Japanese Math Olympiad | A Very Nice Geometry Problem

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[紫瞳-wt]

An easier way:
0. set θ = k, sin(k)>0, observe angle A= 90-2k, angle ADC = 90-k.
1.from ADC: ABC:
2.from eq1: X/sin(3k)=Y/cos(k)+Y/sin(3k) => Y/X
3. from eq2 & eq3: Y/X= sin(2k)/cos(2k)= {cos(k)/[sin(3k)+cos(k)]}
4.from eq4: => cos(4k)=sin(2k)...eq5
5.there are many easy way to solve eq5, one of them: let sin(2k)=u, cos(4k)=1-2u^2 => u=1/2 =>2k=30 => k=15.

[ludmilaivanova]

We can find the angle from triangles built by intersection of CD with the hypotenuse of an isosceles triangle ACE where AC=CE=Y (our construction).
If the point of intersection is O, then Angle AOD = 45degrees +(90-3alfa); at the same time, this is an external angle for the triangle ACO and it is =3alfa+45 degrees.
so, 45 +90-3alfa= 3alfa+45, 90= 6 alfa, alfa=15.

[yuvan]

Can we use simultaneous equations to solve this sir by labelling angle BAC with a variable?

[michaeldoerr]

The angle theta is 15°. Btw Aibuave noticed that a good amount of gemeitey problems in your channel have resulted in the angle being 15°. Is that due to that auxiliary line drawn for the collinear angles???

[oscarcastaneda]

This was challenging.
I took a wild guess at 15 deg and knowing the value of cos(15) I used Law of Sines to confirm my guess.

Some days my intuition for knowing what to Construct is on Vacation : )

[RealQinnMalloryu]

{20°B+20°C+20°D}=60°BCD {30°C+30°D+30°A}=90°CDA {60°BCD+90°CDA}=150°BCDCDA {150°BCDCDA+30}=180°BCDCDA 3^60.3^30^2 3^3^10^2 1^3^2^5^2 3^1^1^2 32(BCDCDA ➖ 3BCDCDA+2).

[jeanmarcbonici]

Nice, there is a trigonometric solution but yours is better!

[bennyhsiao]

x/sin(90°-2☆)=
y/sin2☆
x-y/sin3☆=
y/sin(90°-☆)
x/y=cos2☆/
sin2☆
x-y/y=
sin3☆/cos☆
(x/y) -1=
sin3☆/cos☆
cos2☆/sin2☆-
1=sin3☆/cos☆ solved the equation
we can get
2(sin2☆)^2 +
sin2☆-1=0
(2sin2☆-1)(sin2☆+1)=0
2sin2☆ -1=0
sin2☆ +1=0
sin2☆=1/2
2☆=30°
☆=15°(ok)
sin2☆=-1
2☆=270°
☆=135°（x）

[brettgbarnes]

I guess you could say that the upshot of this problem is that when you draw an angle bisector line from the right angle of a 30:60:90 triangle to its hypotenuse, the length of the section of hypotenuse between the bisector line and the side opposite the 30 degree angle will be equal to the difference of the two perpendicular side lengths.