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Niveau: Secondaire, Lycée
Bac Blanc n°1 S 2010 page 1/4 0 1 1 A D B C M N M' Exercice 1. (5 pts) Question 1. Question 2a. The equation of ( )? is : ( – 2)? ( –1 )? 2x y+ = and the one of the x-axis is : y = 0. We are thus looking for the points M(x ; y) satisfying : ( )( – 3) 1 0( – 2)? ( –1 )? 2 ( – 2)? (0 –1 )? 2 ( – 2)? 1 0 0 0 0 0 x xx y x x y y y y ? =?+ = + = ? =? ? ? ? ? ? ?? ? ? ? = = = =?? ? ? ? The circle ( )? thus intersects the x-axis at two points, the ones associated with the numbers 1 and 3. Question 2b. A is the midpoint of de [BD], ... 3 2 2 B D A D z z z z i+ = ? ? = + Question 3a. ( ) ( )3 6 12 43 2 6 2 1 36 2 205 5 5 5 23 6 2 6 1 3 10 101 5 5 5 5 D M B M i i i i iz z i i i z z ii i + ? ?

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Milne?Thomson circle theorem bac blanc

Bac Blanc n°1 Correction

Exercice 1. (5 pts)D (9)Question 1. M 1 A N Question 2a.M' The equation ofis :(x( –2)²y–1)²1 2B C 0 1 and the one of the x-axis is :y= 0. We are thus looking for the pointsM(x;y) satisfying : (x–2)²#(y–1)²12(x–2)²#(0 –1)²12(x –2)²%110(x –3)x110 Û ÛÛ   y10y10y10y0 1 The circlethus intersects the x-axis at two points, the ones associated with the numbers 1 and 3. z z B D Question 2b.A is the midpoint of de [BD],Û...Ûz13#2iA D 2 3 612 4 3#2i% %i#i z%z6#2i(6#2i! (1#3i!20i 5 55 5 D M Question 3a.1 111 12i3 62 6 z%z1%3i10 10 B M 1% %i%i 5 55 5   z D M Question 3b.is the value of the angleThe argument ofB,MD%z M   z zp D M According to question 3a,2i⇒MB,MD1. Geometrically, it means that the triangle BMD is right-z%z2 B M angled atM and thus M is on the circle with diameter [BD] (1/2 circle theorem), that’s to say. Question 4a.We have proven in question 3 that (BM) and (DM) are perpendicular. The point N belongs to the circle with diameter [AB], then ABN is right-angled at N. The lines (AN) and (DM) are bothperpendicular to (BM), and are thus parallel. Question 4b.In the triangle BMD, we know that : ·A is the midpoint of [BD] ·N belongs to [BM] ·(AN) // (DM) 3 6 1# #i z#z4 3 5 5 B M Thanks to Thales’ theorem, N is the midpoint of [BM].#1 1iN 2 25 5 Exercice 2. (5 pts)D’après bac S juin 2008 Antilles-GuyannePartie A : x We want to solve the differential equation (E) :y'#y1e. x%x%x%x%x x 1.u x1fg'1f'g#g'f1e%xe, sou'(x!#u(x!1e%xe#xe1e, so u is a solution of (E). %x 2.y'y10Ûy'1 %y, whose solutions are all the functions in the formf xKe,KÎℝ. Bac Blanc n°1S 2010page1/4

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