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 $OP$.

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.

Four positive numbers

Let $a_1,a_2,a_3,a_4$ be four positive numbers and let: $$S_1 = a_1 + a_2 + a_3 + a_4$$ $$S_2 = \sum_{i \neq j} a_ia_j$$ $$S_3 = a_1a_2a_3 + a_1a_2a_4 + a_1a_3a_4 + a_2a_3a_4$$ $$S_4 = a_1a_2a_3a_4$$ Given that $$| \frac{S_1-S_3}{1-S_2+S_4} | < 1$$ show that there are two distinct $a_i,a_j$ such that : $$|a_i-a_j| < (\sqrt{2}-1)(1+a_ia_j)$$

Solution 1

If any two $a_i$ are the same then we are done. WLOG, we may now assume that $a_1 < a_2 < a_3 < a_4$. Suppose that for each two $a_i > a_j$ we always have: $$a_i - a_j > (\sqrt{2}-1)(1+a_ia_j)$$ Consider $a_3$ and $a_4$: $$(a_4 - a_3) > (\sqrt{2}-1)(1+a_4a_3)$$ $$(a_4 - a_3)(\sqrt{2}+1) > (1+a_4a_3)$$ $$a_4(\sqrt{2}+1 - a_3) > 1+a_3(\sqrt{2}+1)$$ Because the RHS is positive, in order for the LHS to be positive, we must have $a_3 < \sqrt{2} + 1$. Thus now we have $a_1 < a_2 < a_3 < \sqrt{2}+1$. Next, note that the function $f(x) = \frac{x + \sqrt{2} - 1}{1 - (\sqrt{2}-1)x}$ defined for $0 < x < \sqrt{2}+1$ is definite positive and strictly increasing. The assumption $$a_i - a_j > (\sqrt{2}-1)(1+a_ia_j)$$ for $j=1,2,3$ becomes: $$a_i(1 - (\sqrt{2}-1)a_j) > a_j + \sqrt{2}-1$$ And because $(1 - (\sqrt{2}-1)a_j) > 0 \iff a_j < \sqrt{2}+1$ we can divide both sides to obtain: $$a_i > \frac{a_i + \sqrt{2}-1}{1 - (\sqrt{2}-1)a_j} =f(a_j)$$ So now we have: $$a_4 > f(a_3), a_3 > f(a_2), a_2 > f(a_1)$$ We note that $f(0) = \sqrt{2}-1,f(\sqrt{2}-1) = 1, f(1) = \sqrt{2}+1$. Furthermore, because $f$ is strictly increasing, $$a_2 > f(a_1) > f(0) = \sqrt{2}-1$$ $$a_3 > f(a_2) > 1$$ $$a_4 > f(a_3) > \sqrt{2}+1$$ Also, if $a_2 \geq 1$ then we'd have $a_3 > f(a_2) > f(1) = \sqrt{2}+1$ a contradiction, so we must have $a_2 < 1$. By the same token, if $a_1 \geq \sqrt{2}-1$ we'd have $a_2 > f(a_1) > f(\sqrt{2}-1) = 1$ a contradiction, so we must have $a_1 < \sqrt{2} -1 $. Now we've established that: $$0 < a_1 < \sqrt{2} - 1 < a_2 < 1 < a_3 < \sqrt{2} + 1 < a_4$$ From there it's clear that $a_1a_3 < 1$ and $a_2a_4 > 1$. Next, we assert the following: $$a_1 + a_3 + a_1a_3 > 1$$ which is very easy to see from the bounds above. Also because $a_4 > 0$ and $a_2 < 1$, $$\frac{1}{a_2a_4} + \frac{1}{a_2} + \frac{1}{a_4} > 1$$ $$1 + a_2 + a_4 > a_2a_4$$ Therefore: $$(1-a_1a_3)(1-a_2a_4) > (a_1+a_3)(a_2+a_4)$$