Chinese Math Olympiad Problem | A Very Nice Geometry Challenge

[@RAG981]

CD=5tan(theta) = 1tan(4theta).I decided to go from there. Use tan (4theta) =2tan (2theta)/ (1-tan^2(2theta)), and then tan(2theta)=2tan(theta)/(1-tan^2(theta)), You get 5tan^4(theta) - 26tan^2(theta) +1 =0, which gives tan^2(theta)=(13+/-2rt(41))/5. The + sign in this would make theta too big (79deg) so I continued with the - sign. Now x^2=CD^2+1, so CD= 5tan(theta) = rt(x^2-1). Thus x^2 -1 = 5*(13-2rt(41)). Thus x^2 = 66 - 10rt41 =(rt41 - 5)^2, and then x=rt41-5.

[@ludmilaivanova1603]

Tan theta is CD/AD=CD/5. Tan 4 theta=CD/1, meaning Sin 4theta=Cos4Theta, meaning 4theta=45 degrees. CD=1, DB=1. X= sqrt 2 (1.4)

[@skwest]

Very clever. I used brute force to solve with the formula for tan(4•theta). It's a lengthy solution, based on the initial Pythagorean result that segment CD = sqrt(x² - 1). This yields the two values: tan(theta) = (sqrt(x² - 1))/5, and tan(4•theta) = (sqrt(x² - 1))/1 These can be plugged in to the formula for tan(4•theta), and eventually solved for x = 1.403. Thanks! Nice challenge.

[@lasalleman6792]

I'm getting 1.3054 for x . I ran it through triangle calculator app. And I get a right triangle CDB 50 - 90 - 40 , side CD = .8391 with side DB = 1 and x = 1.3054. Yet the answer here is 1.4031. Small discrepancy I suppose.

[@ConradCorzen]

You are right, according to the drawing, Cosinus 40 = 1/x, so: x = 1/cos40 = 1/0,766 = 1,3054. But the answer is [(root of)41] -5 = 6,403-5 = 1,4031... That is why I don't believe the "problem" is from Chinese math olympiad

[@wolfler_vii]

It’s not 40 degrees it’s 4*theta. I made the same mistake.

[@ConradCorzen]

Why it's not 40 degrees? the summary of angles of every triangel is 180 deg so 180 degrees - 90degrees (right angle) - 50 degrees = 40 degrees

[@wolfler_vii]

@@ConradCorzen it’s 4θ (theta) not 40.

[@muthukumarasamymuthukumara3833]

How you know angle ACE is equal to Theda? E is the middle point of AD. F is the middle of ED. Please clarify

[@muthukumarasamymuthukumara3833]

If it's middle ED=2.5 and also FD=1.25.

[@MathBooster]

We have drawn CE in such a way that angle ACE is θ ( it is our construction). E is not midpoint of AD and F is not midpoint of ED.

[@nantesloire]

Das ist nur eine Annahme von ihm ... das heißt, wenn wir annehmen, dass ACE gleich Teta ist , dann ergibt sich das , was er berechnet hat ich finde korrekt.

[@nantesloire]

Man kann in der Mathematik manchmal durch bloße Annahme zu reale Ergebnisse kommen .

[@howardaltman7212]

A question: At 3:20, how is it known that point F is between points E and D rather than between points D and B?

[@MathBooster]

Because angle CFB is 4θ, and from the figure 4θ is less than 90°. That's why F should be between E and D.

[@howardaltman7212]

Thank you. I also got x=6 as a solution too, but that requires angle B obtuse.@@MathBooster

[@ashugarg5475]

Is it your assumption that angle ACB is greater than 3theta? because, if theta is 25, then CBA is 100 and ACB is 55, from which you can't cut 3*25 =75

[@shatranj]

If you see 4 times theta is an acute angle in the right angle triangle, so theta

[@pedroangarita3412]

Very nice!👍, cheers from Colombia

[@MathBooster]

Thank you very much!

[@yakupbuyankara5903]

X=41^(1/2)-5.

