## Problem 1.

Line $l$ cross curve $y=ax^2$ on points A, B; if $OA \bot OB$, prove that AB passes a fixed point.

### Proof:

$$\begin{gathered} A: (x_A, ax_A^2) \quad B: (x_B, ax_B^2)\\ k_{OA} = ax_A \quad k_{OB} = ax_B \\ OA \bot OB \quad \Rightarrow \quad k_{OA}k_{OB} = a^2x_Ax_B = -1 \\ AB: y-y_B = \frac{y_A-y_B}{x_A-x_B}(x-x_B) \\\end{gathered}$$
\begin{aligned}y &= \frac{y_A-y_B}{x_A-x_B}(x-x_B) + y_B \\ &= \frac{ax_A^2 - ax_B^2}{x_A - x_B}(x-x_B) + ax_B^2 \\ &= a(x_A+x_B)(x-x_B) + ax_B^2 \\ &= a(x_A+x_B)x - ax_Ax_B \\ &= a(x_A+x_B)x + \frac{1}{a}\end{aligned}

Line AB pass $(0, 1/a)$

## Problem 2.

Ellipse $x^2/3 + y^2/2 = 1$ has an inscribed quadrilateral ABCD such that AB passes its left focus point E, and AC, BD passes its right point F. Prove that the slope ratio of line AB and CD $k_{AB}/k_{CD}$ is a fixed value.

### Proof:

We transform the $x-y$ plane by: $(x, y) \rightarrow (x/\sqrt{3}, y/\sqrt{2})$, such that the ellipse becomes a unit circle. We find that

$$\frac{k_{AB}}{k_{CD}} = \frac{(y_B-y_A)(x_D-x_C)}{(x_B-x_A)(y_D-y_C)}$$

keeps unchanged after transformation. The original focus points of ellipse are $(-1, 0), (1, 0)$ now becomes $(-1/\sqrt{3}, 0), (1/\sqrt{3}, 0)$.

Now we are going to prove that after transformation, the $k_{AB}/k_{CD}$ is a fixed value.

