Problem 2: Find all solutions of $\sqrt{x+10}-\dfrac{6}{\sqrt{x+10}}=5$. Solution
Let $y=\sqrt{x+10}$, we have $$y-\dfrac{6}{y}=5$$ Multiplying $y$ on both side of the equation, we have $$y^2-6=5y$$ Moving $5y$ to the left side of the above equation and factorizing it, we have $$(y-6)(y+1)=0$$ Therefore $y=6$, $y=-1$. Obviously, $y=-1$ is invalid as $\sqrt{x+10}>0$. Therefore, $\sqrt{x+10}=6$, which implies $x+10=36$. Therefore $x=\boxed{26}$.
Problem 3: Find all possible values of $(a, b)$, so that $2a+b=12$, and $ab=3$. Solution
Since $ab=3$, we have $2ab=6$. According to Vieta’s theorem, $2a$ and $b$ are the roots of equation $$x^2-12x+6=0$$. Solving the above equation, we have $$2a=6\pm\sqrt{30},\ b=6\mp\sqrt{30}$$ Therefore, there are two solutions of $(a, b)$, as $\boxed{(3+\dfrac{\sqrt{30}}{2}, 6-\sqrt{30})}$ and $\boxed{(3-\dfrac{\sqrt{30}}{2}, 6+\sqrt{30})}$
Problem 4: $4x^2-24x+c$ is a perfect square for all integer $x$. Find the value of $c$. Solution
Let $(ax+b)^2=4x^2-24x+c$, we have $a^2x^2+2abx+b^2=4x^2-24x+c$. Therefore $a^2=4$, $2ab=-24$, and $b^2=c$. This results in $a=\pm 2$, $b=\mp 6$, and $c=b^2=\boxed{36}$.
Problem 5: Find all $z$ such that $9^{z-1}-3^{z-1}-2=0$. Solution
Let $3^{z-1}=x$, we have $x^2-x-2=0$. Solving the equation, $x=2$, $x=-1$. Obviously $x=-1$ is invalid, as $3^{z-1}>0$. Therefore $3^{z-1}=2$, which implies $3^z=6$. Therefore $z=\boxed{\log_{3}6}$.
Problem 6: Find all solutions to the equation $x+\sqrt{x-2}=4$. Solution
Let $y=\sqrt{x-2}$, we have $y^2+y-2=0$. $y=1$, $y=-2$. Obviously, $y=-2$ is invalid, as $\sqrt{x-2}\ge 0$. Therefore $\sqrt{x-2}=1$, which leads $x=\boxed{3}$.
Problem 7: If $\dfrac{a+b}{a}=\dfrac{b}{a+b}$, then (A) No solutions for $a$ and $b$. (B) $a$ and $b$ cannot be both real. (C) Both $a$ and $b$ are imaginary. (D) One of $a$ or $b$ is real, the other imaginary. (E) Both $a$ and $b$ are real. Solution
Obviously $a\ne 0$, $a+b\ne 0$. Multiplying both side with $a(a+b)$, we have $(a+b)^2=ab$. i.e. $a^2+ab+b^2=0$. We have $$a=\dfrac{-b\pm\sqrt{-3b^2}}{2},\ b=\dfrac{-a\pm\sqrt{-3a^2}}{2}$$
If $b$ is real, then $\sqrt{-3b^2}$ is imaginary, and $a$ is imaginary. Similarly, if $a$ is real, then $b$ is imaginary.
If $b$ is imaginary, $a$ could be either real (for example, when $b=1+i\sqrt{3}$, or imaginary (for example, when $b=i$). Similarly if $a$ is imaginary, $b$ could be either real, or imaginary. Therefore the correct answer is (B).
Problem 8: Find all solutions to the equation $\dfrac{3^{x^2}}{27^x}=\dfrac{1}{9}$ Solution
Transforming the equation as $$3^{x^2-3x}=3^{-2}$$, we have $$x^2-3x=-2$$. Therefore $x=\boxed{1}$, or $x=\boxed{2}$.
