Japan Math Olympiad | You should know this trick! | A Nice Geometry Problem

Japan Math Olympiad | A Very Nice Geometry Problem
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[@spacer999]

The square can be divided into 5 triangles the total area of all must add to 16. Area of ADE+AEO+ECO+BCO+ABO = 16 6 + 5r/2 + r/2 + 2r + 2(4-r) = 16 3r + 14 = 16 r = 2/3

[@ferdinandrius6063]

BCO+ABO=ACB=8 (aires)

[@LongZhao-bz9te]

Smart

[@jimlocke9320]

Think outside the box! As done later in the video, label the points of tangency with AE as point M and with BC as point P. The circle's center is point O. Extend AE and BC until they meet and label the intersection as point R. Note that ΔADE and and ΔABR are similar by angle - angle (

[@Irtsak]

Nice solution . My first solution was similar with Math Booster’s Another solution . Use Math booster’s shape in 12:53. *Notice that OM=OQ=r , so OE is bisector of angle AEC.* Bisector theorem in triangle AEC => OA/OC =AE/EC=5/1=5 => 0A/OC=5 (1) OQ//AQ => OA/OC =DQ/QC Thalli theorem => OA/OC=(4-r)/r (2) (1),(2)=> (4-r)/r=5 =>5r=4-r => r=2/3 finish

[@santiagoarosam430]

Lados del triángulo rectángulo de la izquierda 3/4/5. Diagonal del cuadrado =4√2; lado inferior =3+(1-r)+r. Potencia del vértice superior izquierdo del cuadrado respecto a la circunferencia =(4√2-r-r√2)(4√2--r√2+r)=(5-1+r)^2 ; r=2/3. Gracias y un saludo cordial.

[@MarieAnne.]

Using coordinates, we get: A = (0,4), B = (4,4), C = (4,0), D = (0,0), E = (3,0), Center of circle is (with radius R) is tangent to lines BC and CD. Therefore, center O is R units to left of line x = 4 and R units above line y = 0 O = (4−R, R) Line AE has slope = −4/3, and y-intercept = 4, so it has equation: y = −4/3 x + 4 4x + 3y − 12 = 0 Distance formula between point (x₀, y₀) and line Ax + By + C = 0 d = |Ax₀ + By₀ + C| / √(A² + B²) Since circle with radius R and center O(4−R, R) is tangent to line 4x + 3y − 12 = 0, then distance from O to line = R |4(4−R) + 3(R) − 12| / √(4² + 3²) = R |4 − R| = 5R 4 − R = −5R or 4 − R = 5R 4R = −4 or 4 = 6R R = −1 or R = 2/3 Since R > 0, then *R = 2/3*

[@Irtsak]

*For those who are familiar with Analytic Geometry.* Let orthonormal system of axes Dxy and A(0,4),B(4,4) ,C(4,0),D(0,0). ABCD is square and let point E(3,0). Let λ₁, the slope of the straight line AE => λ₁=-4/3 The equation of (AE) : y=-4/3 (x-3) => 4x+3y-12=0 x’x : y=0 Let O(x,y) the center of the circle. d(0,AE)=d(0,x' x) => |4x+3y-12|/√(4²+3² )=|y| => |4x+3y-12|=|5y| => 4x+3y-12=5y or 4x+3y-12=-5y (the other bisector) (EO) : 2x-y=6 (1) (bisector of λ₁=-1 => (AC) : y=-1(x-4) =>x+y=4 (2) Solve the system of (1),(2) and you will find x=10/3 and y=2/3 So radius of circle r=y=2/3

[@realtorn1do]

I want just to say thanks for your exercises I can feel myself improved since I followed you, and I hope you continue thanks❤❤❤❤❤❤

[@SB-7423]

Let CE =b, CB = a. The equation of the line EA (С(0, 0)): aу - (a - b)x - ab = 0. Coordinates of the center of the circle: O(r, r) The distance from the point O to the line EA is equal to the radius: r = [-a∙r + (a - b)∙r + a∙b]/√[a^2 + (a - b)^2]. Hence we get a quadratic equation whose solution is : r = b∙[√[a^2 + (a - b)^2] - b]/[2∙(a - b)]. Substituting a =4, b = 1 gives : r = 1∙[√[4^2 + (4 - 1)^2] - 1]/[2∙(4 -1)] = 2/3 .

[@MrPaulc222]

I had to give this a lot of thought before deciding where to place my right triangle involving r. Before checking the video, I eventually got to this: E to the tangents is 1-r A to the tangent is 5-(1-r) so 4+r AC, which passes through the circle's centre is 4*sqrt(2), so AO (O being circle centre) is 4*sqrt(2)-(r*sqrt(2) Call the tangent near E, T. ATO is a right triangle (4+r)^2 + r^2 = (4*root2 - r*root2)^2 (I'm sure there must be a simpler way, but no matter). Let's go longhand: 16 + 8r + r^2 + r^2 = 32 - 8r - 8r + 2r^2 Start simplifying: 2r^2 + 8r + 16 = 2r^2 - 16r + 32 Again: 24r = 16 r = 16/24 = 2/3 Yes, I see we used pretty much the same method, but I skipped a few steps because I'm not teaching :) I think you did this one very well. Thank you.

