Thailand Math Olympiad Problem | Best Math Olympiad Problems | Geometry Problem
Рет қаралды 20,957
Thailand Math Olympiad Problem | Best Math Olympiad Problems | Geometry Problem
MY OTHER CHANNELS
••••••••••••••••••••••••••••••••
Calculus Booster : www.youtube.com/@CalculusBooster/videos
Math Hunter : www.youtube.com/@mathshunter/videos
--------------------------------------------------------------------------------
Become a member of this channel to get access of ULTIMATE MATH COURSE
Join the channel to become a member
kzfaq.info/love/FYcq-QDZg3lHGmdFcnE3Jwjoin
@Dimaggel1981 6 ай бұрын
We can find x using the law of cosines in the triangle AOB. We need the cosBAO. First we draw the line segments AO=radius AC=√40 and the perpendicular OK to the chord AC. From the orthogonal triangle AOK we calculate cosKAO ,sinKAO From the orthogonal triangle ABC we calculate cosCAB , sinCAB Then we take the type of trigonometry cosBAO=cos(BAC-KAO)=cosBACcosKAO+sinBACsinKAO=7√50/50. FInally from the law of cosines x=√26
@MataniMath 5 ай бұрын
When we find angle BCO = 45º, we can use cosine rule in triangle BCO : x² = BC² + OC² - 2.BC.OC.cos45º x² = 36+50 - 2(6)(sqr.(50))(0.5sqr(2)) x² = 86 - 6(sqr(100)) x² = 86 - 60 = 26 x = sqr(26). Thank you for your sharing
@pwmiles56 6 ай бұрын
Drop a perpendicular from O to M on BC Set MO=a, MB=b In triangle OMB a^2 + b^2 = x^2 [1] In triangle OMC a^2 + (6-b)^2 = 50 [2] In a right triangle on hypotenuse OA (a+2)^2 + b^2 = 50 [3] Subtract [2] from [3] 4a + 4 + 12b - 36 = 0 4a + 12b = 32 a + 3b = 8 a = 8 - 3b Substitute for a in [2] (8-3b)^2 + (6-b)^2 = 50 9b^2 - 48b + 64 + b^2 - 12b + 36 = 50 10b^2 - 60b + 50 = 0 b^2 - 6b + 5 = 0 (b-1)(b-5) = 0 If b=5, a=8-15 = -7, not allowed So b=1, a=5 x^2 = 1 + 25 x = sqrt(26)
@RAG981 6 ай бұрын
Very nice
@user-yx9kr8ur5q 6 ай бұрын
Excellent!
@anandharamang3289 6 ай бұрын
Without using trigonometry!!! Super❤🎉
@patk5724 6 ай бұрын
Great stuff - amazing works, wonderful!!❤
@giuseppemalaguti435 6 ай бұрын
Da carnot abbiamo 50=4+x^2-4xcos(90+OBC) e 50=36+x^2-12xcos(OBC)...da cui risulta x=√74,x=√26(corretta)
@luiso7027 6 ай бұрын
Con Carnot di poteva fare in 4 passaggi manco ahha
@ejrupp9555 6 ай бұрын
Scale of pic is gonna be way off √50 > 6. It's gonna make 1 look more than twice as long as 5.
@iswearillchangemynamesoon 6 ай бұрын
thats why in geometry problems you never take anything to scale or true size
@ejrupp9555 6 ай бұрын
@@iswearillchangemynamesoon or that it is labeled properly.
@iswearillchangemynamesoon 6 ай бұрын
@@ejrupp9555 you really dont need to label it as all math and geometry students know that nothing should be taken to scale in such diagrams. in the first phase of the math olympiad in my country no warning was given for the third question, which was indeed a geometry problem
@ejrupp9555 6 ай бұрын
@@iswearillchangemynamesoon purposely out of scale is like labeling 3 before 2 on the x axis ... or a triangle with sides 9-10-20
@iswearillchangemynamesoon 6 ай бұрын
theres no way that will ever happen cuz thats against the laws of math and geometry. in the picture it may look bigger but it isnt, and thats ok, however if it cant be bigger then its a problem. in this case its not a problem @@ejrupp9555
@kent3563 6 ай бұрын
Draw a circle with line segment AC as its diameter (which also passes through point B). You can solve it more easily.
@jiajun1319 5 ай бұрын
Could you please explain further? I can't get it, thanks😅
@JorgeLopez-kp3rj 6 ай бұрын
Si trazamos op perpendicular a bc, op es igual a raiz de 50 menos 2, angulo en c 46 grados y X= 5.20
@benignofernando8500 17 күн бұрын
If O is the center of the circle and sq. root of 50 is the radius. The segment BC cannot be equal to 6. Because sq rt of 50 = 7.07. Otherwise the drawing is misrepresented.
@user-zf1xh4kk1g 2 сағат бұрын
If we draw the figure correctly, ∠OBC is between 78° and 79°.
@jonathanv.hoffmann3089 6 ай бұрын
❤🎉
@rainierps3238 2 ай бұрын
Why AD = DC
@lastmann1 4 ай бұрын
WTF? how can the radius be shorter than BC? If BC is 6, the radius is 7.07
@someonespadre 4 ай бұрын
His 45° angle looks about 15° and one 5 is about 30% of the other 5.
@dickroadnight 6 ай бұрын
I could not find a quick and easy solution, but this works, and gives me an excuse to use my favourite formula… Extend C0 to P on the circle In triangle ABC, AC^2 = SQRT 40, by Pythagoras in triangle ABC CP IS diameter, so CAP = 90 ACB = atn(2/6) = BAP (right-angle triangles) CosBAP = cos BCA = 3/sqrt10 CP^2 = (2*CO)^2 = 4*50 = 200 AP^2 = 200 - 40 = 160 In triangle APB, by cosine rule 3/sqrt10 = (4+160-y^2)/4*sqrt160 BP = Sqrt116 OP = OC = radius = SQRT50 In triangles BPO and CBO, using the cos (supplement)) rule (see alternative method below) Cos BOP = - Cos COB (x^2 + 50 -BP^2)/2*x*50 = (6^2 - 50 -x^2)/2*x*50 2x^2 = 36 - 100 + 116 x^2 = 26 Alternative method: In triangle BAP, by sin rule 2/sinBPA = BP/sinBAP Angle APB = 3.366 degrees In 90 triangle APC APC = atn(sqrt40/sqrt160) = 26.565 degrees BPO = APC - APB = 23.2 degrees In triangle BOP, by cosine rule Cos BPO = (y^2 +50 -x^2)/2*sqrt50*BP x^2 = 26
@frenchfries-xl3mp 6 ай бұрын
can you explain why BAP is a right angle triangle?
@dickroadnight 6 ай бұрын
@@frenchfries-xl3mpHi, thanks for the feedback. I cannot see that I stated that BAP is a right-angle. CAP is a right-angle as COP is the diameter of the circle, and any angle in a semi-circle (from the diameter of a circle to it’s circumference) is a right-angle.
@frenchfries-xl3mp 6 ай бұрын
@@dickroadnight ah i see, "ACB = atn(2/6) = BAP (right-angle triangles)" I was confused about this part and read BAP as a right-angle triangle, sorry