Problem 9: How many solutions are there for equation $|x^2-5|=4$? Solution
Transforming the equation as $x^2-5=\pm 4$, we have either $x^2-5=4$, or $x^2-5=-4$, which leads to $x^2=9$, or $x^2=1$. Therefore $x=\pm 3$, or $x=\pm 1$. Therefore there are $\boxed{4}$ solutions.
Problem 10: Find the sum of all solutions for the equation $1+\dfrac{2}{x}=x$ Solution
Transforming the equation as $x^2-x-2=0$. According to Vieta’s theorem, the sum of the roots is $-\dfrac{b}{a}=-\dfrac{-1}{1}=\boxed{1}$.
Problem 11: Simplify $\sqrt{53-8\sqrt{15}}$. Solution
The simplified form should be either $x+y\sqrt{15}$, or $x\sqrt{3}+y\sqrt{5}$.
If the answer is in the form of $x+y\sqrt{15}$, then $(x+y\sqrt{15})^2=x^2+15y^2+2xy\sqrt{15}$, which leads to $$x^2+15y^2=53,\ 2xy=-8$$
Solving the above equation, we have $$(x, y)=(\pm 4\sqrt{3}, \mp\dfrac{\sqrt{3}}{3})$$ $$(x, y)=(\pm\sqrt{5},\mp\dfrac{4\sqrt{5}}{5})$$
Verifying the above 4 solutions, only $(x,y)=(-\sqrt{5},\dfrac{4\sqrt{5}}{5})$ and $(x,y)=(4\sqrt{3},-\dfrac{\sqrt{3}}{3})$ are the valid answers, both of them lead to $\sqrt{53-8\sqrt{15}}=\boxed{4\sqrt{3}-\sqrt{5}}$.
If the answer is in the form of $x\sqrt{3}+y\sqrt{5}$, then $(x\sqrt{3}+y\sqrt{5})^2=3x^2+5y^2+2xy\sqrt{15}$, which leads to $$3x^2+5y^2=53,\ 2xy=-8$$
Solving the above equation, we have $$(x, y)=(\pm 4,\mp 1)$$ $$(x,y)=(\pm\dfrac{\sqrt{15}}{3},\mp\dfrac{4\sqrt{15}}{5})$$
Verifying the above solutions, only $(x,y)=(4,-1)$ and $(x,y)=(-\dfrac{\sqrt{15}}{3},\dfrac{4\sqrt{15}}{5})$ are valid answers, both of them lead to $\sqrt{53-8\sqrt{15}}=\boxed{4\sqrt{3}-\sqrt{5}}$.
Problem 12: Find all solutions to the equation $\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}=4$ Solution
Squaring both side, we have $$\left(\sqrt{x+\sqrt{x+11}}+\sqrt{x-\sqrt{x+11}}\ \right)^2=16$$
Simplifying it, we have $$2x+2\sqrt{x^2-x-11}=16$$
Dividing both side by 2 and moving $x$ to the right side, we have $$\sqrt{x^2-x-11}=8-x$$
Squaring both side again, we have $$x^2-x-11=64-16x+x^2$$ Therefore $15x=75$, $x=5$. Verifying the result, we have $x=\boxed{5}$.
Posted inAlgebra|Comments Off on HCS Summer School Exam 1 – 2025
$x^2y^2=1$ forms $4$ hyperbola branches, as $y=\dfrac{1}{x}$ and $y=-\dfrac{1}{x}$ combined. What is the smallest triangle in terms of area that it intersects all $4$ branches.🔑
Solution: Obviously, to be the smallest triangle, the vertices of the triangle must be on the hyperbolas, or in the interior area with edges tangents to the hyperbolas, or any combinations of the above two cases.
Otherwise, if any edge intersects a hyperbolas, or a vertex is outside of the interior area, a smaller triangle can be always found by moving the edge tangent to the hyperbola, or moving the vertex toward the interior region.
Case 1: There is only two vertices on two adjacent branches, and the third one in the interior with its two edges that tangent to the other two branches.
