Very Nice Math Olympiad Geometry Problem | 2 Methods

Very Nice Math Olympiad Geometry Problem | 2 Methods
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[@sarantis40kalaitzis48]

I found a special way of solving in only three lines,without explanations. If we complete the semicircle from the left and extend the CD intersects circle in point K and ED intersects circle in pont M. DM=DG=x by symmetry. DC=a side of square.DK=DC=a by symmetry.Aplply Intersecting Chords Theor. to point D, DE*DM=DC*DK or12*x=a*a=a^2 (1). We calculate a^2=DA^2=DE^2--AE^2=12^2--R^2. By Pythagorean Theorem to (DAE) triangle.Also a^2=DA^2=DC^2=(AC^2)/2 = (R^2)/2,because AC=R from Pythagorean TH. to triangle(DAC). Comparing equalities and having 12^2--R^2=(R^2)/2 or 3*(R^2)/2=144 or (R^2)/2=a^2=144/3 or a^2=48(2). Finally from (1) and (2) we have 12*x=48 or x=48/12 or x=4.

[@abhaypandey2452]

Bro you in which grade

[@sarantis40kalaitzis48]

@@abhaypandey2452 Thank you. Keep trying to find your own way,not the others. I'm proffessor ih Greek high school many years. But this type of advanced thinking called ''thinking out of the box'' i have learnd from Math Booster and similar channels of KZbin.

[@abhaypandey2452]

@@sarantis40kalaitzis48 thank you so much I am currently in 10th grade could you give some suggestions for thinking out of the box questions Or how to improve geometry😔😔😔

[@sarantis40kalaitzis48]

@@abhaypandey2452 Watch the solutions of these types of problems,which are from Mathematical Olympiads or from local pre-Olympiad Mathematical Competitions. Then if you have time try to read coments like mine,with other methods of solving the same problem. Some of them are very interesting and prototype. You'll improve. Be patient, have faith and know that always there is something that someone knows better that you. You can't know everything ,but the way-the journey- to knowledge is everything.

[@abhaypandey2452]

@@sarantis40kalaitzis48 thank you 😊

[@marioalb9726]

Radius R of circle, and Side S of square inside quarter of circle: R² = S²+S² R² = 2.S² S² + R² = 12² S² + 2S² = 12² S² = 12² / 3 S² = 48 cm² (Area of square) Inside the complete circle: The chord theorem states that the products of the lengths of the line segments on each chord, are equal. The intersecting chords are Chord 1 : 12 + x Chord 2 : S + S 12 . x = S² x = S² / 12 = 48/12 x = 4 cm ( Solved √ )

[@spacer999]

Let r=radius of circle, then AD=DC=r/√2. Applying pythogoras theorm on triangle ADE, one can solve r^2=2/3*(144)=96. Mirror DG and DC over AF to DG' and DC', it's apparent that G'DE and C'DC are straight line cords. By intersection cord theorem, 12X=DC^2=r^2/2. So X=r^2/24=96/24=4

[@MarieAnne.]

My solution: AF = AG = AC = AE = R (all are radii of circle) AB = BC = CD = AD = y (all are sides of square) Using Pythagorean theorem in △ABC: y² + y² = R² 2y² = R² y² = R²/2 Using Pythagorean theorem in △ADE: y² + R² = 12² R²/2 + R² = 144 3R²/2 = 144 R² = 96 -➤ R = 4√6 y² = 48 -➤ y = 4√3 Since CD || AB, then ∠AED = ∠DEC = θ In △ADE: sinθ = y/12 = (4√3)/12 = √3/3 cosθ = R/12 = (4√6)/12 = √6/3 Using law of cosines in △ADG AG² = AD² + DG² − 2(AD)(DG) cos(90+θ) R² = y² + x² − 2yx(−sinθ) 96 = 48 + x² + 2(4√3)x(√3/3) 96 = 48 + x² + 8x x² + 8x − 48 = 0 (x − 4) (x + 12) = 0 Since x > 0, then x = 4

[@marioalb9726]

