Given \(a,b,c,d\) are all real numbers, consider the quadratic equation:
 \[
 x^2 + (132a^2 + 66b)\,x + (66a^2 + 33b)^2 = 0
 \]
 Let's complete the square to rewrite this in the form:
 \[
 (x + c)^2 = d
 \]
 If \(a=1, b=-1\), what is \(c^2- \sqrt{d}\)?

Given \(a,b,c,d\) are all real numbers, consider the quadratic equation:

 \[
 x^2 + \left(132a^2 + 66b\right)x + 8099a^4 + 1090b^2 + 4356a^2b = 0
 \]

 Let's complete the square to rewrite this in the form:

 \[
 (x + c)^2 = d
 \]

 If \(a=1, b=-1\), what is \(c^2- \sqrt{d}\)?

Through calculation, it can be determined that the discriminant is negative, so \(d\) is negative. Therefore \( \sqrt{d} \) is invalid.

Answer: 1089


Find the solution set of the inequality $2a^2+8a\le-6$.

The solution set of the inequality \(x^2+x+2 \le 2^b\) is \( a \le x \le 3\). Determine the value of \(a+b\).

\(x^2+x+2=(x+\tfrac12)^2+\tfrac74\ge \tfrac74\) for all real \(x\). Hence \(x^2+x+2+2^a ;\ge; \tfrac74+2^a ;>;0\) for every \(x\). The inequality (x^2+x+2+2^a\le 0) has no real solution.

Answer: -2


Suppose \( a \) and \( b \) are complex numbers satisfying the conditions \( |2a - b| = 25 \), \( |a + 2b| = 5 \), and \( |a + b| = 2 \). Determine the value of \( |a| \).

Suppose \( a \) and \( b \) are complex numbers such that \( |2a - b| = 25 \), \( |a + 2b| = \sqrt{435} \), and \( |a + b| = 2 \). Determine the value of \( |a|^2 \).

Let
 \[
 X=|a|^2,\;Y=|b|^2,\;Z=\Re(a\overline b).
 \]
 Then
 \[
 |2a-b|^2=4X-4Z+Y=625,\quad
 |a+2b|^2=X+4Z+4Y=435,\quad
 |a+b|^2=X+2Z+Y=4.
 \]
 Solving gives \(X=-1\), which contradicts \(X=|a|^2\ge0\). Therefore no such \(a,b\) exist and \(|a|^2\) is undefined.

Answer: 9


Solve for this inequality
 \[
 \frac{y^2}{y+1} \ge \frac{2}{y-1} + \frac{5}{4}.
 \]

The solution of the inequality \( \frac{y^{2}}{y + 1} \geq \frac{2}{y - 1} + c \) is \( -1<y<1) \). Find the value of \( c \).

Let
 \[
 f(y)=\frac{y^2}{y+1}-\frac{2}{y-1}-c.
 \]
 As \(y\to+\infty\),
 \[
 \frac{y^2}{y+1}=y-1+o(1),\quad \frac{2}{y-1}=o(1),
 \]
 so
 \[
 f(y)\sim y-1-c\to+\infty.
 \]
 Thus \(f(y)\ge0\) for all sufficiently large \(y\), making the solution set unbounded above. No finite \(c\) can restrict the solutions to exactly \((-1,1)\).

Answer: (-1, 1) \cup [3, \infty)


Let \( p \) and \( q \) be the two real roots of the equation \( y^2 - 4y + 7 = 19 \) and \( p \geq q \). Compute \( 2p + q \).

Let \( p \) and \( q \) be the two real roots of the equation \( y^2 - ay + 7 = 5 \). For which real values of \(a\) is \( \sqrt{(pq)^2 - (p + q)^2} \) a real number?

