Canada Math Olympiad | A Very Nice Geometry Problem

[@xualain3129]

We can solve it only using the law of cosine for three angles - theta,angle(ABP)=a and angle(CBP)=b. Let AB=x cos(a)=(1+x^2-sqrt(7)^2)/(2*1*x)=(x^2-6)/(2* x) cos(b)=(1+x^2-3^2)/(2*1*x)=(x^2-8)/(2*x) a+b=pi/2 cos(b)=sin(a) from cos(a)^2+sin(a)^2=1 we can get an equation of x X^4-16*x^2+50=0 from which x^2=8+sqrt(14) . With x known, we can then calculate cos(theta)=(1+sqrt(7)^2-(8+sqrt(14))/(2*1*sqrt(7))=-sqrt(2)/2 Which means theta=3*pi/4 or 145 degrees.

[@hongningsuen1348]

I like solution without construction. Good construction is like a hidden entrance to an easy way but finding a good entrance is really an art.

[@xualain3129]

@@hongningsuen1348 I totally agree with you. I have difficulty finding the right construction to solve geometry problems. Thanks to trigonometry, I can solve some of them without construction at all.

[@davidellis1929]

You don't need the Law of Cosines. Once we know P'P=sqrt(2), we can see that triangle P'PA is a right triangle, since P'P^2+PA^2 = 2+7 = 9 = AP'^2. I've seen the ingenious rotation trick in other problems, sometimes creating an equilateral triangle, and it's far from obvious but works great!

[@michaeldoerr5810]

That is probably the first time that I have heard of ever rotating a triangle. And I am not surprised that the Law od Cosines had to be used. When is rotating triangle necessary may I ask? And also theta is an obtuse angle. Would the domain of cosine allow for the existence of acute angles a la Law of Cosines? I think so, but I want to check.

[@oscarcastaneda5310]

Here's a solution without the Law of Cosines using "Unfolding" : Unfold each of the the three "inner" triangles using the right triangle's sides as hinges. Once done, you'll find two new Isosceles right triangles and can find their hypotenuses. The sides of length "one" become "lined" up perfectly for a straight length of "2". Now, notice the triangle with lengths 2, root(14) and 3root(2), they're a "Pythagorean" triple. From here just add the 45 and 90 degrees for the answer : )

[@OverclockingCowboy]

I think you meant the triangle with lengths 2, root 14, and 3 root 2. Would there still be a Pythagorean Triple if AB is not equal to BC?

[@oscarcastaneda5310]

@@OverclockingCowboy Gracias, I've made the typing correction . Good Question, it would be fun to investigate this : ) Perhaps not, but looking at it in this manner still gives a triangle whose sides are known and we can calculate the requisite angle to then add 45 degrees to : )

[@OverclockingCowboy]

⁠@@oscarcastaneda5310 If you shorten segment AP, angle theta becomes bigger. So it is apparent that a Pythagorean Triple is not guaranteed even under the condition AB=BC. After unfolding, a new straight line is formed at B with length 2 BP. A new isosceles triangle is formed at A (side length = AP) with side angles = 90 degree - angle A. Thus, the isosceles side angle equals 45 degree only if angle A = 45 degree. Unfolding is still a nice concept with some unique properties.

[@oscarcastaneda5310]

@@OverclockingCowboy Gracias,... We think Alike.

[@giuseppemalaguti435]

Io ho usato il teorema dei coseno..dopo i calcoli risulta 2√7(cosθ)^3-8(cosθ)^2-√7cosθ+4=0...1 soluzione è cosθ=1/√2,ma non va bene perché θ è>90...altra soluzione è cosθ=((8-√14)-√(78+16√14))/4√7=-1/√2...θ=135..la cubica diventa (cosθ-1/√2)(cosθ+1/√2)(2√7cosθ-8)=0...quindi l'unico angolo ottuso è 135

[@اقرء]

Tanks for watching this video

[@52soccerstar]

Stewarts theorem might fail (or work)but If it's too tough don't crack it

[@Istaphobic]

Good problem.

[@murdock5537]

φ = 30°; ∎ABCD → AB = AF + BF = (a - y) + y = BC = BG + CG = x + (a - x) = CD = AD = AE + DE = x + (a - x) = a = EG = EP + GP = (a - y) + y; BP = 1; BPA = θ = ? CP = 3; AP = √7 → (a - y)^2 + x^2 + 2 = (a - x)^2 + y^2 = 9; y = √(1 - x^2) → (a^2 - 6)/2a = √(1 - x^2); x^2 = ((a^2 - 8)/2a)^2 → (a^2 - 6)^2 + (a^2 - 8)^2 = 4a^2 → z ∶= a^2 → z = 8 ± √14 → a ≥ 3 → z = 8 + √14 → z = 8 - 2(√7)cos⁡(θ) → cos⁡(θ) = -√2/2 → θ < 6φ → θ = 3φ + 3φ/2 = 9φ/2 btw: ∆ BCP → BCP = α; BC = a; BP = 1; CP = 3 → cos⁡(α) = (a/150)(57 - 4√14) → α ≈ 16,22° ∆ ABP → PAB = β; AB = a; AP = √7; BP = 1 → cos⁡(β) = (√7/350)(√(8+ √14))(49 - 3√14) → β ≈ 11,9°

[@RAG981]

Cunning