\begin{aligned} & x^2 + y^2 = 1, c = 1/\sqrt{3}\\ & E:(-c, 0), F:(c, 0) \\ & A:(x_B, y_B), B:(x_B, y_B)\\\\ & AB: y=k_1(x+c)\\ & (1+k_1^2)x^2 + 2k_1^2cx + k_1^2c^2-1=0\\ & x_A + x_B = \frac{-2k_1^2c}{1+k_1^2}\\ & x_Ax_B = \frac{k_1^2c^2-1}{1+k_1^2}\\\\ & BD: y=k_2(x-c), x=y/k_2+c, k_2 = y_B/(x_B-c)\\ & (1+k_2^2)x^2 - 2k_2^2cx + k_2^2c^2-1=0\\ & (1+1/k_2^2)y^2+2cy/k_2+c^2-1=0 \\ & x_B + x_D = \frac{2k_2^2c}{1+k_2^2} = \frac{2y_B^2c}{(x_B-c)^2+y_B^2}=\frac{2k_1^2c(x_B+c)^2}{1+c^2-2cx_B}\\ & y_B + y_D = \frac{-2c/k_2}{1+1/k_2^2} = \frac{-2ck_2}{1+k_2^2} = \frac{-2c(x_B-c)y_B}{1+c^2 - 2cx_B} = \frac{-2ck_1(x_B^2-c^2)}{1+c^2-2cx_B}\\\\ & AC: y=k_3(x-c), k_3 = y_A/(x_A-c)\\ & x_A + x_C = \frac{2k_1^2c(x_A+c)^2}{1+c^2-2cx_A}\\ & y_A + y_C = \frac{-2ck_1(x_A^2-c^2)}{1+c^2-2cx_A}\\\\ & k_{CD} = \frac{y_C-y_D}{x_C - x_D} = \frac{y_C+y_A - (y_B+y_D)+y_B-y_A}{x_C+x_A-(x_B+x_D)+x_B-x_A}\end{aligned}
\begin{aligned} & y_C+y_A - (y_B+y_D)+y_B-y_A \\ =& \frac{-2ck_1(x_A^2-c^2)}{1+c^2-2cx_A} - \frac{-2ck_1(x_B^2-c^2)}{1+c^2-2cx_B} + y_B - y_A\\ =& -2ck_1\frac{(x_A^2-c^2)(1+c^2-2cx_B)-(x_B^2-c^2)(1+c^2-2cx_A)}{(1+c^2-2cx_A)(1+c^2-2cx_B)} + y_B-y_A\\ =& -2ck_1\frac{(1+c^2)(x_A^2-x_B^2)-2c(x_Bx_A^2-c^2x_B-x_Ax_B^2+c^2x_A)}{(1+c^2)^2-2c(1+c^2)(x_A+x_B)+4c^2x_Ax_B} + y_B -y_A\\ =& -2ck_1(x_B-x_A)\frac{-(1+c^2)(x_A+x_B)-2c(-x_Ax_B-c^2)}{(1+c^2)^2-2c(1+c^2)(x_A+x_B)+4c^2x_Ax_B} + y_B - y_A \\ =& k_1(x_B-x_A)\frac{2c(1+c^2)(x_A+x_B)-4c^2(x_Ax_B+c^2)}{(1+c^2)^2-2c(1+c^2)(x_A+x_B)+4c^2x_Ax_B} + y_B - y_A\end{aligned}
\begin{aligned}& x_C+x_A-(x_B+x_D)+x_B-x_A \\=& \frac{2k_1^2c(x_A+c)^2}{1+c^2-2cx_A} - \frac{2k_1^2c(x_B+c)^2}{1+c^2-2cx_B} + x_B - x_A\\=& 2k_1^2c\frac{(x_A+c)^2(1+c^2-2cx_B)-(x_B+c)^2(1+c^2-2cx_A)}{(1+c^2)^2-2c(1+c^2)(x_A+x_B)+4c^2x_Ax_B} +x_B - x_A\\=& 2k_1^2c\frac{(1+c^2)(x_A^2+2cx_A-x_B^2-2cx_B) -2c(x_Bx_A^2+c^2x_B-x_Ax_B^2-c^2x_A^2)}{(1+c^2)^2-2c(1+c^2)(x_A+x_B)+4c^2x_Ax_B} + x_B-x_A\\=& (x_B-x_A) \frac{-2ck_1^2(1+c^2)(x_A+x_B+2c) + 4c^2k_1^2(x_Ax_B-c^2)}{(1+c^2)^2-2c(1+c^2)(x_A+x_B)+4c^2x_Ax_B} + x_B - x_A\\\end{aligned}
\begin{aligned}k_{CD} &= \frac{y_C+y_A - (y_B+y_D)+y_B-y_A}{x_C+x_A-(x_B+x_D)+x_B-x_A} \\&= \frac{k_1(x_B-x_A)\frac{2c(1+c^2)(x_A+x_B)-4c^2(x_Ax_B+c^2)}{(1+c^2)^2-2c(1+c^2)(x_A+x_B)+4c^2x_Ax_B} + y_B - y_A}{(x_B-x_A) \frac{-2ck_1^2(1+c^2)(x_A+x_B+2c) + 4c^2k_1^2(x_Ax_B-c^2)}{(1+c^2)^2-2c(1+c^2)(x_A+x_B)+4c^2x_Ax_B} + x_B - x_A} \\&= k_1\frac{2c(1+c^2)(x_A+x_B)-4c^2(x_Ax_B+c^2)+P}{-2ck_1^2(1+c^2)(x_A+x_B+2c)+4c^2k_1^2(x_Ax_B-c^2)+P}\\&= k_1\frac{-4c^2(1+c^2)k_1^2-4c^2(2k_1^2c^2+c^2-1)+P(k_1^2+1)}{-2ck_1^2(1+c^2)(2c)-4c^2k_1^2(c^2+1)+P(k_1^2+1)}\\&= k_1\frac{-16k_1^2-8(k_1^2-1)+36k_1^2+4}{-16k_1^2-16k_1^2+36k_1^2+4}\\&= k_1\frac{12k_1^2+12}{4k_1^2+4}\\&= 3k_1 \\&= 3k_{AB}\end{aligned}
\begin{aligned} \frac{k_{CD}}{k_{AB}} =& \frac{-4c^2(1+c^2)k_1^2-4c^2(2k_1^2c^2+c^2-1)+(k_1^2+1)(1+c^2)^2+4c^2k_1^2(1+c^2)+4c^2(k_1^2c^2-1)}{-8c^2k_1^2(1+c^2)+(k_1^2+1)(1+c^2)^2+4c^2k_1^2(1+c^2)+4c^2(k_1^2c^2-1)} \\ =& \frac{(-4c^2-4c^4-8c^4+(1+c^2)^2+4c^4+4c^2+4c^4)k_1^2-4c^2(c^2-1)+(1+c^2)^2-4c^2}{(-8c^2-8c^4+(1+c^2)^2+4c^4+4c^2+4c^4)k_1^2+(1+c^2)^2-4c^2} \\ =& \frac{(-3c^4+2c^2+1)k_1^2-3c^4+2c^2+1}{(c^4-2c^2+1)k_1^2+c^4-2c^2+1} \\ =& \frac{-(3c^2+1)(c^2-1)}{(c^2-1)^2} \\ =& \frac{1+3c^2}{1-c^2}\end{aligned}

## Problem 3.

Ellipse $x^2/3 + y^2/2 = 1$ intersects line $l$ on points $P(x_p, y_p), Q(x_q, y_q)$ such that the area of $\triangle OPQ$ is $\sqrt{6}/2$, or $S_{\triangle OPQ} = \sqrt{6}/2$.

1. Prove $x_p^2 + x_q^2$ and $y_p^2 + y_q^2$ are fixed values
2. Let $M$ be the midpoint of $PQ$, find maximum value of $|OM|\cdot|PQ|$
3. Let $\triangle DEG$ inscribes on the ellipse such that $S_{\triangle ODE} = S_{\triangle OEG} = S_{\triangle OGD} = \sqrt{6}/2$, check the existence of $\triangle DEG$ and its shape if it exists.