The equation \(y^2 - a y + 7 = 5\) simplifies to \(y^2 - a y + 2 = 0\), so
 \[
 p + q = a,\quad pq = 2.
 \]
 Real roots require \(a^2 - 8 \ge 0\), i.e.\ \(|a|\ge2\sqrt2\). Meanwhile,
 \[
 (pq)^2 - (p+q)^2 = 4 - a^2,
 \]
 which is negative whenever \(|a|\ge2\sqrt2\). Therefore \(\sqrt{(pq)^2 - (p+q)^2}\) is not real while \(pq\) is, making any real comparison impossible.

Answer: 10


Find the value of \( y \) that satisfies the equation
 \[
 \sqrt[3]{20y + \sqrt[3]{20y + 13}} = 13.
 \]

For \( y \) that satisfies the equation
 \[
 \sqrt{\sqrt{20y + \sqrt{20y + 0.5}}} = \frac{\sqrt{2}}{2},
 \]
determine the radius of the circumcircle of a triangle with side lengths \( y \), \( y^{2} \), and \( y + 1 \).

Squaring step by step gives the single real root
 \[
 y=-\frac1{80}.
 \]
 The supposed side‐lengths would be
 \[
 y=-\tfrac1{80},\quad y^2=\tfrac1{6400},\quad y+1=\tfrac{79}{80},
 \]
 but a triangle cannot have a negative side. Hence no such triangle exists and its circumcircle radius is undefined.

Answer: \frac{546}{5}


Calculate the total entries in the first 20 rows of Pascal's Triangle, counting from the 0th row up to the 19th row.

In the first 8 rows of Pascal’s Triangle (from the 0th row to the 7th row), there are \( a \) total entries, among which \( b \) entries are equal to 6. Let \( x \) be the least common multiple of \( a \) and \( b \). Then, what is the smallest possible value of \( y \) such that \( y \) regular \( x \)-sided polygons can seamlessly tile a finite, closed 3D polyhedron in Euclidean space?

Rows 0–7 of Pascal’s Triangle have
 \[
 a=\sum_{n=0}^7(n+1)=36
 \]
 and the entry 6 appears once in row 4 and twice in row 6, so
 \[
 b=3,\quad x=\mathrm{lcm}(36,3)=36.
 \]
 A regular 36‑gon has interior angle \(\tfrac{(36-2)\cdot180°}{36}=170°\). Any finite closed polyhedron (non‑degenerate) requires at least three faces meeting at each vertex, but \(3\times170°=510°>360°\). Hence no such tiling is possible.

Answer: 210


Points \( P \), \( Q \), \( R \), and \( S \) lie on a straight line in that order. Given \( PQ = 2 \) units, \( QR = 5 \) units, and \( PS = 14 \) units, find the ratio of \( PR \) to \( QS \). Express your answer as a simplified fraction.

Points \( P \), \( Q \), \( R \), and \( S \) lie in order on a straight line. If \( PQ = 2 \) units, \( QR = 5 \) units, and \( PS = 14 \) units, what is the area of the convex quadrilateral whose side lengths are \( PQ \), \( RS \), \( QS \), and \( QR \)?

From collinear P–Q–R–S with PQ=2, QR=5, PS=14, we get RS=7 and QS=12, so the quadrilateral’s sides are 2, 5, 7, 12. Four side lengths alone do not fix a convex quadrilateral’s area—there is a continuum of possible shapes and areas unless an angle or cyclic condition is given—so the problem is underspecified.

Answer: \frac{7}{12}


Calculate the sum
 \[
 \frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}+\frac{1}{3^{5}}+\frac{1}{3^{6}}
 \]
 and express your answer as a common fraction.

\(\sum_{k = 1}^{9}\frac{k}{3^{k}} = \frac{p}{q}\), where \(p\) and \(q\) are positive integers with \(\gcd(p, q) = 1\). If \(p\) leaves a remainder of 0 when divided by 3, determine the value of \(p\).

