Sweden Math Olympiad | A Very Nice Geometry Problem

Рет қаралды 10,381

Күн бұрын

Пікірлер: 35

@harikatragadda 18 күн бұрын

Draw GF parallel to DC and FK perpendicular to DC. By proportionality, G is the midpoint of AD and the Right Triangle ∆EGF is Congruent to the Right Triangle ∆CKF, with EG = 1= KC and FG = FK = 7 and X being their hypotenuse. X = √50

@kylekatarn1986 18 күн бұрын

The triangles ABE and DCE are congruent for the second consistency criteria for right triangles: Two right triangles are equals if they have one side and the two adjacent angles in common.

@oscarcastaneda5310 18 күн бұрын

Excellent! From this, DC is 8 and AC is 10. So it follows that X = 5sqrt(2).

@billnoyce5026 18 күн бұрын

This is true even if the triangles were not right triangles. The ASA (Angle, Side, Angle) congruency theorem says if one side and the angles touching that side are equal to the corresponding parts of another triangle then the triangles are congruent.

@marioalb9726 18 күн бұрын

Pytagorean theorem, twice: 2x² = 8²+6² = 100 x = √50 = 5√2 cm ( Solved √ ) Right triangles ABE and EDC are congruents!!!

@GabrieleIris-is7bg 16 күн бұрын

ABEF is equal to CDEF, they have all the angles and 2 adjacent side equal, so we have EB=EC=hypotenuse of a 6-8-10 (3-4-5) triangle ABE and CDE=10 so X=5√2

@sakurayayoi-p2r 17 күн бұрын

別解法　直角三角形の合同から上辺は６下辺は8と求まる。台形の面積を(上辺＋下辺)*高さ/2とxｘ/2*2+8*6/2*2が等しくなるｘを求める。5√2となる。

@kateknowles8055 18 күн бұрын

Thank you for a pleasant geometry problem. Initial observations: Angle BCE will be 90 degrees because triangles BEF and EFC are congruent rightangled triangles Angle BCE is 90 degrees because 45 +45 = 90 AB is parallel to DE Main content : BE.BE = 2.X.X ~~~~~ P is the mipoint of BE. It is the centre of a circle which passes through the cyclic points F,E,A, B O is the midpoint of EC. It is the centre of a circle which passes through the cyclic points F,C,D,E FEAB and FCDE are congruent quadrilaterals. DCF being theta - > DEF= 180 - theta - > AEF = theta - > ABF = 180 - theta So AE =DC =8 and ED = AB = 6 In triangle ABE: 10.10 = 8.8 +6.6 (3,4,5 triangle with BE equal to 10 by Pythagoras' theorem) Putting BE.BE =100 ~~~~~ into BE.BE = 2.X.X gives 100 = 2.X.X X.X =50 X = 5 . root(2) Answer

@MdShahriarHossain-l1n 18 күн бұрын

I solved it and feeling confident for my math exam tomorrow! Thanks!

@MdMontasir-b9o 17 күн бұрын

Where are you form

@MdShahriarHossain-l1n 17 күн бұрын

@MdMontasir-b9o Bangladesh. You?

@raghavansadagopan3905 15 күн бұрын

The required side EF is 5 times Sq root of 2. As EC =EB = 10.

@giuseppemalaguti435 18 күн бұрын

Dal trapezio calcolo la base minore e base maggiore, bm=√(2x^2-64)..BM=√(2x^2-36)... Pitagora 4x^2=14^2+(BM-bm)^2...x=5√2

@ناصريناصر-س4ب 18 күн бұрын

The triangle EBC is right and isosceles, so EB=EC=x√2, and according to the Pythagorean theorem AB=√(2x²-64) and CD=√(2x²-36), let H be the midpoint of AD, so FH=(√(2x²-64)+√(2x²-36))/2=√(x²-1), and this equation is equivalent to 4x⁴-200x²=0, i.e. 4x²(x²-50)=0, so x=5√2

@oscarcastaneda5310 18 күн бұрын

We think Alike : )