Step 1 – tangent line to $xy=k$ For $xy=k$, the tangent at $(x_0,y_0)$, with $x_0y_0=k$, is $$y_0x+x_0y=2k$$
Step 2 – find tangent parameters For $y=−\dfrac{1}{x}$ (i.e. $xy=−1$), let the tangent point be $(t,−\dfrac{1}{t})$, where $t>0$. The tangent line is: $$−\dfrac{1}{t}x+ty=−2\tag{1}$$
This line must pass through $B=(−b,−\dfrac{1}{b})$, where $b>0$. Substituting gives $$\dfrac{b}{t}−\dfrac{t}{b}=−2$$
so $t$ satisfies $$t^2−2bt−b^2=0$$ The root with $t<0$ is $t=(1−\sqrt{2})b$. Let $\gamma=\sqrt{2}-1$, we have $t=-b\gamma$.
For $y=\dfrac{1}{x}$ (i.e. $xy=1$), let the tangent point be $(s,\dfrac{1}{s})$, where $s>0$. The tangent line is: $$\dfrac{1}{s}x+sy=2\tag{2}$$
This line must pass through $C=(c,−\dfrac{1}{c})$, where $c>0$. Substituting gives $$\dfrac{c}{s}-\dfrac{s}{c}=2$$
so $s$ satisfies $$s^2+2cs−c^2=0$$ The root with $s>0$ is $s=c(\sqrt{2}−1)=c\gamma$.
Step 3 – intersection $A=(u,v)$ of the two tangents Solving the two linear tangent equations $(1)$ and $(2)$, for value of $(x,y)=(u,v)$ gives $$u=\dfrac{2st(s+t)}{s^2+t^2}\ \ \ \ \ \ \ v=\dfrac{2(s−t)}{s^2+t^2}$$
Simplifying the above with $s=c\gamma$, $t=-b\gamma$, we have $$u=\dfrac{2bc(b−c)\gamma}{b^2+c^2}\ \ \ \ \ \ v=\dfrac{2(b+c)}{(b^2+c^2)\gamma}$$
Step 4 – area of $\triangle{ABC}$ Use the determinant formula $$[\triangle{ABC}]=\dfrac{1}{2}∣det([B−A, C−A])∣$$
Carrying out the algebra, by substituting the A,B,C coordinates above, yields the closed form $$[\triangle{ABC}]=\dfrac{((b−c)^2\gamma+2bc)((b+c)^2+2bc\gamma)}{2bc(b^2+c^2)\gamma}$$
where $\gamma=\sqrt{2}−1$.
Replacing $\gamma=\sqrt{2}−1$ gives the following: $$[\triangle{ABC}]=\dfrac{((\sqrt{2}−1)(b−c)^2+2bc)((b+c)^2+2(\sqrt{2}−1)bc)}{2(\sqrt{2}-1) bc (b^2+c^2)}$$
Simplifying the above formula, we have $$[\triangle{ABC}]=\dfrac{(b^2+c^2+2\sqrt{2}bc)^2}{2bc(b^2+c^2)}$$
Step 5 – smallest area of $\triangle{ABC}$
Let $x=\dfrac{b}{c}$, where $x>0$, we have $$[\triangle{ABC}]=f(x)=\dfrac{(x^2+2\sqrt{2}x+1)^2}{2x(x^2+1)}$$
Because $$f'(x)=\dfrac{x^6-7x^4+7x^2-1}{2x^2(x^2+1)^2}=\dfrac{(x^2-1)(x^4-6x^2+1)}{2x^2(x^2+1)^2}$$
Let $f'(x)=0$, we find the critical points as $$x^2=1, \ \ \ \ \ \ x^2=3\pm2\sqrt{2}$$ Because $x>0$, we have $$x=1,\ \ \ \ \ \ x=\sqrt{2}-1,\ \ \ \ \ \ $x=\sqrt{2}+1$$
Evaluating $f(1)$, $f(\sqrt{2}-1)$ and $f(\sqrt{2}+1)$, we have $$f(1)=3+2\sqrt{2}\ \ \ \ \ \ f(\sqrt{2}-1)=4\sqrt{2}\ \ \ \ \ \ f(\sqrt{2}+1)=4\sqrt{2}$$
As $t\rightarrow\infty$, $f(x)\rightarrow \infty$, the minimum value of $[\triangle{ABC}]=4\sqrt{2}$ when $x=\dfrac{b}{c}=\sqrt{2}\pm 1$.
