Sweden Math Olympiad | A Very Nice Geometry Problem

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

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Sweden Math Olympiad | A Very Nice Geometry Problem | 2 Different Methods
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@jimlocke9320 4 ай бұрын

In method #2, also drop a perpendicular from D to BC and label the intersection as point F. ΔCDF is a 30°-60°-90° special right triangle. By ratios of sides, DF = x/2 and CF = x(√3)/2. ΔCAE is also a 30°-60°-90° special right triangle. By ratios of sides, AE = x and CE = x(√3). ΔBDF is a 15°-75°-90° right triangle, so has a short side : long side ratio of 1:(2 + √3). So BF is (2 + √3) times as long as DF, or (x/2)(2 + √3). BE = BC - CE = BF + CF - CE = x(2 + √3)/2 + x(√3)/2 - x(√3) = x. So BE = AE = x and ΔABE is an isosceles right triangle,

@Irtsak 4 ай бұрын

*Same solution as Math Booster did (without using Trigonometry)* In orthogonal triangle AEC : ED is median => *ED=AD=DC=x* (1) In orthogonal triangle AEC : AE=AC/2=2x/2=x => *AE=x* (2) (1),(2)=> AE=ED=AD=x => *triangle AED is equilateral* . Hence

@devondevon4366 4 ай бұрын

30 Angle D = 135 (180 - (15 + 30) Draw a perpendicular line from line BC to D to form a 30-60-90 right triangle, CDP (with P being a new point). This also forms triangle BDP ) Since DC =5 (the hypotenuse ) , then DP = 2.5 since in a 30-60-90 right triangle the shortest leg is one-half the hypotenuse Angle D in triangle BD P =75 ( as 75 + 60 = 135 , recall angle D=135) But angle D is triangle ADB = 45 degrees (since 45 + 75 + 60 =180 degree , a straight line) Hence, triangle BDP angles are 75, 15, and 90 Notice that angle 90 faces line BD in triangle ABD (to get back to this later) Using the Law of Sines, let's calculate line BD using angles 75 degrees, 90 degrees, and length 2.5. This yields 9.659. Hence, in triangle, ADB the two of the sides are 5, 9.659, and one of the angles is 45 degrees (note that since DC=5, then AD =5 as they are equal) Using the Law of Cosine for triangle ADB yields 105 degrees, 30 degrees, and 45 degrees. Hence theta =30 degrees Answer

@Irtsak 4 ай бұрын

*Another solution with inscribed and concentric angles* Consider the circle with diameter AC. Extend BD and let M,N are the points which BD intersects the circle. arc(MK)=15° . Hence *BK=DK* (1) arc(AK)=arc(AM)+arc(MK)=45°+15°=60°. Triangle AKD isosceles (AD=DK=R) and ΔADK is equilateral => *AK=DK* (2) (1),(2) => AK=BK => ΔABK is orthogonal and isosceles and at last θ=30° As Math Booster estimated.

@guyhoghton399 4 ай бұрын

_∠ADB = 45°_ (external to _ΔBCD)._ Draw _DE_ to meet _BC_ at _E_ such that _|DE| = |DC|._ _ΔCDE_ is isosceles, ∴ _∠CED = 30°_ and _∠CDE = 120°._ ∴ _∠BDE = 180° - (120° + 45°)= 15°._ ∴ _ΔBED_ is isosceles. ∴ _|BE| = |DE|._ _∠ADE = 15° + 45° = 60°_ and is the apex angle of isosceles _ΔADE._ ∴ _ΔADE_ is equilateral. ∴ _|AE| = |DE|_ ∴ _|AE| = |BE|_ and _∠AED = 60° ⇒ ∠AEB = 180° - (60° + 30°) = 90°._ ∴ _ΔAEB_ is a right angled isosceles triangle. ∴ _∠ABE = 45°_ i e. _θ + 15° = 45°_ ⇒ *_θ = 30°_*

@MarieAnne. 2 ай бұрын

Here is my solution, which has some similarities to 2nd method, except I used law of sines to determine that △ABE is an isosceles right triangle. Let AD = CD = x. ∠BDC = 180° − 15° − 30° = 135° Using law of sines in △BCD we get: BC/sin(135°) = x/sin(15°) BC = x sin(135°)/sin(15°) = x (1/√2) / ((√3−1)/(2√2))= 2x/(√3−1) *BC = (√3+1)x* Drop perpendicular from A to BC at point E. △ACE is a 30°-60°-90° triangle. So ratio of sides is: AE : CE : AC = 1 : √3 : 2 AC = 2x → *AE = x, CE = √3 x* In △ABE, we have AE = x BE = BC − CE = (√3+1)x − √3 x = x Since ∠AEB = 90° and AE = BE = x, then ∠ABE = ∠BAE = 45° θ + 15° = ∠ABE = 45° *θ = 30°*

@michaeldoerr5810 4 ай бұрын

It might just be me but I think that BOTH methods make sense. That is probably because I am catching up on most vids. And I am going to make use of that as practice. So that it becomes easy for me!