Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods

[oscarcastaneda]

Knowing the value of the sine of 15deg one can find the length of AB.
From this one can find the length of AC which will be 6, thus the answer is 30deg.

[ناصريناصر-سب]

For the first time, the algebraic method outperforms the geometric method.😊😊😊

[ludmilaivanova]

it is interesting:
if we draw a line from the pont B under 45 degrees and make DE=a, then we will have an isosceles triangle BED. Mirror this triangle to the right.
FC=6-2a, AE=3-a, we have got a trapecium AEFC, both sides are equal. Then the angles at the base are 45 degrees. Correct me, please, if i am wrong.

[imetroangola]

If I'm not mistaken, this question has already been asked on this channel. Every now and then there are repeated questions.

[michaeldoerr]

The second method is similar to several problems on this channel that showed a 15°-15° isosceles triangle and goodness me, this is where clever geometry and algebra combine for the best!!!

[jimlocke]

Take an educated guess that AC and BC are the same length. Then, ΔABC is isosceles and <CAB = <CBA = 75°, so Θ = <ACB = 180° - 75° - 75° = 30°. ΔACD is a 30° - 60° - 90° right triangle, so, given AD = 3, AC = 6, which agrees with our guess, and CD = 3√3. Then BD = 6 - 3√3 = 3(2 - √3). AD/BD = 3/(6 - 3√3) = 1/(2 - √3). ΔABD is given as a 15° - 75° - 90° right triangle, which appears quite often in geometry problems, so I recommend that all of us be familiar with its ratio of sides, short : long : hypotenuse, is (√3 - 1):(√3 + 1):(2√2). So AD/BD = (√3 + 1)/(√3 - 1). If we multiply 1/(2 - √3) by (√3 + 1)/(√3 + 1), we get (√3 + 1)/((√3 + 1)(2 - √3)) = (√3 + 1)/(2√3 + 2 -(√3)(√3) - √3) = (√3 + 1)/(2√3 + 2 - 3 - √3) = (√3 + 1)/(√3 - 1), which equals the ratio long side: short side for the 15° - 75° - 90° right triangle. The 75° angle is opposite the long side, AD in this case. So, the givens are correct for our guess.

We really don't need the educated guess. Compute length BD from ratio of sides for a 15° - 75° - 90° right triangle. Subtract from 6 to find CD. Write the ratio AD/CD. It should reduce to 1/(√3), which is the ratio short side/long side for a 30° - 60° - 90° right triangle. So, Θ = 30°.

[lasalleman]

Using given angle of 75 degrees at point B, find line BD. Do this by going Pythgorean, and the cosine of angle ABD, line BD, works out to .8038 . Subtract line BD from line BC. You get 5.1961 for line DC, which is the cosine of the unknown angle. Then, the sine of the unknown angle, AD, which is 3, divided by the cosine of the unknown angle which is, 5.1961, gives you a raw tangent of .5773. Then just enter that number on arctan on your calculator and the angle turns out to be 30 degrees.

[ducduypham]

From C draw a perpendicular to AB intersect AB at E. From E draw a perpendicular to BC intersect BC at F. As BEC is right triangle with angle EBC=75° and EF is altitude of hypotenuse BC, we have EF=BC/4=3/2.In Triangle BAD, EF//AD and EF=3/2=AD/2 then E is midpoint of AB.In triangle ABC, CE is an altitude of AB but also a median therefore ABC is isosceles triangle with base AB. Then CE is bisector of angle ACB then theta= Angle ACB=2*angle ECB=2*15°=30°

[marioalb]

BD = x = 3 / tan75° = 0, 80345 cm
tan θ = 3 / (6-x) --> θ=30° (Solved √)

[alainpeugny]

Thank you for the video. A pure geometrical solution without any calculation: build an equilateral triangle AOB based on segment AB.
Quite easy to show that triangle BOC is isosceles with angle OCB = 15°. Then you demonstrate the triangles BOC and AOC are congruent and so θ=15°+15°=30°

To show that BOC is isosceles, define H between B and C so that angle BHO = 90°. Triangles ADB, BHO and CHO are all congruent.

[jairoeveliogordillomarin]

Tan 75°=3/BD. BD= 0.804 u
BC= 6-BD= 5.196 u.
Arctan ángulo BCA= 3/5.196.
Ángulo BCA= 30°.

[ناصريناصر-سب]

We have BD=3/tan(75) and CD=BC-BD=6-3/tan(75) and from it so θ=30🎉

[murdock]

φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = k; BCA = θ = ? BC = BD + CD = x + (6 - x)
sin⁡(BDA) = 1; DAB = φ/2 → tan⁡(φ/2) = 2 - √3 = x/3 → x = 3(2 - √3) → 6 - x = 3√3 → θ = φ

[RealQinnMalloryu]

(3)^2(6)^2={9+36}=45 {75°B+75°A+30°D}=180°BAD /45=4BAS 2^2 (BAD ➖ 2BAD+2).