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Thailand Math Olympiad Problem | Best Math Olympiad Problems | Geometry Problem
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- Опубликовано: 29 дек 2023
- Thailand Math Olympiad Problem | Best Math Olympiad Problems | Geometry Problem
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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
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)
Very nice
Excellent!
Without using trigonometry!!! Super❤🎉
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
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.
If we draw the figure correctly, ∠OBC is between 78° and 79°.
Scale of pic is gonna be way off √50 > 6. It's gonna make 1 look more than twice as long as 5.
thats why in geometry problems you never take anything to scale or true size
@@iswearillchangemynamesoon or that it is labeled properly.
@@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
@@iswearillchangemynamesoon purposely out of scale is like labeling 3 before 2 on the x axis ... or a triangle with sides 9-10-20
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
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)
Con Carnot di poteva fare in 4 passaggi manco ahha
Great stuff - amazing works, wonderful!!❤
Draw a circle with line segment AC as its diameter (which also passes through point B).
You can solve it more easily.
Could you please explain further? I can't get it, thanks😅
❤🎉
Si trazamos op perpendicular a bc, op es igual a raiz de 50 menos 2, angulo en c 46 grados y X= 5.20
Why AD = DC
WTF? how can the radius be shorter than BC? If BC is 6, the radius is 7.07
His 45° angle looks about 15° and one 5 is about 30% of the other 5.
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
can you explain why BAP is a right angle triangle?
@@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.
@@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