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

[@StilHOT1970]

This is the shortest way: Just in less than 15 seconds. Draw another equilateral triangle CDE: Side CD=side DE=side CE=c. Angle CDE=60 deg. (equilateral triangle). Side BE=a. Opposite to BE=90 deg. side BE=a is the longest side and BDE is a right triangle with BD=b, DE=c, and BE(hypotenuse)=a. Therefore, theta = 60 deg+90 deg=150 deg. Or refer to VITAMATH 94 in my YT channel.

[@montynorth3009]

Let line AD be vertical, bisecting 60 degree angle BAC into 2 x 30. Then b will equal c. Let b & c = 1. Then a^2 = 1^2 + 1^2. as the requirement. a^2 = 2. a = sq.rt. of 2. In new triangle ABD. b / sin 30 = a / sin ABD. 1 / 0.5 = sq.rt.2 / sin ABD. Sin ABD = 0.5 x sq.rt.2 = 0.7071. Angle ABD = 45 degrees. By external angle formula. Theta / 2 (ref. extended line AD) = 30 + 45 = 75. So Theta = 2 x 75 = 150.( New triangles ABD & ADC congruent)

[@幕天席地-w9c]

Constructing equilateral triangle BDE and then connecting EC is equivalent to rotating triangle BDA clockwise 60 degree with B as pivot. So BEC must be congruent to BDA

[@josleurs4345]

yes , and then it is clear , the nice idea is is the conjection that EC is a , and this inside or idea is the solution ... in all the details one can lose the very good idea from the beginning... anyway I find it also to slow as it is explained and one can lose the idea from the beginnig but the author wants also to be educational ... the very good idea from the beginning is nice ... may it would be nice why the author got that idea ...what did inspire him... because the rest is just some easy details ... but the first idea is very good ...

[@josleurs4345]

may by the idea for the first solution came from the second solution ... I mean the second solution was actually the first ...

[@georgexomeritakis2793]

There is faster way to solve this. Suppose E is area of large equilateral triangle, Ea, Eb, Ec areas of equilateral triangles with sides a, b, c and Eabc area of triangle with sides a, b , c, then 2E = Ea + Eb + Ec + 3Eabc. This is because E = Eab + Ebc + Eac, the sum of areas of the 3 inner triangles. Now if you combine Eab with Ebc, it is sum of Eb and Eabc and so on. Using this principle and the fact Eabc is right triangle I found that L^2 = b^2 + c^2 + √3bc, where L is the side of big equilateral triangle. That means cos θ = -√3/2 or θ = 150°. Remember area of equilateral triangle with side x is √3 x^2 /4

[@Sam10099]

How did you find out that L^2=… If you only did math with the areas of triangles. Also, you are right about the formulas, but how did you find out 2E=…? I saw the idea behind calculations, but the total picture is unclear for me

[@vkr122]

1:22 DBE=60 explain please!

[@MathBooster]

Because DBE is equilateral triangle.

[@vinitkumarpriyadarshi3924]

Superb and Fantastic ❤❤❤❤❤❤❤🎉🎉🎉🎉🎉Sir

[@saddatsabanature4307]

thanks sir

[@zdrastvutye]

i have found a different result but i don't know why: 10 ls=1:dim x(2),y(2):sw=ls/1E2::goto 120 20 dis=lb^2-ys^2:if dis0 then ys1=ys else ys2=ys 100 if abs(dg)>1E-10 then 90 110 return 120 lb=sw 130 gosub 60:if disls then stop 160 goto 130 170 print dis,dis2 180 lb1=lb:dg1=dg:lb=lb+sw:if lb>ls then stop 190 lb2=lb:gosub 60:if dg1*dg>0 then lb1=lb else lb2=lb 200 lb=(lb1+lb2)/2:gosub 60:if dg1*dg>0 then lb1=lb else lb2=lb 210 if abs(dg)>1E-10 then 200 220 print xs,ys:x(0)=0:y(0)=0:x(1)=ls:y(1)=0:x(2)=ls/2:y(2)=ls*sqr(3)/2 230 mass=1E3/ls:goto 250 240 xbu=x*mass:ybu=y*mass:return 250 xba=0:yba=0:gcol8:for a=1 to 3:ia=a:if ia=3 then ia=0 260 x=x(ia):y=y(ia):gosub 240:xbn=xbu:ybn=ybu:gosub 270:goto 280 270 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 280 nexta:x=xs:y=ys:gosub 240:xba=xbu:yba=ybu:for a=0 to 2: 290 x=x(a):y=y(a):col=8:gcol col:if a=2 then col=7 300 gcolcol:gosub 240:line xba,yba,xbu,ybu:next a 310 zx=(x(0)-xs)*(x(1)-xs):zy=(y(0)-ys)*(y(1)-ys) 320 nxy=sqr((x(0)-xs)^2+(y(0)-ys)^2)*sqr((x(1)-xs)^2+(y(1)-ys)^2) 330 w=acs((zx+zy)/nxy):print "der gesuchte winkel=";deg(w) 340 a$=inkey$(0):if a$="" then 340 else cls 350 lb=lb+sw:goto 130 0.00575145227 0.84832334 0.0839942195 0.0542675879 der gesuchte winkel=143.743564 run in bbc basic sdl and hit ctrl tab to copy from the results window

[@rohitmadashri7250]

Yo man, this surely deserves a PhD.