Use the identity
 \[
 \sum_{k=1}^n kx^k=\frac{x\bigl[1-(n+1)x^n+nx^{n+1}\bigr]}{(1-x)^2}
 \]
 with \(x=\tfrac13\), \(n=9\), to get
 \[
 \sum_{k=1}^9\frac{k}{3^k}=\frac{4919}{6561},
 \]
 so \(p=4919\). But \(4919\equiv2\pmod3\), not \(0\), contradicting the given condition—hence no solution exists.

Answer: \frac{364}{729}


Two concentric circles share the same center \( P \). The larger circle has a radius of \( 10 \), while the smaller circle has a radius of \( 6 \). FDetermine the area of the annular region lying between the two circles.

The area of the annular region between two concentric circles is \(64\pi\), and the radii of both circles are integers not greater than 15. After decreasing both radii by \(x\), the area of the resulting annulus becomes \(8\pi\). Determine the value of \(x\).

Answer: 64\pi


Let \(n=2018\) and define
\[
p(y)=\prod_{k=1}^n (y+b_k).
\]
From the problem statement, for each \(i\) we have \(p(a_i)=2018\). Note that \(p(y)\) is a monic polynomial of degree \(n\). Therefore, \(p(y)-2018\) is also a monic polynomial of degree \(n\), and its roots are exactly \(a_1,\dots,a_n\). Thus,
\[
p(y)-2018=\prod_{i=1}^n (y-a_i).
\]

Now fix some \(j\). Since \((-b_j)-a_i=-(a_i+b_j)\) and \(n\) is even, it follows that
\[
\prod_{i=1}^n (a_i+b_j)=\prod_{i=1}^n\bigl((-b_j)-a_i\bigr)
=\Bigl[\prod_{i=1}^n (y-a_i)\Bigr]_{y=-b_j}
=p(-b_j)-2018.
\]

But we have
\[
p(-b_j)=\prod_{k=1}^n(-b_j+b_k)=0
\]
(since the factor with \(k=j\) equals zero). Therefore,
\[
\prod_{i=1}^n (a_i+b_j)=-2018<0.
\]

Hence \(\sqrt{\prod_{i=1}^n (a_i+b_j)}\) does not exist in the real numbers, so it cannot be compared with the real number \(2025\).

Answer: 729


A certain species of sloth absorbs only \(25\%\) of the fiber it consumes. One sloth absorbed 10.5 ounces of fiber in a single day. Calculate how many ounces of fiber it ingested that day.

For a certain sloth species, the relationship between the amount of food \(u\) consumed and the amount of fiber \(v\) it absorbs is given by \(v=(c - 3)u^{2}+\frac{u}{3}-7\). The parameter \(c\) is an integer and remains constant across all sloths. The maximum \(u\) for male sloths is 50, while for female sloths, it is strictly less than that. A female sloth named Lila absorbed 10.5 ounces of fiber in one day after eating \(u\) ounces of food. The value of \(c\) for Lila satisfies \([2025.3c]=6075\). A male sloth named Max ate \(u' = 0.88u\) ounces of food that day. Calculate how many ounces of fiber Max ingested that day.

\([2025.3\,c]=6075\) means \(\lfloor2025.3\,c\rfloor=6075\). For integer \(c\), only \(c=3\) gives \(\lfloor2025.3\times3\rfloor=\lfloor6075.9\rfloor=6075\).
 Then \(v=(c-3)u^2+\tfrac u3-7=\tfrac u3-7\).- Setting \(v=10.5\) ⇒ \(\tfrac u3-7=10.5\) ⇒ \(u=52.5\).
 But female sloths have \(u<50\), so \(u=52.5\) is invalid ⇒ no valid \(u\) ⇒ cannot find \(u'\) or Max’s \(v\).

Answer: 42


Determine the leading coefficient of the polynomial expression \(-3(y^4 - y^3 + y) + 7(y^4 + 2) - 4(2y^4 + 2y^2 + 1)\).

Rewrite the polynomial \(-3(y^4 - y^3 + y) + 7(y^4 + 2) - 4(2y^4 + 2y^2 + 1)\) in the form \(p(y^5 - y^3) + q(y^4 + 2) + ry^2 + s\), where \(p, q, r, s\) are constants, and determine the value of \(s\).

