Find the length X | A Nice Geometry Problem | Math Olympiad 

Using coordinate geometry: D = (0, 2a), E = (2b, 0), A = (0, 2a+2), C = (2b+4, 0) Since DF = EF, then F = midpoint of DE = ((0+2b)/2, (2a+0)/2) = (b, a) Since AG = CG, then G = midpoint of AC = ((0+2b+4)/2, (a+2+0)/2) = (b+2, a+1) x = distance between F (b,a) and G (b+2, a+1) x = √((b+2−b)² + (a+1−a)²) = √(2² + 1²) = √5

Rotate a copy of the triangle by 180° about the center G. This makes a rectangle with a parallelogram inside, one of whose the sides is 2X, which is also the hypotenuse of the right triangle at the corner with sides 2 and 4. Hence 2X = 2√5 X = √5

Bhai there is a short solution as follows Let BD= b, and BE=a so coordinates of F is ( a/2, b/2 ) . AB= b+2 and BC= a+4 so coordinates of G are ( a/2+2 , b/2+1 ) so by distace formula we get FG equal to root5

AD is vector (0,2), EC is vector (4,0). Average them to get vector FG is (2,1). |(2,1)| = Sqrt[5].

BD->0, BE->0, F->B, FG - constant => AB=2, BC=4 => AC=2sqrt5 => x=BG=AC/2=sqrt5

Roughly, if the solution is not depend on the triangle DBE, the convergence limit with BE towards zero might be also the same value.

By coordinate method, let F=(a,b), so G=(a+2,b+1), thus FG=sqrt(2^2+1^2)=sqrt(5).😊

You solved it very well

It looked pretty difficult, thanks for the solution.

Hello, please tell me which country is the source of this Olympiad question? Similar to this question was asked in Iran's math test this year

Drop perpendiculars from F & G on AB & BC giving the horizontal & vertical distances as 'b', 'B' & 'h', 'H' respectively. Now from similarity AB=2H & DB=2h as AB-DB=2 H-h=1 Also BC=2B & BE=2b as BC-BE=4 B-b=2 But x²=(H-h) ² + (B-b)² =1²+2²= 5 or x=5^½

Lengths DB and BE are not given, so the problem statement implies that the solution will be valid for all valid lengths for DB and BE. So, we choose a special case which is straightforward to solve. Let ΔABC be an isosceles right triangle with sides of length 6. Then, DB4 = 4 and BE = 2. Using coordinate geometry and making point B our origin, it is fairly straightforward to find that F is (1 , 2) and G is (3 , 3) and the distance between the two points is √((3 - 1)² + (3 - 2)² = √5. If this were a multiple choice problem, we would be done! If we try to generalize our special case, let BE = 2b and DB = 2a, as Math Booster did. then F is (b , a) and G is ((2b + 4)/2 , (2a + 2)/2) = ( (b + 2) , (a + 1)) When calculating distance, the values of b and a subtract out, leaving us with √((3 - 1)² + (3 - 2)² = √5, same as our special case.

√5

Hello, following my simplyfied and shortened way to find the answer. I used a view based on coordinates in u and v with B being (0,0). We are looking for the length x of FG. I used c as the length of BD, and this is also the value for BE. So AB has a length of c+2, and BC has a size of c+4. The lower left point F is located at (v1,u1) = (c/2,c/2). For (v2,u2) we can see the following. (c+4-u2):CG = (c+4):AC (c+4-u2) = (c+4)/2 u2 = (c+4) - (c+4)/2 u2 = (c+4)/2 v2:CG = (c+2):AC v2 = (c+2)/2 x² = ((c+4)/2 - c/2)² + ((c+2)/2 - c/2)² = 2² + 1² = 5 . Therefore, x is the square root of 5.

I still prefer black text on a white background.But it doesn’t matter, it’s your choice!