Blow of sight

I was calculating like crazy, but the answer was right in front of my nose the whole time

$\measuredangle MAB = 90^{\circ}$ , so arc $MCB$ must be a semicircle, and therefore $MB$ is a diameter. The same reasoning can be applied to $NB$ . Therefore, when connecting these two segments, $\bigtriangleup MNB$ and $\bigtriangleup PQB$ are created, and $\bigtriangleup MNB \simeq \bigtriangleup PQB$ .

Since $PB$ is a radius of $P$ and $MB$ is a diameter, the scale factor between the two triangles is $2$ , so $MN = 3*2 = \boxed{6}$ .

Consider the special case where the circles with centers P and Q are tangent, and the answer is obvious.

Its actually so easy if you simplify it down to squares

GOOD GRIEF!!! There are so many geometric "solutions", but the answer is intuitive. WHAT IF the intersecting points were ON PQ? The radius of Circle P plus the radius of Circle Q would be the length of PQ and MN would be an extension of PQ, also known as the DIAMETERS of Circles P and Q, or TWICE the length of PQ. THEREFORE MN = 2PQ is the base length equation, and MN = 6 where PQ = 3. DON'T MAKE IT SO HARD, PEOPLE!

An intuitive solution is examining the special case when the two circles are tangential. I know for a perfect solution you should be sceptical by not accepting the implicitly suggested rule that the length of the line segment MN is independent of the radii, and suspecting incomplete data in the problem.

Think of a special case where the 2 circles are identical and just tangentially touch at one point. Then PQ = 2 R, R being the radius of the circles. Then, MN is colinear with PQ and its length is 2 (2R). This gives MN = 2*3 = 6!

2x+2y=MN (equation 1), x+y=3 (equation 2). common factor: 2(x+y) =MN. (inserting equation 2 in 1): 2*3=MN, MN=6

PQ=3; MP+QN=3;MP+QN=PQ;2MP=3-PQ;2MP=3-3=0! => MP=0 , so MN = 6 because then the 2 circles are concentric.

You'll sometimes come across a problem like this, where if you trust the setters to know what they are doing you can solve it instantly on inspection. Notice that they haven't said anything about the distance between the lines MN and PQ, so it's reasonable to assume the answer does not depend on this distance. So make the distance zero, as others have said, when the circles are tangent and the lines are collinear. Then it's obvious that MN is twice PQ.

I used the theorem of intersecting lines :

Let $X_1$ and $X_2$ be the intersection points of the two circles; $X_1$ lies on $MN$ .

From symmetry, we see that $PQ$ is a bisector of $X_1X_2$ . Let $R$ be the midpoint of $X_1X_2$ , i.e its intersection with $PQ$ .

Now consider the triangles $MX_1X_2$ and $PRX_2$ . They are similar triangles (here we use the fact that $MN$ is parallel to $PQ$ ), with ratio $X_1X_2:RX_2 = 1:2$ . Thus $MR:PR = 1:2$ . Similarly, $RN:RQ = 1:2$ ; and so $MN:PQ = 1:2$ .

Conclusion: $MN = 2PQ = 2\cdot 3 = \boxed{6}.$

I simply measured both parallel lines to determine that Line MN was twice line PQ.

Almost no math required. Line MN looked twice as long as line PQ.

1.5^2+1.5^2=x^2 x=2.12 3^2=1.5^2+x^2 x=2.59 2.59^2+2.59^2=x^2 x=3.67 MN=3.67+2.12=5.79 or fucking 6 !!!!!

If you regard the other intersection point of the circles as a vertex, R, than triangle PQR is similar to triangle MOR. Since line segment MR is the diameter of the circle to which MP is the radius, MR must be twice the measure of PR. (The same goes for OR and QR.) Since they share vertex R and are similar, triangle MOR is necessarily twice the size of triangle PQR, and so line segment MO is twice the length of line segment PQ.

Since the question says "as shown" I have to assume that the dimensions and angles are exact. Thus I just used a ruler to measure PQ on my screen (4.5cm) and then measured MN (9cm) and multiplied by the same scale factor (2/3).

If we increse the diameter of Q and decreese the diameter of P mantening the size of PQ, we will see the line MN decending to the center of the circules and and his size will be the sum of diametrr P and Q. And as we did, diameter P = Q, MN =6. Sorry about gramatic errors, I'm Brazilian.

As we connect the two intersecting points with a line we see that PQ cut it in the middle then for the suggested line from N through Q according to Thales' theorem it must end with its double which mean the diameter of the circle that would not intersect the line connecting the two intersecting points in a point other than the lower intersecting point that forms a triangle with the line MN and by applying Thales' theorem scale fraction it is clear that MN=2PQ=2(3)=6 😊

A very nice, very clear solution is this:

The intersection point will be denoted as T. Take PQ's midpoint, and take the line crossing it, perpendicular to PQ. T's reflection to the line will be called point S. Obviously, S is on MN.

Since P's reflection to the line is Q, PT's reflection is QS, PT=QS. But PT is the radius of the bigger circle, so QS=PM, therefore PQSM is a paralelogram, and MS=PQ. Same bway can be shown that QPSN is also a paralelogram, so SN=PQ.

MN=MS+SN=PQ+PQ=6

At tangency, the equation is PQ=R+r and MN=2PQ, giving a general equation of MN=2(R+r) where R is the radius of the left hand circle and r is the radius of the right hand circle.

Let's take an special case of the two intersecting circles, touching each other: Now, if r1 and r2 are the radii of the circles, PQ = r1 + r2 = 3. Or in terms od diameters, d1/2 + d2/2 = 3 => d1+d2 = MN = 6

John Knox. The intersection of the circles is not fixed so the problem asks for a ratio or we can move the circles as long as they have a common point of intersection. By moving the second circle to the right so that it is tangent to the first circle it becomes obvious that MN equals 2XPQ. In this instance or configuration of circles PQ equals 3, the sum of the radii of the two circles, while it is obvious from this configuration that MN equals PQ plus the sum of the radii of the two circles or 6. The fact that MN equals 2XPQ holds as long as the circles have a common point although I have not given a rigoruos proof. Proof anyone?

Let P be the center of coordinates with coordinates (0,0).. The coordinates of Q are (3,0). Let (x,y) be the coordinates of the intersection.point R. Then MR = 2x and RN = 2(3-x). so MN = MR + RN = 6. Ed Gray

Let R1 be the radius of the left circle and R2 be the radius of the right circle. Let a be the length of the horizontal distance between the overlapping circles on PQ. We can see that MN = 2(R1-a/2) + 2(R2-a/2) = 2(R1 + R2 -a) However, PQ = R1 + R2 -a = 3 Therefore, MN = 2 x PQ = 6

The answer can be easily arrived at by using the converse of midpoint theorem. Draw diameters of both circles intersecting the POI of both circles and lying on MN. By applying the coverse of midpoint theorem ( PQ parallel MN) (PQ bisects common chord) Thus, MN=2PQ

As PQ = 3, I thought of the special case where the centers are tangent. With this in mind, the answer is obvious.

Add an identical circle the size of Q on the other side of P that is also 3 units away. Construct a line that is perpendicular to PQ that passes through both intersection points. Repeat for the circle on the other side. Similar constructions can also be done on the other side of Q with a circle of size P. These constructions show that |MN| = 2|PQ|.