Chinese Math Olympiad Problem | A Very Nice Geometry Challenge

[@davide5420]

I think there is little mistake at minute 8:30. (2√2r)² is 8r and not 8r². But in the end the solution is the same. A lucky error 😂

[@MathBooster]

Yes, sorry for the mistake. We will get 16r = 16 , so r will be equal to 1.

[@MYldrm]

U r obsuletely right bro! Congratulations for your attention👏

[@jimlocke9320]

At 10:35, Math Booster has determined that r = 1, so ∆MNO has side MN = 1, side NO = 2√2 and hypotenuse MO = 3. Later in the video, the intersection of AB and XO is labelled E. ∆MNO and ∆OEA are similar by angle-angle (

[@andydaniels6363]

This is quite a bit simpler if we invert relative to the large (semi) circle: the image of the medium-sized circle is a horizontal line through the point called X in the video and the extension of diameter PQ maps to itself. Inversion preserves incidence relationships and tangency, so the images of the two smaller circles will be tangent to these parallel lines and externally tangent to the large semicircle at A and B, respectively. Our problem is reduced to finding these externally tangent circles, which is much simpler. These circles clearly have radius 2 and their centers are halfway between the two lines. Also, their centers must be at a distance of 4+2=6 from O. Calling the horizontal distance from O to one of these centers d, we have d^2+2^2=6^2, i.e., d=4√2. Using similar triangles, the distance EB is then 4√2*4/6 and AB is twice that, or (16/3)√2.

[@HeywoodUmanoff]

I'm not sure what and how you are inverting. Do you mean to rotate the 3 circles, but not the semi-circle, so that they end up back in the plane but above the semicircle (such that PQ is not moved)? And how does the inversion reduce the problem? And why do you say the smaller circles have radius = 2?

[@andydaniels6363]

@@HeywoodUmanoff Look up “circular inversion” or “inversive geometry.” The key thing here is that the image of the medium-sized circle under inversion w/r to the large one is a line tangent to the large (semi)circle and parallel to the given diameter. So, instead of having to find a circle tangent to two others and a line, the problem is transformed into that of finding a circle tangent to two parallel lines and a circle, which is much easier. The radius of such a circle should be obvious from inspection.

[@HeywoodUmanoff]

Thank you for the elegant solution! How do we know that O, M and A are colinear? Also, for OM^2=MN^2+ON^2, 16-8r=8r (not 8r^2). But r still =1.

[@zsboya]

If we have the result r = 1 (11:05), we can find the solution in other ways: A.) 2 right triangles with angular functions: 1. MOD triangle: cos(MOD) = 1/3 2. AOE triangle: sin(AOE)^2 = 1 - 1/9 = 8/9 and sin(AOE) = AE / 4 = 2*2^0.5 / 3 -> AE = 8/3 * 2^0.5 -> AB = 2*AE = 16/3 * 2^0.5 B.) With the help of a cosinus theorem (AOB): cos(AOE) = cos(MOD) = 1/3 -> cos(AOB) = cos(2*AOE) = 2*cos(AOE)^2 - 1 = 2/9 - 1 = -7/9 -> AB^2 = 2*4^2 - 2*4^2*cos(AOB) = 2*4^2 * (1 + 7/9) = 2*4^2 * 16/9 -> AB = 16/3 * 2^0.5 C.) Using ratio pairs: YE : (AY/2) = YC : MY = 2/r = 2 -> YE = AY -> AB = 4*YE ∆NMO and ∆CEY are similar, because AB = 16/3 * 2^0.5

[@soroushneyestani]

I am just a programmer and i don't have any knowledge of math olympiad, but I think every time there's a fast trick if you use it, it will be solved just in 10 seconds I have made this formula, for fast and easy to calculate questions like this : Before starting you must know the circle units for cos 45 degree for half 30 degree for lower than half 60 degree for higher than half Formula: Circle Diameter multiplied 2 multiplied cos of degree (upside down only numbers) 8*2(radical 2 divide 3) You can change positions and values, the answers is same If you can find any error i can fix it.

[@ritwikgupta3655]

Interesting to note that the sum of diameters of the 3 inner circles equals the dia of the semicircle.