[@52soccerstar]

You can extend the line AE to some point T. Then, using similarity, calculate CT with this equation: CT=> 4/(4+x) = 1/x => x is solved to be CT = 4/3. Then, you draw a line from CB to AE that touches the circle at point S(on BC) and F (on AE). Clearly, TS equals 4/3 + 2r. Using triangle similarity again, you calculate SF as follows: SF => (4/3 + 2r) / y = (4/3+4)/4 implying y = 1+3r/2 and thus SF = 1 + 3r/2. Name the point from SF onto the circle to be J. Using circle tangent theorem we see that (1+3r/2 - r) = FJ = 1+ r/2. Without going too much into detail on the (individual) points we get the equation (1+ 3r/2)^2 + (4/3 + 2r)^2 = (7/3 + 3r/2)^2. Solving and using positive r we get r = 2/3.

[@quigonkenny]

Let the center of the circle be O and the three points of tangency to circle O be M (on EA), N (on EC), and T (on CB). Triangle ∆ADE: DE² + AD² = EA² 3² + 4² = EA² EA² = 9 + 16 = 25 EA = √25 = 5 DC = AD = 4, and DE = 3, so EC = 1. As ON and OT are radii and DC and CB are tangent to circle O at N and T respectively, ∠ONC = ∠CTO = 90°. As ∠NCT = 90° as well, ∠TON must also equal 90° and ONCT is a square of radius r. As NC = r, EN = 1-r. As ME and EN are tangent to circle O and intersect at E, ME = EN. As AE = 5, AM = 5-(1-r) = 4+r. Draw OA. Draw OP, where P is the point on BA where OP is perpendicular to BA. As NP = CB = 4 and ON = r, OP = 4-r. As BA = 4 and BP = OT = r, PA = 4-r. Triangle ∆OPA: OP² + PA² = OA² (4-r)² + (4-r)² = OA² OA² = 2(4-r)² OA = √(2(4-r)²) = (4-r)√2 Triangle ∆AMO: OM² + AM² = OA² r² + (4+r)² = (2(4-r)²) r² + 16 + 8r + r² = 2(16 - 8r + r²) 8r + 16 = 32 - 16r 24r = 16 r = 16/24 = 2/3

[@AmirgabYT2185]

r=2/3

[@michaeldoerr5810]

I think that I know the trick and I think that the trick is much more simple than the " Think outside the box" comment. The trick is to construct a right angled square in the circle so that-starting with the triangle, you can iterate the Pythagorean Theorem in relation with delta. And delta is def the radius. And the further iterations decontrust a right angle into a 45 degree so that the final and third iteration is solved through algebra. I hope that it is sufficient.

[@michaeldoerr5810]

I probably clarify that the other trick is comparing the second and third right triangles which are the second and third iterations of Pythagoras Theorem. Does that make sense?

[@robertloveless4938]

I found a much easier way to find the answer..... just skip to the end of the video.

[@sergeyvinns931]

Продлим AE до пересечения с ВС в точке К, и закроем окружность катетом FM параллельным ЕС. Получим прямоугольный треугольник FKM, c прямым углом MFK, подобный треугольнику ADE. Заметим, что ЕС, является средней линией треугольника FKM, и FM = 2EC=2, 3/4=FM/FK, 3/4=2/FK, откуда FK=8/3=4R. R=2/3! Перевести на английский

[@rabotaakk-nw9nm]

С какого бодуна "ЕС является средней линией ΔFKM"?! 😡

[@sergeyvinns931]

@@rabotaakk-nw9nm Малшик, ты не знаешь геометрии, медиана. проведенная из прямого угла, делит гипотенузу пополам, и сама равна половине гипотенузы. Линия проведенная через середину гипотенузы параллельно катету АВ, в два раза меньше этого катета. АВ=6, средняя линия 3! Точно так же, линия проведенная через середину катета АВ, параллельно катету ВС, равна половине катета ВС, который равен 5+3=8, то есть 4!

[@rabotaakk-nw9nm]

​@@sergeyvinns931Да-а-а... Диагноз товарища Саахова явно подтверждается!

[@sergeyvinns931]

@@rabotaakk-nw9nm Диагноз "товарищу" Саахову поставил суд самый гуманный в мире. Даже здесь, ты пролетел, как фанера над Парижем, по которому теперь бегают крысы, клопы и украинские нацисты, а правит там Микрон - Наполеон, нетрадиционной сексуальной ориентации!