### Solution:

We transform the $x-y$ plane by: $(x, y) \rightarrow (x/\sqrt{3}, y/\sqrt{2})$, such that the ellipse becomes a unit circle. Since, $S_{\triangle OPQ} = \frac{1}{2}(|x_Q|-|x_P|)(|y_Q|+|y_P|)$ or $S_{\triangle OPQ}=\frac{1}{2}(|x_Q|+|x_P|)(|y_Q|-|y_P|)$, $S_{\triangle OPQ} \rightarrow S_{\triangle OPQ}/\sqrt{6}$, which means after transformation $S_{\triangle OPQ} = 1/2$.

We found that $S_{\triangle OPQ} = \sin \angle POQ /2 = 1/2$, therefore $\sin\angle POQ = 1$, which means $OP\bot OQ$.

We can set $P(\cos \alpha, \sin \alpha), Q(-\sin \alpha, \cos \alpha)$ as they are on unit circle after transformation.

### Solution 3.1:

We can see that the value of $x_p^2 + x_q^2 = \cos^2\alpha + \sin^2\alpha = 1$ is $1/3$ of its original and $y_p^2+y_q^2 = \sin^2\alpha + \cos^2\alpha=1$ is $1/2$ of its original. Therefore on original ellipse,

$$x_p^2 + x_q^2 = 3, \quad y_p^2+y_q^2=2$$

### Solution 3.2:

The original points on ellipse are $P(\sqrt{3}\cos\alpha, \sqrt{2}\sin\alpha), Q(-\sqrt{3}\sin\alpha, \sqrt{2}\cos\alpha)$, therefore the original value

\begin{aligned} |OM|^2\cdot|PQ|^2 &= (3(\frac{\cos\alpha - \sin\alpha}{2})^2+2(\frac{\sin\alpha + \cos\alpha}{2})^2)(3(\cos\alpha+\sin\alpha)^2+2(\sin\alpha-\cos\alpha)^2)\\ &= \frac{1}{4}(3-6\cos\alpha\sin\alpha+2+4\cos\alpha\sin\alpha)(3+6\cos\alpha\sin\alpha+2-4\cos\alpha\sin\alpha) \\ &= \frac{1}{4}(5-\sin2\alpha)(5+\sin2\alpha) \\ &= \frac{1}{4}(25-\sin^22\alpha) \leq \frac{25}{4}\end{aligned}

Therefore the maximum value of $|OM|\cdot |PQ|$ is $5/2$.

### Solution 3.3:

If $S_{\triangle ODE} = S_{\triangle OEG} = S_{\triangle OGD} = \sqrt{6}/2$, then $OD \bot OE, OE \bot OG, OD \bot OG$, which is not possible on $xy$ plane, therefore $\triangle DEG$ does not exists.

## Problem 4.

A circle $O$ has an inscribed triangle $\triangle ABC$ such that $AB > BC$. Let $E$ be the midpoint of arc $\overset{\frown}{AC}$, $EF \perp AB$ on F. Prove $|AF| = |FB| + |BC|$.

### Proof:

Let circle's radius be $r$, and

\begin{aligned} & A:(r\cos\alpha, r\sin\alpha) \\ & B:(r\cos\beta, r\sin\beta) \\ & C:(r, 0) \\ & \pi > \alpha - \beta > \beta > 0 \\\end{aligned}

Then:

\begin{aligned} & E:(r\cos\alpha/2, r\sin\alpha/2) \\ & |BC| = 2r\sin\frac{\beta}{2} \\ & |AB| = 2r\sin\frac{\alpha-\beta}{2} \\ & |EB| = 2r\sin\frac{\alpha-2\beta}{4} \\ & |AE| = 2r\sin\frac{\alpha}{4} \\ & |BF| = |EB| \cos\angle EBA \\ & \cos\angle EBA = \frac{|EB|^2 + |BA|^2-|EA|^2}{2|EB||AB|} \\ & |AF| = |FB| + |BC| \Leftrightarrow |AB| = 2|BF| + |BC| \Leftrightarrow \cos\angle EBA = \frac{|AB| - |BC|}{2|BE|} \\ & \Leftrightarrow \cos\angle EBA = \frac{|EB|^2 + |BA|^2-|EA|^2}{2|EB||AB|} = \frac{|AB| - |BC|}{2|EB|} \\ & \Leftrightarrow |EB|^2 + |BA|^2-|EA|^2 = |AB|(|AB|-|BC|) \\ & \Leftrightarrow |EB|^2 + |AB||BC| = |EA|^2 \\ & \Leftrightarrow \sin^2\frac{\alpha-2\beta}{4} + \sin\frac{\alpha-\beta}{2}\sin\frac{\beta}{2} = \sin^2\frac{\alpha}{4}\\ & \Leftrightarrow 1 - \cos\frac{\alpha-2\beta}{2} - \cos\frac{\alpha}{2}+\cos\frac{\alpha-2\beta}{2} = 1 - \cos\frac{\alpha}{2} \\ & \Leftrightarrow 0 = 0\end{aligned}