Find the angle X | Japan Math Olympiad Geometry Problem

GET MY EBOOKS
•••••••••••••••••••••••

OTHER CHAPTERS : COMING SOON.....
--------------------------------------------------------------------------------

Join the channel to become a member

[jimlocke]

At 2:15, note that <PQB is an exterior angle of ΔCPQ and equals the sum of the 2 opposite interior angles, 2Θ + 2Θ = 4Θ. <AQB = <PQB - <PQA = 4Θ - Θ = 3Θ. We use this information later in the video to reduce the number of steps. At 9:05, we find that the other two angles of ΔANQ are 90° and 60°, leaving 30° for <AQN, same angle as <AQB. So 3Θ = 30° and Θ = 10°. From <BAC, x = Θ + 60° + Θ = 80°, as Math Booster also found.

[DB-lgsq]

شكرا لكم على المجهودات
يمكن استعمال
AB=1
BQ=b
ACB=180-2x
CQP=4x-180
45<x<90
sin(180 -x)=sinx / 1+b
Sin(180-x)=sin (4x-180)

cos3x =- 1/2 ; 135<3x<270
3x=240
x=80

[michaeldoerr]

This is really similar to your previous videos. It is basically using isosceles triangles and HL congruency and this seems like the kind of procedure that you can only get good at through memorization or practice. Also the theta is 10°.

[phungcanhngo]

<AQN=30, <AQN =3theta, Hence 3theta=30 degree, 1theta-=10 degree.

[jiangmingdar]

AP+PC=CQ+BQ
⇒△ABC is an isosceles triangle
Passing P do PR = PQ intersects BC at point R
Let ∠C=a ⇒∠CPQ=a
⇒∠PQR=∠PRQ=2a
⇒∠APR=2a+a=3a
⇒∠ARP=(180°-3a)/2

⇒△ABR is an isosceles triangle
∴ AB=AR=AP=PR
⇒△APR is an equilateral triangle
∴ 3a=60° ⇒a=20°

⇒△ABC is an isosceles triangle
∴ ∠B=(180°-a)/2

[epsom]

make congruent triangles BQR (outside)
AC parallel BR
AP parallel BR, AP=BR ⇒ AB=AP
PR=AB ⇒ ΔARB is an equilateral triangle
Let ∠C=a ⇒∠PQR=2a+a=3a=60°
⇒∠PQR=∠PRQ=2a
⇒∠APR=2a+a=3a=60° ⇒a=20°
x=(180°-20°)/2=80°

[hanswust]

Very tricky problem, what a relief to get to the solution!

[alexbayan]

This problem stumped me but for those who are good with trigonometry but without Math Booster's geometric insight here is a solution.
From the right triangle with theta and X, cos X = (1/2)a /(a+b)
b=2a cos 2theta
Let t = sin theta
then cos X = t
b = 2a (2(1-t^2)-1)=a(2-4t^2)
t = (1/2)/(1+ (2-4t^2)) = (1/2)/ (3-4t^2)
3t-4t^3= 1/2
sin 3theta = 1/2
3 theta = 30 or 150 degree
but angle BPQ = 4 theta < 180 so 3 theta cannot be 150.
3 theta = 30 degree
theta =10 degree
Not very elegant but if you had a bad day and didn't have the geometric insight, this works.

[ناصريناصر-سب]

Assuming that PC=BQ=a and AP=PQ=AB=QC=b and the measure of angle ACB is α, applying Cachy's theorem in triangle ABC we find cosα=(2(a+b)²-b²)/2(a+b)², and applying it in triangle CPQ we find cosα=(a²+b²-b²)/2ab=a/2b, from which after comparison we find a³-3ab²-b³=0, and dividing by b³ we get (a/b)³-3(a/b)-1=0, assuming a/b=2cos(θ), then 8cos³θ-6cosθ-1=0, from which 4cos³θ-3cosθ=1/2, so cos(3θ)=1/2, from which 3θ=60°, so θ=20° Hence a/b=2cos20° and thus cοsα=a/2b=cos20 so α=20° and since triangle ABC is isosceles then B=A=X=80°

[RealQinnMalloryu]

{60°A+60°B+60°C}=180°ABC 3^60 3^6^10 3^6^2^5 3^2^3^2^2^3 1^1^1^1^2^3 2^3(ABC ➖ 3ABC+2).

[EfoVip]

Why didn't you just subtract 90-∅ from 180⁰-2∅ (i.e <ABC +<APQ = 180⁰ ) in order to find the value of ∅ which is 30⁰ from my calculations. Then straight forward you use X = 90 - ∅ to find the value of x which is 60⁰