAB = 1
lenAC = lenAB*cos20
lenBC = lenAB*sin20
lenAF = lenAC/cos10
lenCF = lenAF*sin10
lenBE = lenBC/cos30
lenCE = lenBE*sin30
lenCF/lenCE == sin40/cos40

The result (printed as True) will show that $\angle FEC=40°$. From the angle sum in $\triangle BCE$ one can conclude $\angle BEC=180°-90°-30°=60°$. So you get $\angle BEF=\angle BEC-\angle FEC=60°-40°=20°$ as the experiment suggested.

MvG
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  • Perhaps there's another way to compute the value of x without a software. Anyway, thanks ! – Joel Dec 31 '13 at 13:14
  • There is indeed! I'm going to give my solution now. – user21820 Dec 31 '13 at 13:14
  • @Joel: The computations I wrote down for sage could have been performed using a scientific pocket calculator, albeit with reduced precision. Taking the atan of that lenCF/lenCE fraction would give the angle of $40°$, which would yield the final result as described in the proof. The above approach is not well suited for computation without any technical aid at all, since these trigonometric functions don't yield easy values. But an easy solution might still exist, and I'm looking forward to what user21820 has come up with. – MvG Dec 31 '13 at 13:25
  • It's a purely geometric solution. Give me a few more minutes to finish typing it up and including a diagram. =) – user21820 Dec 31 '13 at 13:31
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Construction: Construct $\triangle$ PQR $\cong$ $\triangle$ ABC. Construct $\angle$s PST and PSU both equal to 30 intersecting sides PR and PQ at points T and U respectively.

Proof: $\angle$ USQ=70. US=UQ. $\triangle$ PTS $\cong$ $\triangle$ PUS. $\triangle$ POT $\cong$ $\triangle$ POU.
$\angle$ POT = $\angle$ POU=90. Therefore TS=US. $\triangle$ UST is an equilateral triangle and $\angle$s PTO and PUO are both equal to 80. Therefore, $\angle$s UTQ and UQT are both equal to 40. Therefore, $\angle$ QTS=20 and $\angle$ RQT=30 Now $\triangle$ BCE $\cong$ QRT and $\triangle$ DCE $\cong$ SRT. That proves x=20.

enter image description here

Iftikhar
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  • Looks good! Welcome to Math SE! If you want an account on Math SE, and you still can access your profile page, you can register. Otherwise you may lose access to your unregistered account. – user21820 Aug 24 '17 at 08:42