Answer: -4


Suppose \(a,b, c\) are real numbers satisfying \(a + 2b + c = 4.\) Determine the maximum of \( ab + ac + bc. \)

Suppose \(a, b, c\) are positive real numbers satisfying \(a + 2b + c = 4.\) Determine the maximum of \(ab + bc + ca.\)

Answer: 4


How many distinct unordered combinations of three positive perfect squares add up to \(81\)?

There are multiple ways to express \(81\) as the sum of two positive perfect squares \(p^{2}+q^{2}\). What is the largest possible value of \(q\)?

Answer: 3


For all real numbers \(a\) and \(b\) with \(ab \neq 1\), the function \(g(a,b)\) satisfies the relation \(g(a,b)=a+b\,g(b,a)\). Compute the value of \(g(1,2)\).

For all real numbers \(a\) and \(b\), the function \(g(a,b)\) satisfies the relation \(g(a,b)=a+b\,g(b,a)\). Compute the value of \(g(1,2)\).

Substituting \(a=b = 1\) gives \(g(1,1) = 1 + g(1,1)\), so no such function \(g(a,b)\) exists.

Answer: -5


 Let
\[
g(x)=|k-6|\,x^2+(k-6)\,x+10.
\]
Suppose the minimum value of \(g\) (over all real \(x\)) is \(6\). Determine the minimum integer \(k\) among those for which
\[
g(x)-k>0\quad\text{for all real }x.
\]

The minimum value of the function
\[
g(x) = -\lvert k - 6\rvert x^{2} + \lg\!\Big(e^{k - 6} + k^{3} - 3k^{2} - \tfrac{63}{4}k - \tfrac{27}{2}\Big)x + 6
\]
is 6. Determine the minimum integer \(k\) such that \( g(x) - k > 0.\)

Write
\[
g(x) = -|k-6|\,x^2 + A\,x + 6,\quad
A = \lg\!\bigl(e^{k-6} + k^3 - 3k^2 - \tfrac{63}{4}k - \tfrac{27}{2}\bigr).
\]

The leading coefficient is \(-|k-6|\le0\), so \(g\) is either constant or a concave-down parabola.  A nonconstant concave-down quadratic on \(\mathbb R\) has no finite global minimum. To have a “minimum value” of 6, it must be constant ⇒ \(|k-6|=0\) ⇒ \(k=6\), giving \(g(x)\equiv6\).

Then
\[
g(x)-k = 6-6=0,
\]
so \(g(x)-k>0\) has no solutions.

Answer: -10


Given two vectors \(\mathbf{u} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}\) and \(\mathbf{v} = \begin{pmatrix} -5 \\ 2 \end{pmatrix}\), determine the area of the triangle formed by the points \(\mathbf{0}\), \(\mathbf{u}\), and \(\mathbf{v}\).

Given two vectors \(\mathbf{u} = (3, 1)\) and \(\mathbf{v} = (-5, 2)\), let the x-coordinate of the centroid of the triangle with vertices \(\mathbf{O} = (0, 0)\), \(\mathbf{u}\), and \(\mathbf{v}\) be \(q\). Define \(\mathbf{u}' = 2\mathbf{u}\), \(\mathbf{v}' = 2\mathbf{v}\), and \(\mathbf{w} = \big(0, \frac{11}{4}\big)\). Let the x-coordinate of the centroid of the triangle with vertices \(\mathbf{w}\), \(\mathbf{u}'\), and \(\mathbf{v}'\) be \(q'\). Find the value of \(q - q'\).