[@harikatragadda]

Draw a horizontal line MT perpendicular to OC. If K is the midpoint of chord AY, and L is the midpoint of the middle circle chord, then ∆ALO, ∆MKY, ∆ONM and ∆MTC are Similar. AL/AO = YL/YC AL/4 = YL/2 AL = 2YL Hence if KY = x, then AL = 4x and YL = 2x Also, KY/MY = CL/YC x/r = 2x/2 r = 1 In the Right Triangle ONM, MN² + NO² = MO² MO = 4-2r = 2 MN = r = 1 NO = 3x This gives x = ⅔√2 AB = 8x = (16/3)√2

[@zsboya]

Hello! You've made a typo here: "Draw a horizontal line MN perpendicular to OC." - But MN and OC are parallel! Or is it to be understood that point D in the video is point N for you, i.e. MN for you is perpendicular to the MN section in the video? But I don't understand this at all: "KY/MC = CL/YC" -> YC/MC = CL/KY ?! ∆CLY and ∆MKY, ∆MNC are similar -> CL : YC = KM : YM = CN : MC or CY : MC = CL : CN -> CN = KY , why ?! And how does it follow that if KY = x, then YL = 2x, so AY = YL ? And here: "MO = 4-2r" ?! - The MO = 4 - r = 3 ! MN^2 = (3x)^2 = M0^2 - NO^2 = 3^2 - 1^2 = 8 -> x = (8/9)^0.5 = 2/3*2^0.5

[@harikatragadda]

@@zsboya Corrected, thank you!

[@tunneloflight]

Ah! The key is in recognizing that OMA are colinear. That arises from a line from the contact point through their centers must form a right angle. ergo the line AM ans AO must coincide, making AMO a straight line. Or, as others noted, the general rule, used already with exterior contact relying on CYM being linear, is that there is a linear connection between the centers of teo circles and there tamgent point of contact - whether they are in exterior or interior contact.

[@spafon7799]

This just looks wrong to me. First of all, I don't believe point M lies on OA. Therefore OM+MA is not equal to OA. Secondly, I don't believe AB goes through point y. Therefore the use of Pythagorean on triangle AEy appears to me to be incorrect. The way the problem is shown it appears to me that line AB is defined to run between the intersection of the small circle on the left and the large semicircle, and the other small circle on the right and the large semicircle. There is no justification to assume it also runs through the tangent points between the small and medium sized circles. OK, after reading some of the comments I see that indeed M does lie on OA. I still question whether AB runs through point y.

[@User-jr7vf]

I would answer "yes" to both of your questions, the reason being that I solved the problem by a different method after 3 hours of trial and error, and I got the same answer as "MathBooster".

[@ritwikgupta3655]

Y is the point of contact of small circle and large circle AB is defined to run through that point. As for your other question, already answered in the presentation.

[@RAG981]

Quite an involved problem, with many things to work out on the way. Pity about your error at 8:33, but luckily you got the correct answer anyway!

[@mariopopesco]

Why do you assume that A, B and Y are colinear ?

[@shalinichawla2198]

First straight line de rkhi hai ab

[@papomanronpecabesaospina8913]

r=2

[@papomanronpecabesaospina8913]

Hay un error al elevar al cuadrado

[@Zine-EddineHachlafi]

Yes it's 8r and r=1. it's no thing Thank you very much for all the work

[@zg0885]

How do you know? point(M) is on line(OA)

[@zg0885]

line(OM)+line(MA) may be not straight line (OA)

[@MathBooster]

If you join the centre of 2 touching circle (internally or externally) they it will pass through common point of contact.

[@jarikosonen4079]

Or how to show point y is on line AB.. if its needed.

[@howardaltman7212]

Nice, as this proves that segments OA and MA are both perpendicular to the same tangent line at the point of tangency on semicircle O, which forces them to be colinear.@@MathBooster

[@zg0885]

What will happen? If the line (OA) does not pass through the center of the circle(M)

[@ivanhuertas5307]

Why you assume that OMA are colinear???

[@mariopopesco]

Two tangent circles have colinear centers

[@eromascanu]

your solution assumes points A, y, E colinear

[@ДжорджЛукавич]

При возведении в квадрат на 8:26 ошибка, но результат всё-равно правильный😂

[@Stanislaw2344]

Velice zajímavý příklad.

[@aboubakrboubker-qg7sy]

In around mn 7 who guarantee that OA will pass through M center of small circle...???!!!!

[@MathBooster]

If you join centre of 2 touching circle (intenally or externally touching), then it will pass through common point of contact. Hence OM will pass through A.

[@howardaltman7212]

It's amazing to me that AY = (1/4)AB

[@giuseppemalaguti435]

Io ho usate le equazioni delle 3 circonferenze,ponendo la condizione di tangenza(b^2-4ac=0)...non ho trovato di meglio,ah ah... comunque,risulta AB=16/3√2

[@Touijarsoulaimane]

Why o is the middle of pq

[@papomanronpecabesaospina8913]

Perdón, r=1, pero el se equivoco al elevar la última cantidad al 😮cuadrado, 2raiz de (dos por r), al cuadrado = 8r, 8r más 8r = 16r=16, donde r =1

[@motogee3796]

Too easy to be olympiad problem

[@papomanronpecabesaospina8913]

No existe ecuación de segundo grado

[@lwels49]

It is over not by