answer:If each car was selected exactly thrice, and each client selects 3 cars, then the total number of selections made by all clients is equal to the number of cars multiplied by the number of times each car was selected. Total selections = Number of cars × Number of times each car was selected Total selections = 15 cars × 3 selections per car Total selections = 45 selections Since each client makes 3 selections, we can find the number of clients by dividing the total number of selections by the number of selections per client. Number of clients = Total selections / Number of selections per client Number of clients = 45 selections / 3 selections per client Number of clients = 15 clients Therefore, boxed{15} clients visited the garage.

answer:To solve the given problem, we need to use the following lemma: Let the intersection points of the circles k and k' be E and F. Any arbitrary line e passing through the point E intersects the circle k at point P and the circle k' at point Q. We need to prove that the midpoint O of the segment PQ lies on a fixed circle passing through points E and F. 1. First, consider the triangle triangle FPQ. - The angle angle OPF = alpha is constant (an inscribed angle). - Similarly, the angle angle PQF = beta is also constant, implying that the angles are independent of the positions of points P and Q. 2. Thus, the angle angle QOF = varepsilon is also constant. Therefore, the midpoint O lies on a circle with angle varepsilon inscribed over the segment EF. - If O lies inside the circle k and k separates P and Q, then O is on the arc of the circle containing EF within k (refer to Figure 1). - If O lies inside the circle k' and k separates P and Q, then O is on the complementary arc of the circle containing EF within k' (refer to Figure 2). This conclusion is evident from the diagram, as now the segment EF is viewable from O at an angle of 180^circ - varepsilon. 3. We also need to consider cases where k does not separate points P and Q. - Suppose both P and Q lie within the circle k. In this scenario, the triangle triangle PFQ still has angles alpha and beta constant, ensuring the inscribed angle of segment EF remains varepsilon and passes through O. - If P and Q lie within the circle k', the proof is similar to the previous case, showing that the geometric locus of O remains on the complementary arc. Thus, the fixed circle indeed exists. 4. With this lemma established, we proceed to the assertion in the problem. The Feuerbach circle of a triangle passes through the midpoints of its sides, the feet of its altitudes, and the midpoints of the segments connecting the vertices to the orthocenter (a total of 9 points, hence often called the 9-point circle): - Any three of these nine points uniquely determine the circle. 5. For triangle triangle ABC, the altitudes are BE and CF (refer to Figure 4), and their intersection point is M. The circumcircle of the cyclic quadrilateral AEMF is k', while the circle with diameter BC is k. 6. The points E and F, together with the midpoint O of the segment BM, define the Feuerbach circle of triangle triangle ABC. The line BM and line PQ are equivalent to line e used in the lemma, so by the lemma, the midpoint of PQ, denoted as R, lies on the same circle as E and F, forming the Feuerbach circle of triangle ABC. 7. If the triangle is obtuse, the roles of points A and M are exchanged, but the proof remains the same. This validates the proposition. Conclusion. boxed{}

answer:To solve for the value of a given that left{begin{array}{l}{x=2}{y=a}end{array}right. is a solution to the linear equation 2x-3y=5, we substitute the values of x and y into the equation: [ begin{align*} 2x - 3y &= 5 2(2) - 3a &= 5 4 - 3a &= 5 -3a &= 5 - 4 -3a &= 1 a &= frac{1}{-3} a &= -frac{1}{3} end{align*} ] Therefore, the value of a is -frac{1}{3}. Hence, the answer is boxed{C}.

answer:1. **Inversion Transformation:** Let ( A^*, B^*, C^*, D^* ) be the images of points ( A, B, C, D ) under an inversion centered at ( O ) with radius ( R ). The inversion simplifies the problem by transforming the original configuration into a new one where the claim becomes: "There exists a circle tangent to the sides of the quadrilateral ( A^*B^*C^*D^* )." 2. **Properties of Inversion:** Under inversion, lines through the center of inversion are mapped to themselves, and circles passing through the center of inversion are mapped to lines. The angles between intersecting lines or circles are preserved. 3. **Parallelism and Similarity:** Let ( O' ) be a point such that ( OA parallel O'D ) and ( OB parallel O'C ). Since ( AB parallel CD ), triangles ( AOB ) and ( DO'C ) are similar. Also, since ( AD parallel BC parallel OO' ), we have that ( AD parallel BC parallel OO' ). 4. **Cyclic Quadrilateral:** Given ( angle AOB + angle COD = 180^circ ) and ( angle AOB = angle DO'C ), quadrilateral ( DOCO' ) is cyclic. This follows from the fact that the sum of opposite angles in a cyclic quadrilateral is ( 180^circ ). 5. **Angle Chasing:** By angle chasing, we find: [ angle ABO = angle DCO' = angle DOO' = angle ADO ] Since ( angle AOB = angle DO'C ), we have: [ angle OA^*B^* = angle ABO = angle ADO = angle OA^*D^* ] 6. **Cyclic Quadrilaterals under Inversion:** By the properties of inversion, quadrilaterals ( A^*ABB^* ) and ( AA^*DD^* ) are cyclic. Therefore, the angles ( angle OA^*B^* ) and ( angle OA^*D^* ) are equal. 7. **Existence of the Incircle:** Since the angle bisectors of ( A^*B^*C^*D^* ) meet at point ( O ), there exists a circle centered at ( O ) that is tangent to the sides of the quadrilateral ( A^*B^*C^*D^* ). This circle is the incircle of the quadrilateral ( A^*B^*C^*D^* ). 8. **Conclusion:** The existence of the incircle in the inverted configuration implies that there exists a circle ( k ) tangent to the circumscribed circles of the triangles ( triangle AOB, triangle BOC, triangle COD, triangle DOA ) in the original configuration. (blacksquare)