For the second set of vertices:

 - \(\mathbf{u}' = (6,2)\)
 - \(\mathbf{v}' = (-10,4)\)
 - \(\mathbf{w} = \big(0,\frac{11}{4}\big)\)

 We check collinearity:
 \[
 (-10-6)\bigg(\frac{11}{4}-2\bigg) = (-16)\bigg(\frac{3}{4}\bigg) = -12
 \]
 \[
 (4-2)(0-6) = (2)(-6) = -12
 \]
 Both sides are equal, so the three points are collinear. Because collinear points do not form a valid triangle, the centroid of the triangle is undefined. Thus, \(q'\) does not exist, and the value \(q - q'\) cannot be determined.

Answer: \frac{11}{2}


Find the minimum value over \(y\) fo \(|y - 1| + |y - 1.5| + |y - 2|\).

The minimum value over \(y\) for \(|y - 1|+|y - x|+|y - 2|\) is 1. Determine the range of the function \(g(x)=\lg(x - 2)+x^{2}\). What values can \(g(x)\) take?

The condition “the minimum of \(\lvert y-1\rvert+\lvert y-x\rvert+\lvert y-2\rvert\) equals \(1\)” forces \(x\in[1,2]\), but
\[
g(x)=\lg(x-2)+x^2
\]
requires \(x>2\) (so that \(\lg(x-2)\) is defined). There is no \(x\) satisfying both.

Answer: 1


Let \(n\) be a positive integer. The square of \(3n\) minus \(n\) equals \(2010\). Find \(n\)

Let \(n\) be a positive integer. Three times the square of \(n\) minus three times \(n\) equals \(2025\). Find \(n\).

\((3n)^2 - 3n = 3n(3n - 1)\). Since \(n\) and \(3n - 1\) have different parity, \((3n)^2 - 3n\) is even, so it cannot be 2025.

Answer: 15


What is the number of solutions to the equation \(x^{2}-13[x]+11 = 0\)?

The number of solutions to the equation
 \(x^{2} - 13\lfloor x \rfloor + 11 = 0.\) is \(y\)
 Then, compute the value of
 \[
 \lg(y - 4) + e^{2y}.
 \]

On one hand, \(x^2 = 13[x] - 11 \in \mathbb{N}\), so \(x > 1\).

 On the other hand, since \(-1 < x - 1 < [x] \leq x\), we have
 \[
 \begin{cases}
 x^2 - 13(x - 1) + 11 > 0 \\
 x^2 - 13x + 11 \leq 0
 \end{cases}
 \]
 At this time, \(-24 < x^2 - 13x \leq -11\). Solving gives
 \[
 x \in \left[\frac{13 - 5\sqrt{5}}{2}, \frac{13 - \sqrt{73}}{2}\right) \cup \left(\frac{13 + \sqrt{73}}{2}, \frac{13 + 5\sqrt{5}}{2}\right],
 \]
 and \([x]\) can take values \(1, 2, 10, 11, 12\).

 Substituting in, we get
 \[
 x = \sqrt{2}, \sqrt{119}, \sqrt{132}, \sqrt{155}.
 \]
 Therefore, \(y = 4\) which is undefined for \(\lg(y-4)\)

2024 Peking University Strong Foundation Program

Answer: 4


In \(\triangle ABC\), let \(AD\) be the angle bisector of \(\angle A\), with point \(D\) lying on side \(BC\). Given that the incenter of \(\triangle ADC\) coincides with the circumcenter of \(\triangle ABC\), determine a possible value of \(\angle A\).

In \(\triangle ABC\), let \(AD\) be the angle bisector of \(\angle A\), with point \(D\) lying on side \(BC\). Given that \(\angle C = \frac{\pi}{8}\) and the incenter of \(\triangle ADC\) coincides with the circumcenter of \(\triangle ABC\), determine a possible value of \(\angle A\).