DGE is a right triangle !!! Cosine theorem is not necessary Right triangle DGE: cos 2θ = x / 12 x = 12 cos 2θ x = 4 cm. ( Solved √ )

[@quigonkenny]

Draw the length r radius AC. As AC is also the diagonal of the square, BC = DC = r/√2, as are their opposite sides AB amd AD. Triangle ∆DAE: AD² + AE² = ED² (r/√2)² + r² = 12² r²/2 + r² = 144 3r²/2 = 144 r² = 144(2/3) = 96 r = √96 = 4√6 Mirror the entire diagram about AF so that it is a semicircle. New points will be labeled as prime versions of their original (G', E', etc.). As ∠GDC = ∠CDE, G, D, and E' will be colinear. Draw GE. As E and E' are opposite ends of a diameter and G is on the corcumference, ∆E'GE = 90°. This means that both ∆E'GE and ∆DGE are right triangles. Triangle ∆DGE: DG² + GE² = ED² x² + GE² = 12² GE² = 144 - x² ---- [1] Trisngle ∆E'GE. GE² + E'G² = EE'² 144 - x² + (x+12)² = (2(4√6))² 144 - x² + x² + 144 + 24x = 64(6) 288 + 24x = 384 24x = 384 - 288 x = 96/24 = 4

[@sumithpeiris8440]

Let the square be of side a and the radius of the quarter circle be R. Let t= theta tan^2(t )= a^2/R^2 = 1/2, so cos^2 = 2/3 Hence x = 12 cos(2t) = 12 (2cos^2 t - 1) = 12 X (4/3-1) =4 Sumith Peiris Moratuwa Sri Lanka

[@sumithpeiris8440]

Errata - cos^2 (t) = 2/3.

[@ramprasadp4027]

Hi Sumith, why x = 12 × cos2t?

[@santiagoarosam430]

AB=a → AE=AC=Radio r=a√2 → a²+r²=12² → a²+(a√2)²=3a²=12² → a²=48 La figura es simétrica respecto al radio vertical → Potencia del punto D respecto a la circunferencia: 12X=a²=48 → X=4 Gracias y saludos cordiales.

[@akifbaysal9141]

This and one similar some other conyributor pointed is the fastest in my opinion and should be added to two other solutions author has described. It makes use of the symmetry very well too.

[@СтасМ-ъ8б]

Точно также решил. Применил произведение отрезков пересекающихся хорд.

[@marioalb9726]

Square inside quarter of circle: S = side of square S²+S²=R² 2S²=R² S² = ½ R² Right triangle with legs R and S: S² + R² = 12² ½R² + R² = 12² R² = 2 . 12² / 3 R = 9,798 cm cos θ = R / 12 θ = 35,26° Right triangle with leg 'x' and hyp.12 cos 2θ = x / 12 x = 12 cos 2θ x = 4 cm. ( Solved √ )

[@اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب. شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين

[@MathBooster]

Thank you 🙂

[@DB-lg5sq]

شكرا لكم يمكن استعمال المثلثADG

[@maqboolahmad9301]

Line A B is 28m.m.form point B is tilted 30° at C, distance from AC is 34m.m what's the length of BC? Can you solve it please

[@rcdmiranda]

a really good explanation!

[@MathBooster]

Thank you 🙂

[@claudeabraham2347]

Very good! I love it!

[@MathBooster]

Thank you!

[@joshuapeng831]

we can easily know 12*x = cd*cd。 then x can get.

[@daddykhalil909]

Why is your voice so low???

[@monkeblazer3154]

is there any specific trick for solving olympiad geometry ? if so pls let me know

[@MathBooster]

No, there is no any specific trick. If you practice a lot of geometry problems then you will be able to think faster that how can you start solving the problem.

[@monkeblazer3154]

@@MathBooster alright tysm

[@kaushikbasu9707]

very good problem

[@MathBooster]

Thank you 🙂

[@krishnamoyghosh6047]

Very easy problem. Generally not given by mathbooster❤

[@duongtranthanh1461]

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