Case 2: All three vertices on three branches, and one edge tangents to the remaining branch, two vertices are the tangent points of other two edges
Let $A=(-a,\dfrac{1}{a})$, $B=(b, \dfrac{1}{b})$, $C=(-c,-\dfrac{1}{c})$, and $a,b,c>0$. The equation of line $BC$ is $$\dfrac{y-\frac{1}{b}}{-\dfrac{1}{c}-\dfrac{1}{b}}=\dfrac{x-b}{-c-b}$$ Simplifying it, we have $$y=\dfrac{1}{bc}x+\dfrac{1}{b}-\dfrac{1}{c}$$
Because $BC$ intersects to the branch in the 4th quadrant, we have $b>c$ and $$-\dfrac{1}{x}=\dfrac{1}{bc}x+\dfrac{1}{b}-\dfrac{1}{c}$$ i.e. $$x^2+(c-b)x+bc=0\tag{3}$$ Because $BC$ is tangent to the branch in the 4th quadrant, Equation (1) has only one root. Therefore $$\Delta=(c-b)^2-4bc=0$$ i.e $$b^2-6bc+c^2=0$$ Solving $b$ in the above equation, we have $$b=(3\pm2\sqrt{2})c$$ Given $b>c$, we have $$b=(3+2\sqrt{2})c$$
Use shoelace formula, $$[\triangle{ABC}]=\dfrac{1}{2}(x_ay_c-x_cy_a+x_cy_b-x_by_c+x_by_a-x_ay_b)$$ $$=\dfrac{1}{2}((-a)\cdot(-\dfrac{1}{c})-(-c)\cdot\dfrac{1}{a}+(-c)\cdot\dfrac{1}{b}-b\cdot(-\dfrac{1}{c})+b\cdot\dfrac{1}{a}-(-a)\cdot\dfrac{1}{b})$$ $$=\dfrac{1}{2}(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{b}{c}-\dfrac{c}{b})$$
Because $b=(3+2\sqrt{2})c$, we can simplify the above equation as $$[\triangle{ABC}]=(2+\sqrt{2})\dfrac{c}{a}+(2-\sqrt{2})\dfrac{a}{c}+2\sqrt{2}\tag{4}$$
By AM-GM inequality, we have $$(2+\sqrt{2})\dfrac{c}{a}+(2-\sqrt{2})\dfrac{a}{c}\ge 2\sqrt{(2+\sqrt{2})\dfrac{c}{a}\cdot(2-\sqrt{2})\dfrac{a}{c}}\ge 2\sqrt{2}\tag{5}$$
The equality holds when $$(2+\sqrt{2})\dfrac{c}{a}=(2-\sqrt{2})\dfrac{a}{c}$$
Because $a,b,c>0$, solve the above equation, we have $$a=(\sqrt{2}+1)c$$
Therefore, combining $(4)$ and $(5)$, the minimum value of $[\triangle{ABC}]$ is $4\sqrt{2}$, when $$a=(\sqrt{2}+1)c$$ $$b=(3+2\sqrt{2})c$$ or $$b=(\sqrt{2}+1)a$$ $$c=(\sqrt{2}-1)a$$
which is a covered by Case 1.
Therefore, the smallest area of triangles intersecting all 4 hyperbola branches of $x^2y^2=1$ is $\boxed{4\sqrt{2}}$.
Two opposite edges of a unit square are folded along one of the diagonal to form a parallelogram. Then, those two opposite edges on the parallelogram are folded together to form a trapezoid. Find the area of the trapezoid.🔑
Solution: Edge $AD$ and $CB$ of unit square $ABCD$ are folded along diagonal $AC$, forming parallelogram $AECF$. Paralleogram $AECF$ is folded along $GH$, forming trapezoid $AIJB$.