@gelbkehlchen 17 күн бұрын

Solution: I draw a line from E to B and from E to C and see that triangle EFB is congruent to triangle ECF because of the equality of 2 sides and the interior angle of 90°. In addition, triangle EFB and triangle ECF are isosceles and the base angles are 45°, as are angles FEB and CEB, so angle BEC is 90°. Let angle BEA = α, then angle DEC = 180°-90°-α = 90°-α, so the complementary angle to α and triangle EBA and triangle EDC are congruent. Therefore: EB = EC = √(8²+6²) = 10 and x²+x² = 10² ⟹ 2x² = 100 |/2 ⟹ x² = 50 |√() ⟹ x = √50 ≈ 7.0711

@ina-j2p 18 күн бұрын

6：8：10の直角三角形だから、 ∴x=10/√2=5√2

@quigonkenny 18 күн бұрын

Draw BE and EC. As CF = FB = x and EF is common, then ∆EFB and ∆CFE are congruent. Additionally, as EF = CF = FB = x and ∠EFB = ∠CFE = 90°, then ∆EFB and ∆CFE are isosceles right triangles and ∠FBE = ∠BEF = ∠FEC = ∠ECF = 45°. Let ∠EBA = α and as ∠BAE = 90°, then ∠AEB = 90°-α = β. As ∠BEC = 45°+45° = 90°, then ∠CED = 180°-(90°+β) = 90°-β = α. As ∠EDC = 90°, then ∠DCE = 90°-α = β. As ∠EBA = ∠CED = α, ∠AEB = ∠DCE = β, and CE = EB, then ∆BAE and ∆EDC are congruent. As DC = AE = 8 = 4(2) and BA = ED = 6 = 3(2), then ∆BAE and ∆EDC are 3-4-5 Pythagorean triple right triangles and CE = EB = 5(2) = 10. Triangle ∆CFE: CF² + FE² = EC² x² + x² = 10² 2x² = 100 x² = 100/2 = 50 [ x = √50 = 5√2 units ]

@cesarelai 14 күн бұрын

L'angolo BEC = 90° quindi gli angoli AEB e CED sono complementari, quindi i triangoli BAE e EDC sono congruenti e hanno l'ipotenusa = 10. Il triangolo BCE è metà quadrato di lato = 10. x è la semidiagonale. x = (10sqrt(2))/2 = 5sqrt(2)

@RAG981 18 күн бұрын

Save some time. In CDE sin a = 6/xrt2, and in ABE sin a = AB/xrt2, so AB = 6. Then 6^2 +8^2 =2x^2, and x = 5rt2.

@santiagoarosam430 18 күн бұрын

BEC es triángulo rectángulo de lados X√2b ; X√2 ; 2X---> Los triángulos BAE y EDC son congruentes y sus lados miden 6 ; 8 y X√2---> EB=√(6²+8²)=10 =X√2---> X=5√2. Gracias y saludos.

@alexniklas8777 16 күн бұрын

Decided orally, in your opinion x=5√2. Thanks sir!

@GuohaoLiu-b2f 15 күн бұрын

Turn it 180 degrees and put it to right side. We get a square. Each side is 14. F is the center. Make a horizontal line. We get right triangle with the sides: 1,7,X. So, I get the answer without pen.

@中西康記 18 күн бұрын

f は直角三角形の斜辺の中点。

@juanalfaro7522 17 күн бұрын

I got the same answer by the same method. Clearly EB=EC=x√2. Since EF=BF=CF=x, BEF=CEF=45. Then BEC=45+45=90. Let AEB = α and CED = β, then α + 90 + β = 180 --> α + β = 90. Since AEB = α then ABE = β. Similarly, since CED = β then DCE = α. Now look at EAB and CDE: (1) AEB = DCE = α. (2) EB=EC=x√2. (3) ABE = CDE = β. Then EAB and CDE are congruent triangles, thus AB=6 and CD=8. Since both triangles are right triangles, then x√2 = 10. Thus x= 10/√2 = 5√2.