Given a circle $(O,R)$ and a point $P$ in its interior. For any two points $A,B$ on its perimeter, we define $f(A,B)$ to be the point $C$ such that $APBC$ is a parallelogram.
Determine the image of $f$ when $A,B$ vary over all possible pairs of points on the perimeter. That is, determine the set $S = \{f(A,B) | AO = BO = R\}$
Solution Let $O'$ be a point on the extension of $PO$ such that $OO' = 2PO$. We shall show that $S$ is a circle (including the interior) centered at $O'$ with radius $2R$.

Outline of proof:

Let $T$ to be the circle we claim. Note that $T$ is a 2x dilation of the original circle with center $P$. We show that the boundary of $S$ is boundary of $T$ as well, by letting $A=B$ in the opration

Next we show that any point in $T$ is an image of a specific application of $f$. Let $X' \in T$, then $X'$ is an image of a 2x dilation. Suppose $X$ is the original point. We claim that there exists $A,B$ such that $PX$ is a median of $PAB$. Indeed, if $OAB$ is a triangle such that $OX$ is its median, then $PX$ is median of $PAB$.

Then it is routine to show that $f(A,B) = X'$.

Note: this problem can be extended to any point $P$, and any convex shape $V$ in $R^n$. The sum of vectors $PA + PB$ where $A,B$ are any two points on its boundary is equivalent to the 2x dilation of $V$ with center $P$. However, $V$ must be bounded.

Challenge: prove this generalization. (Hint: use the connectedness of $S^{n-1}$ in $R^n$).

Challenge: give a counter example of this result of $V$ is unbounded

Tangents and secant

On a circle $L_1$ centered at $O$, we draw two points $A$ and $B$ that are not diametrically opposite. The tangents at $A$ and $B$ meet at $P$. Circle $L_2$ are created with $OB$ as the diameter. $L_2$ and $AB$ intersect at $B$ and $Q$. Show that $Q$ lies on $OP$.

four tangent circles

Given four circles: $L_1,L_2,L_3,L_4$ such that $L_1$ is tangent to $L_2$ at $X$, $L_2$ is tangent to $L_3$ at $Y$, $L_3$ is tangent to $L_4$ at $Z$ and $L_4$ is tangent to $L_1$ at $W$. All tangencies are outside (the circles do not overlap each other other than at the points of tangency).

Show that $X,Y,Z,W$ all lie on a circle.

Technology Review March/April 2011 Problem 3

The figure contains three semicircles and one circle. The semicircles all have horizontal diameters, and their centers are shown. Their radii are $R, r_1, r_2$ with $r_1 + r_2 = R$. The circle is constructed tangent to the three semicircles. Find $p$ the radius of the circle, and show that the distance from its center to the baseline is $2p$.

(Note: I couldn't draw them as semicircles and I was too lazy to crop the image. But you get the idea).

Solution

Suppose the circle centered in $A$ has radius $r_1$, and the circle centered in $B$ has radius $r_2$. We can readily see that $BC = r_1$ and $AC = r_2$. Also because the third circle is tangent to all circles:
$OA = p+r_1$
$OB = p+r_2$
$OC = r_1 + r_2 -p$

From Stewart's Theorem:

$$OC^2 AB = OA^2 BC + OB^2 AC - AB.BC.AC$$
$$(r_1 + r_2 - p)^2 (r_1+r_2) = r_1(p+r_1)^2 +r_2(p+r_2)^2 - r_1 r_2 (r_1 + r_2)$$

The coefficient of $p^2$ on both sides of that equation is $r_1 + r_2$, thus this is not a quadratic equation, but merely a linear equation in $p$. Expanding and canceling both sides gives us:

$$ p = \frac{r_1r_2(r_1+r_2)}{(r_1+r_2)^2-r_1r_2} = \frac{r_1r_2R}{R^2-r_1r_2}$$

To compute the distance $h$ from $O$ to $AB$, first we calculate the area of triangle $OAB$. Since $OA = p+r_1, OB = p+r_2, AB = r_1+r_2$, we can use Heron's formula:

$$S_{OAB} = \sqrt{(p+r_1+r_2)pr_1r_2} = \frac{1}{2} h AB = \frac{1}{2} h R$$
$$4(p+r_1+r_2)pr_1r_2 = h^2 R^2$$
$$h^2 = \frac{4(p+r_1+r_2)pr_1r_2}{R^2}$$

Now note that $p+r_1+r_2 = p+R = \frac{r_1r_2R}{R^2-r_1r_2} + R = \frac{R^3}{R^2-r_1r_2}$

Plug this into the equation above:

$$h^2 = \frac{4(p+R)pr_1r_2}{R^2} = \frac{4}{R^2} \frac{R^3}{R^2-r_1r_2} \frac{r_1r_2R}{R^2-r_1r_2} r_1r_2 = 4 \frac{(r_1r_2R)^2}{(R^2-r_1r_2)^2}$$
$$h = 2p$$

Circumcircle and Incircle Tangency

Given a triangle $ABC$, its circumcircle $C_O$ and incircle $C_I$. Suppose $X,Y,Z$ are points on $C_O$ such that $XY$ and $XZ$ are tangent to $C_I$, prove that $YZ$ is also tangent to $C_I$.

Solution

Let us denote $R$ to be the radius of $C_O$ and $r$ to be the radius of $C_I$. Also let $O$ and $I$ to be the centers of the circle. Let also Suppose $AB$ is tangent to $C_I$ at $L$. And suppose $AI$ meets $C_O$ at $M$ and $MO$ meets $C_O$ at $N$.

Now suppose $a = \angle MAZ = \angle MAB = \angle MNC = \angle MCB$. Because $\angle ALI = \angle MCN = 90^o$ then by similarity of $\triangle ALI$ and $\triangle MNC$ we have: $$AI.MC = MN.LI = 2Rr$$

Now $\angle MIC = \angle IAC + \angle ACI = x + \angle ACI = \angle MCB + \angle ICB = \angle MCI$ so that we know $MC = MI$. Thus: $$AI.MI = 2Rr$$

Now, we draw similar figures to the triangle $X,Y,Z$. That is: $XY$ is tangent to $C_I$ at $L'$. And suppose $XI$ meets $C_O$ at $M'$ and $M'O$ meets $C_O$ at $N'$.

In our analysis above, we showed that $AI.MC = 2Rr$ but we didn't use the fact that $BC$ is tangent to $C_I$, so we can repeat the argument to claim that $$XI.M'Z = 2Rr$$.

On the other hand, $XI.IM' = AI.IM = 2Rr$ so we have $IM' = M'Z$ which means $IM'Z$ is an isosceles. $$\angle M'IZ = \angle M'ZI$$ $$\angle M'XZ + \angle IZX = \angle M'ZY + \angle YZI$$ Because $XY$ and $XZ$ are both tangent to $C_I$ then $XM$ is a bisector of $\angle YXZ$ so that $\angle M'XZ = \angle M'XY = \angle M'ZY$. So our equation above becomes: $$\angle IZX = \angle YZI$$ which means $ZI$ is also internal angle bisector, which means $YZ$ is tangent to $C_I$.

Cyclic quadrilateral

In a cyclic quadrilateral $ABCD$ such that $AB = AD$ and $AB+BC < CD$, show that $ABC > 120^o$

Solution: http://dharmath.blogspot.com/2010/05/solution-cyclic-quadrilateral.html

Circles and Coloring

Several circles are drawn on a piece of paper to form several regions on the paper. We wish to color these regions such that no two regions that share a common boundary would have the same color.

(It is okay if two regions that share a common point to have the same color, but not boundary)

How many colors minimum are needed?