Poland Math Olympiad Problem | A Very Nice Geometry Challenge

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@davidellis1929 4 ай бұрын

You don't need the Law of Cosines to find the length of BE, because you have already established that EBC is isosceles.

@lusalalusala2966 4 ай бұрын

Alternative solution: Draw the height h of the triangle from the vertex A with foot you can name G between D and C. We get then 3 right triangles with the right angle at G, namely ABG, ADG and ACG. Let x be the length of DG, then lhe lenghts of BG and GC are 1+x and 2-x, respectively. It follows that tan 45=h/(1+x); tan60=h/x and tan(theta)=h/(2-x). We have 1=h/(1+x) and sqrt(3)=h/x, or h=1+x and h=sqrt(3)*x. Solve this to get x=(sqrt(3)+1)/2 and h=(3+sqrt(3))/2. Therefore tan(theta)=h/(2-x) =((3+sqrt(3))/2)/(2-(sqrt(3)+3)/2)=(3+sqrt(3))/(3-sqrt(3))=2+sqrt(3). Finally, Theta=5Pi/12.

@oscarcastaneda5310 4 ай бұрын

Knowing that angle BAD is 15 deg and that sin(15) = sqrt(2)(sqrt(3) - 1)/4 one can determine the length of AD via the Law of Sines. From this one can determine the length of AC via the Law of Cosines. Finally using the Law of Sines one can determine the size of the requested angle to be 75 deg. PS: It helps to have sin(15) = sqrt(2)(sqrt(3) - 1)/4 & cos(15) = sqrt(2)(sqrt(3) +1)/4 memorized : )

@TheAlavini 4 ай бұрын

Good demonstration. Congrats

@quigonkenny 4 ай бұрын

As ∠ADC = 60° and an exterior angle to ∆ABD at D, ∠BDA = 180°-60° = 120° and ∠DAB = 60°-45° = 15°. By the law of sines: BD/sin(15°) = DA/sin(45°) = AB/sin(120°) 1/((√3-1)/2√2) = DA/(1/√2) = AB/(√3/2) DA = (1/√2)/((√3-1)/2√2) DA = 2/(√3-1) DA = 2(√3+1)/(√3-1)(√3+1) DA = 2(√3+1)/(3-1) = √3 + 1 AB = (√3/2)/((√3-1)/2√2) AB = √6/(√3-1) AB = √6(√3+1)/(√3-1)(√3+1) AB = (3√2+√6)/2 = (3+√3)/√2 AB/sin(θ) = BC/sin(135°-θ) ((3√2+√6)/2)/sin(θ) = 3/sin(135°-θ) 3sin(θ) = (3√2+√6)sin(135°-θ)/2 6sin(θ) = √6(√3+1)(sin(135°)cos(θ)-cos(135°)sin(θ)) √6sin(θ) = (√3+1)(cos(θ)+sin(θ))/√2 sin(θ)/(sin(θ)+cos(θ)) = (√3+1)/2√3 1 + cot(θ) = 2√3/(√3+1) cot(θ) = 2√3/(√3+1) - 1 cot(θ) = (2√3-(√3+1))/(√3+1) cot(θ) = (√3-1)/(√3+1) tan(θ) = (√3+1)/(√3-1) tan(θ) = ((√3+1)/2√2)/(√3-1)/2√2) tan(θ) = sin(75°)/cos(75°) = tan(75°) θ = 75°

@danmimis4576 4 ай бұрын

I dropped the perpendicular AE from A on BC and I worked on the 2 right triangles: one isosceles (AEB) and the other 60 and 30 degrees (AED). Did you notice that the angles are in an arithmetic progression: 45 -- 60 -- 75?

@florianbuerzle2703 4 ай бұрын

Trig solution 😉 Drop the altitude from A onto BC, call the intersection point E. Then AE = h, DE = x and CE = 2 - x. So tan(60º) = √3 = h/x tan(45º) = 1 = h / (1 + x) tan(𝜃) = h/(2 -x) From the first 2 equations, we get x = 1 / (√3 - 1) and h = (3 + √3) / 2 and finally tan(𝜃) = h/(2 -x) = (3 + √3) / (3 - √3) = 2 + √3 and therefore 𝜃 = 75º.

@jimlocke9320 4 ай бұрын

You can also recognize that the long side, AE = h in this case, of a 15°-75°-90° right triangle is (2 + √3) times as long as the short side, CE or 2 - x in this case, so ΔACE is a 15°-75°-90° right triangle and the long side is opposite the 75° angle, making Θ = 75°. The 15°-75°-90° right triangle shows up quite often in geometry problems, making it a good idea to keep notes of its side ratios and other properties handy, or remember for closed book exams.

@florianbuerzle2703 4 ай бұрын

@@jimlocke9320 Great observation 🙂 I know the 15º-75º-90º right triangle, but failed to observe that ∆ACE is one. However, as I'm on the other side of the exam process, I won't be taking one any time soon 😅

@jimlocke9320 4 ай бұрын

@@florianbuerzle2703 I'm well on the other side of the exam process, also, but I enjoy doing these problems, anyway! The ratio of sides for a 15º-75º-90º right triangle short : long : hypotenuse is (√3 - 1):(√3 + 1) : 2√2. The ratio long side to short side is (√3 + 1)/(√3 - 1) which equals (2 + √3). You got the ratio (3 + √3)/(3 - √3). Divide both numerator and denominator by √3 and the result is (√3 + 1)/(√3 - 1). Furthermore, it is useful to know ratio short : long can be expressed as 1:(2 + √3), as you did, and long : short as 1:(2 - √3). The area is 1/8 the square of the hypotenuse. I suggest including these properties of the 15º-75º-90º right triangle in your notes.

@AmirgabYT2185 4 ай бұрын

75°

@RAG981 4 ай бұрын

BE = 2x 1cos30= rt3. No need for cosine rule, and actually we know it is isos already.( 6 min mark)

@giuseppemalaguti435 4 ай бұрын

2/sin(60+θ)=t/sinθ...1/sin15=t/sin45...risulta ctgθ=(2sin15/sin45-cos60)/sin60...θ=75

@duantran-bk4mv 4 ай бұрын

Góc (ACB) = 75°

@заводмихельсона 4 ай бұрын

Ни единого отрезка считань не надо.Просто все треугольники равнобедренные по углам при основании.

@RealQinnMalloryu4 4 ай бұрын

{60°A+60°B+60°C}= 180°ABC (1)^2=1(2)2=4 {1+4}= 5 180°ABC/5 = 30ABC 5^6 5^1^3^2 1^1^3^2 3^2 (ABC ➖ 3ABC+2).

@professorrogeriocesar 4 ай бұрын

Fiz usando a Lei dos Senos. Deu trabalho.

@michaeldoerr5810 4 ай бұрын

The answer is 45 degrees. I have noticed that this is similar to the other problem from Canada, from the day before yesterday. Where if there is a 60° angle and no other angle subtended, your make the 30-60-90 conteiction and after that the exterior triangle theorem is used to justifty that the triangle ABD has the third angle 120°. Then you use the Law of Cosines. This just to make sure and that is why you need the Law of Cosines. To make sure that there are no similar triangles. I hope that I have noticed a similarity. And I shall use that as practice!!!