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.

Excircle and circumcircle

In $\triangle ABC$, the $L_1$ be the excircle with respect to $AB$. It touches $CA$ and $CB$ and $M$ and $N$. The circle centered at $C$ and passing through $M$ and $N$ intersects $AB$'s extension at $P,Q$. Show that $L_2$ the circumcircle of $\triangle CPQ$ is tangent to $L_1$, and that the point of tangency between $L_1$ and $L_2$, the point of tangency between $L_1$ and $\triangle ABC$, and $C$ all lie on one line.

Solution Let $X$ be the tangency point of $L_1$ and $AB$. Let $CX$ meets $L_1$ in $Y_1$ and $CX$ meets $L_2$ in $Y_2$. We shall establish that $Y_1 = Y_2$.

Let $C'$ be the diameter of $L_2$ and let $CC'$ meets $AB$ (or its extension) at $D$. Because $CP = CQ$ then it's easy to see that $CD \perp PQ$. Let $s = CM = CN = CP = CQ$.

By power point on $L_1$: $$CX.CY_1 = CM^2 = s^2$$.

By similarity of $\triangle CXD \sim \triangle CC'Y_2$ and $\triangle CC'P \sim \triangle CDP$ : $$CX.CY_2 = CD.CC' = CP^2 = s^2$$

Thus $CY_1 = CY_2$ which means $Y_1 = Y_2$. Now we know that $L_1$ meets $L_2$ in at least one point $Y = Y_1 = Y_2$. Suppose they meet in another point $Z \neq Y$

Let $CZ$ meets $L_1$ in $CZ$ and $X_1 \neq X$. By power point we again have:$CX_1.CZ = s^2$.

Let $CZ$ meets $AB$ in $X_2$. Similar as above, because $Z \in L_2$, we can repeat the similarity argument to establish: $CX_2.CZ = s^2$. This means that $CX_1 = CX_2$, thus $X_1 = X_2 \in L_1, AB$ but that is a contradiction because $L_1$ is only tangent to $AB$, and can only meet $AB$ at $X$.

Now that we've established the tangency between $L_1$ and $L_2$, it's clear that $C,X,Y_1/Y_2$ all lie on one line from the proof.

Circumcenter and 3 circumcircles

Given an acute triangle $ABC$ and its circumcircle is centered at $O$ and has radius $R$.

$AO$ meets the circumcircle of $\triangle OBC$ at $X$, $BO$ meets the circumcircle of $\triangle OAC$ at $Y$, and $CO$ meets the circumcircle of $\triangle OAB$ at $Z$.

Show that $$AX + BY + CZ \geq 9R$$

Determine when equality occurs.

Solution

Let $AX$ meets $BC$ at $D$. And let $OO'$ be the diameter of the circumcircle of $\triangle OBC$, and let $M$ be the intersection of $OO'$ and $BC$.

$$\frac{OX}{OO'} = \frac{OM}{OD}$$ $$OX.OD = OO'.OM = OC^2 = R^2$$ $$AX = AO + OX = R + \frac{R^2}{OD} = \frac{R(R + OD)}{OD} = R\frac{AD}{OD}$$

If we define $E$ as the intersection of $BY$ and $AC$, and $F$ as the intersection of $CZ$ and $AB$, then similarly: $$BY = R \frac{BE}{OE}$$ $$CZ = R \frac{CF}{OF}$$

So: $$AX + BY + CZ = R(\frac{AD}{OD} + \frac{BE}{OE} + \frac{CF}{OF})$$ $$\geq \frac{9R}{\frac{OD}{AD} + \frac{OE}{BE} + \frac{OF}{CF}}$$ But $OD / AD = A_{\triangle OBC} / A_{\triangle ABC}$ and so on, so: $$\frac{OD}{AD} + \frac{OE}{BE} + \frac{OF}{CF} = 1$$ $$AX + BY + CZ \geq 9R$$

and equality happens when the area of $\triangle OBC$ is exactly $\triangle ABC / 3$. And this happens only when $O$ is the centroid of $\triangle ABC$, which means $ABC$ is an equilateral.