Let $x=BE$, Because $B’E=BE$, and $\triangle{AB’E}$ is an isosceles right triangle, and $AE=AB-BE=1-x$, by Pythagorean Theorem, we have $$x^2+x^2=(1-x)^2$$
Solving the above equation and ignoring the invalid $x$ value, we have $x=\sqrt{2}-1$. Therefore
Let $P(x)$ is a polynomial with integer coefficients so that $P(d)=\dfrac{2025}{d}$, where $d$ is a positive divisor of $2025$. Find $P(x)$.🔑
Claim: There is no $P(x)$ to satisfy $P(d)=\dfrac{2025}{d}$, where $d$ is a positive divisor of $2025$.
Lemma: If $P(x)$ is a polynomial with integer coefficients, and $a$ and $b$ are different integers. Then $P(a)-P(b)$ is a multiple of $a-b$.
Let $P(x)=\sum_{i=0}^{n}a_ix^i$. Then $P(a)-P(b)=\sum_{i=0}^{n}a_i(a^i-b^i)$. Since $a^i-b^i$ is a multiple of $a-b$, and $a_i$ are integers, therefore $P(a)-P(b)$ is a multiple of $a-b$.
Nine light bulbs are equally spaced around a circle. When the power is turned on, each of the nine light bulbs turns blue or red, where the color of each bulb is determined randomly and independently of the colors of the other bulbs. Each time the power is turned on, the probability that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle is $\dfrac{m}{n}$, where m and n are relatively prime positive integers. Find $m+n$. (PurpleComet 2023, Middle School, Problem 20) Click here for the solutions.
Solution1: Overall, there are total $2^9=512$ combinations when the light bulbs are turned blue or red randomly. The problem can be divided into following cases, depending on the number of light bulbs turning blue, for satisfying the requirement that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle.
Case 1: The number of the blue bulbs is $0$. There is only $1$ way to arrange the bulbs to satisfy the requirement:
Case 2: The number of the blue bulbs is $1$. There is $0$ way to arrange the bulbs to satisfy the requirement, as the color of the two adjacent bulb of the blue bulb is red.
Case 3: The number of the blue bulbs is $2$. These two blue bulbs must be next to each other, and there are $9$ ways to arrange the bulbs to satisfy the requirement:
Case 4: The number of the blue bulbs is $3$. These three blue bulbs must be next to each other, and there are $9$ ways to arrange the bulbs to satisfy the requirement:
Case 5: The number of the blue bulbs is $4$. This can be divided into the following sub-cases:
Case 5.1: If the $4$ blue bulbs are arranged to be next to each other and there are $9$ ways to arrange the bulbs to satisfy the requirement:
Case 5.2: If there are only $3$ blue bulbs are arranged to be next to each other, There is $0$ way to arrange the bulbs to satisfy the requirement, as the color of the two adjacent bulb of the remaining blue bulb is red.
Case 5.3: If the $4$ blue bulbs are arranged to $2$ groups, and each group contains $2$ blue blubs that are arrange to be next to each other. There must be $2$ or $3$ red blubs in-between these $2$ groups. There are $9$ ways to arrange these $5$ red bulbs in order to satisfy the requirement:
Case 6: If there are $5$, $6$, $7$, $8$, and $9$ blue bulbs, there are $4$, $3$, $2$, $1$, and $0$ red bulbs, which would be same for case 1, 2, 3, 4, and 5.
Therefore, the number of arrangement of $9$ light bulbs satisfying the requirement is $$2\cdot(1+9+9+9+9)=74$$
The probability that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle is $$\dfrac{74}{512}=\dfrac{37}{256}$$ Therefore $m=37$, $n=256$, and $$m+n=37+256=\boxed{293}$$
Solution 2: Let $n$ be the number of light bulbs on the circle. There will be $2^n$ combinations when the light bulbs are turned blue or red randomly. We try to examine combinations $f(n)$ satisfying the requirement that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle.
If a red blub is assigned as bit $0$, and a blue blub as bit $1$. There will be $2^n$ $n$-bit binary numbers. In order to satisfy the requirement, the binary number meets all the following conditions:
1. It cannot contain a sequence of $101$, nor $010$
2. It cannot start with $01$ and end up with a $1$
3. It cannot start with $10$ and end up with a $0$
4. It cannot starts with a $1$ and end up with $10$
5. It cannot starts with a $0$ and end up with $01$
