China Math Olympiad Problem | A Very Nice Geometry Challenge | 4 Different Methods

Very Nice Geometry Problem From China | 4 Different Methods
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[@ducduypham7264]

Construct a perpendicular from A to BC, this perpendicular intersect BC at M. We can easily prove that AMD is equilateral triangle. Therefore, angle MAD is 60°. Angle CAB is 15° so angle MAC is 45°. In right triangle MAC (angle M is 90°), angle MAC equal to 45° so MA=MC and angle MCA=45°. In triangle MDC, MC=MD(=MA) so MDC is isosceles triangle with base DC. In this isosceles triangle, angle DMC=30° (as MDB is isosceles triangle with base MB, angle DBM=30°) so angle MCD should be 75°. Also, angle MCA=45° then angle ACD=30°. In triangle ACD, theta=angle CDB = angle CAD + angle ACD = 15°+ 30°=45°

[@jimlocke9320]

Method 3, alternatively, avoiding trigonometry, we can recognize that ΔACE is a 15°-75°-90° right triangle, so the ratio of sides, short to long, is (√3 - 1):(√3 + 1). Therefore, CE/AE = (√3 + 1)/(√3 - 1). If we multiply by (√3 + 1)/(√3 + 1) and simplify, CE/AE = 2 + √3 and CE = x(2 + √3). ΔBCE is a 30°-60°-90° right triangle, so the ratio of sides, short to long, is 1:√3 and BE/CE = √3, making BE = x√3. Pick up at 18:20. Once we find that CE = DE = x, we can recognize that ΔCDE is an isosceles right triangle, so Θ = 45°.

[@1ciricola]

Only angular measurements are given in this problem. By assigning a number value to the two equal [base] line segments - I chose 5 - the value of angle θ can fairly easily be determined; using familiar trigonometric and geometric methods . I realize this odd approach in no way compares with mathematical rigor of the three solutions presented in the video. But if obtaining the value of angle θ is the objective, I prefer the less rigorous [cheating?😉] approach to determine θ.

[@Aravind_Vishwakarma]

Sir your videos are really good. BTW can I know which state you are from?

[@User-jr7vf]

I used the third method (before looking at the solution) but I wish I had the intelligence for solving it by method 2 or method 4 .I wonder if one day I will be like Math Booster?

[@ybodoN]

It looks simple but it's not really easy to solve 🤯
The second method is the one I like the most 🤩

[@quigonkenny]

Drop a perpendicular from C to E on AB. Let AD = DB = x. Let CE = y.
tan(30°) = CE/EB
1/√3 = y/EB
EB = √3y
tan(15°) = CE/AE
tan(45°-30°) = y/AE
(tan(45)-tan(30))/(1+tan(45)tan(30)) = y/AE
(1-(1/√3))/(1+(1/√3)) = y/AE
(√3-1)/(√3+1) = y/AE
(√3-1)AE = (√3+1)y
AE = (√3+1)y(√3+1)/(√3-1)(√3+1)
AE = y(3+2√3+1)/(3-1)
AE = (2+√3)y
√3y + (2+√3)y = 2x
2y(√3+1) = 2x
x = (√3+1)y
DE = AE - AD
DE = (2+√3)y - x
DE = (2+√3)y - (√3+1)y
DE = 2y + √3y - √3y - y
DE = y
tan(θ) = CE/DE = y/y = 1
θ = tan⁻¹(1) = 45°

[@windofchange-pg5jk]

💚🤍❤️

[@shivachaturvedhi8840]

mathematical solution
sin(alpha) = sin( 180 - (t+30)) = sin(t+30)
sin(beta) = sin(180 - (15 + (180-t)) = sin(t-15)
common = a / sin(alpha) * sin(30) = a / sin(beta) * sin(15)
sin(30)/sin(alpha) = sin(15)/sin(beta)
sin(alpha)/sin(beta)=sin(30)/sin(15) = sin(t+30) / sin(t-15)
p sin(t+30) sin(30) st c30 + ct s30
--- = --------- = --------- = ----------------
q sin(t-15) sin(15) st c15 - ct s15
p s15 + q s30 st sin(15+30) 2 s15 s30
-------------- = ----------------- = ----------------- = 1
p c15 - q c30 ct cos(30+15) sin(30-15)
tant = 1 => theta = 45 degrees.

[@shivachaturvedhi8840]

mathematical solution
sin(alpha) = sin( 180 - (t+30)) = sin(t+30)
sin(beta) = sin(180 - (15 + (180-t)) = sin(t-15)
common = a / sin(alpha) * sin(30) = a / sin(beta) * sin(15)
sin(30)/sin(alpha) = sin(15)/sin(beta)
sin(alpha)/sin(beta)=sin(30)/sin(15) = sin(t+30) / sin(t-15)
p sin(t+30) sin(30) st c30 + ct s30
--- = --------- = ------------ = ---------------------------
q sin(t-15) sin(15) st c15 - ct s15
this gives, for random m and n,
p = m s30, q = m s15, p = n st c30 + n ct s30, q = n st c15 - n ct s15
which gives below equation to keep st and ct only onside of division
p s15 + q s30 n st (c30 s15 + s30 c15) n st sin(30+15) m 2 s15 s30
----------------------- = ----------------------------------------- = ---------------------------- = ---------------------- = 1
p c15 - q c30 n ct (c30 c15 - s30 s15) n ct cos(30+15) m sin(30-15)
tant = 1 => theta = 45 degrees.