[@holyshit922]

cos(4theta) = 8cos^4(theta) - 8cos^2(theta)+1 sin(4theta) = (8cos^4(theta)-4cos^2(theta))tan(theta) tan(4theta) = (8cos^4(theta)-4cos^2(theta))tan(theta)/(8cos^4(theta) - 8cos^2(theta)+1) tan(4theta) = (8-4/cos^2(theta))tan(theta)(8-8/cos^2(theta)+1/cos^4(theta)) tan(4theta) = (8-4(1+tan^2(theta)))tan(theta)/(8-(8+8tan^2(theta))+(1+2tan^2(theta)+tan^4(theta))) tan(4theta) = (4-4tan^2(theta))tan(theta)/(1-6tan^2(theta)+tan^4(theta)) Here I used Chebyshev U and Chebyshev T polynomials to expand tan(4theta) T_{n}(x) = c_{n}(x^{n}+\sum\limits_{k=1}^{\lfloor\frac{n}{2} floor}\frac{(-1)^k}{2^{n}}\cdot\frac{n}{n-k}\cdot {n-k \choose k}\cdot (2x)^{n-2k}) where c_{n}(1 + \sum\limits_{k=1}^{\lfloor\frac{n}{2} floor}\frac{(-1)^k}{4^{k}}\cdot\frac{n}{n-k}\cdot {n-k \choose k}) = 1 U_{n}(x) = 1/(n+1) \frac{d}{dx} T_{n+1}(x) And if we know Legendre polynomials Chebyshev polynomials can be found as follows \sum\limits_{k=0}^{n}P_{k}(x)P_{n-k}(x) = U_{n}(x) \sum\limits_{k=0}^{n}P_{k}(x)T_{n-k}(x) = (n+1)P_{n}(x) Once we have tan(4theta) in terms ot tan(theta) we have equation tan(theta) = h/5 tan(4theta) = h

[@anthony17mapoy46]

Hi! 12:58 - Why did you say one and not five here?

[@MathBooster]

By mistake. Sorry for it.

[@anthony17mapoy46]

@@MathBooster Thank you!

[@marcelomatos2584]

tanθ=CD/5 and tan4θ=CD ⇒︎ tan4θ=5tanθ →︎ 2tan2θ/(1-tan²2θ)=5tanθ →︎ 4tanθ/(1-tan²θ) / (1-4tan²θ/(1-tan²θ)² →︎ 4(1-tan²θ)/(1-6tan²θ+tan⁴︎θ) =5 →︎ 5tan⁴︎θ-26tan²θ+1) ∴︎ tan²θ=(13±2√︎41)/5 →︎ tanθ=(13+2√︎41)/5 ⇒ θ≈︎66° (Reject) or tanθ=(13-2√︎41)/5 ⇒ θ≈︎11°. Recall tanθ=CD/5 ⇒︎ CD²=25tan²θ. In the ◣︎BCD x²=CD²+1² →︎ x²=25tan²θ+1 →︎ x²=5(13-2√︎41)+1 →︎ x=√︎(66-10√︎41) →︎ x=√︎(√︎41-5)² →︎ x=√︎41-5

[@abdulnasarinjikkal9788]

How you get √164/2 =√41

[@purnajinananandaavadhuta8605]

,√164=√4*√41=2*√41

[@harrymatabal8448]

Xcos 4thetha😂

[@zdrastvutye]

the drawing is not accurate: 10 l1=5:l2=1:sw=l1/100:dim x(2),y(2):wt=sw:@zoom%=@zoom%*1.4:goto 50 20 dgu1=tan(wt):dgu2=l2/l1*tan(4*wt):dg=dgu1-dgu2:return 50 gosub 20 60 wt1=wt:dg1=dg:wt=wt+sw:if wt>pi/2 then stop 65 wt2=wt:gosub 20:if dg1*dg>0 then 60 70 wt=(wt1+wt2)/2:gosub 20:if dg1*dg>0 then wt1=wt else wt2=wt 80 if abs(dg)>1E-10 then 70 90 print "der gesuchte winkel ist=";deg(wt);"°":x(0)=0:y(0)=0:x(1)=l1+l2 100 y(1)=0:x(2)=l1:y(2)=l1*tan(wt):mass=1E3/(l1+l2):goto 120 110 xbu=x*mass:ybu=y*mass:return 120 xba=0:yba=0:for a=1 to 3:ia=a:if ia=3 then ia=0 130 x=x(ia):y=y(ia):gosub 110:xbn=xbu:ybn=ybu:goto 150 140 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 150 gosub 140:next a:gcol9:x=l1:y=0:gosub 110:xba=xbu:yba=ybu 150 x=x(2):y=y(2):gosub 110:xbn=xbu:ybn=ybu:gosub 140 der gesuchte winkel ist=11.1363413° > run in bbc basic sdl and hit ctrl tab to copy from the results window

[@skwest]

The drawing is never assumed to be accurate. It's the labels that are accurate.