In ∆ACO drop \(OH\perp AC\) at \(H\) and in ∆BCO drop \(OK\perp BC\) at \(K\). Since \(\angle OHC=\angle OKC=90^\circ\), \(OC\) is common, and \(\angle OCH=\angle OCK=90^\circ-\tfrac{C}{2}\), we have \(\triangle OCH\cong\triangle OCK\), implying \(CH=CK\). Then \(\triangle AHO\cong\triangle BKO\) by right angles at \(H,K\), \(OH=OK\), and \(OA=OB\), which gives \(AH=BK\). Thus \(AC=AH+HC=BK+CK=BC\).  Since \(AD\) bisects \(\angle A\), in \(\triangle ADC\) point \(O\) lies on that bisector, giving \(\angle OAC=\angle OCA=\tfrac{A}{4}\); but \(\angle OCA=\tfrac{C}{2}\), hence \(A=2C\). With \(A=B\) and \(A+B+C=\pi\), we get \(5C=\pi\) ⇒ \(C=\tfrac{\pi}{5}\neq\tfrac{\pi}{8}\), hence no solution.

Answer: \frac{\pi}{5}


Consider a semi-circular region with radius \(p\). In this region, a triangle \(\triangle XYZ\) is drawn such that its base \(XY\) coincides with the semi-circle's diameter \(XY\). The third vertex \(Z\) is placed such that \(Z\) is neither at \(X\) nor at \(Y\). The combined length of \(XZ\) and \(YZ\) is denoted by \(t\). Find the maximum possible value of \(t^2\) in terms of \(p\)

Consider a semicircular region with radius \(p\).  In this region, a triangle \(\triangle XYZ\) is drawn such that its base \(XY\) coincides with the semi-circle's diameter \(XY\). The third vertex \(Z\) is placed such that \(Z\) is neither at \(X\) nor at \(Y\). Let \(t\) be the sum of the lengths of \(XZ\) and \(YZ\). Given that \(\sqrt{t^2 - 9p^2} + 2t = 4\), find the length of \(XZ\).

Observe that \(\angle XZY=90^\circ\), we have \(XZ^2+YZ^2 \;=4p^2,\) so it’s natural to set
\[
XZ = 2p\sin\theta,\quad YZ=2p\cos\theta,
\]
for some \(\theta\in(0,\tfrac\pi2)\), which is \(\angle XYZ\). Then
\[
t = XZ + YZ = 2p\bigl(\sin\theta+\cos\theta\bigr),
\]
and since \(\sin\theta+\cos\theta\) attains its maximum \(\sqrt2\) at \(\theta=45^\circ\), \(t_{\max} = 2p\sqrt2\), \(t_{\max}^2 = 8\,p^2\). Then we have \(t^2-9p^2< 8p^2-9p^2<0\), so it is not valid to take the square root.

Answer: 8p^2


There is a sequence of \(58\) elements; each element is represented by \(Q+n\), where \(Q\) represents the product of all prime numbers starting from \(2\), up to and including \(61\), and \(n\) represents sequential values \(2, 3, 4, \ldots, 59\). How many prime numbers does this series have?

There is a sequence of elements; each element is represented as \( Q + n \), where \( Q \) represents the product of all prime numbers less than or equal to 12, and \( n \) takes integer values from 2 to \( \overline{abc} \) (3-digit number) in sequence. Define \( M \) as the total number of prime numbers in this sequence. When the value of \( M \) is 0, what is/are the possible value(s) of \( a + b + c \)?

\( Q = 2310 \). For any \( \overline{abc} \geq 2 \), there exists at least one \( n \) in \( 2 \) to \( \overline{abc} \) such that \( Q + n \) is prime (e.g., \( Q + 11 = 2321 \) is prime). Thus, \( M = 0 \) is impossible for any 3-digit \( \overline{abc} \).

Answer: 0

The original question in MATH does not include answer; according to its solution, the answer should be 0.


Consider a square \(WXYZ\) with center \(M\). The length of side \(WZ\) is \(900\) units. Points \(P\) and \(Q\) are located on line segment \(WZ\), such that \(WP<ZQ\) and point \(P\) lies between points \(W\) and \(Q\). The measure of angle \(\angle QMP\) is \(45^\circ\) and the distance between points \(P\) and \(Q\) is \(400\) units. If \(ZQ = s+t\sqrt{u}\), where \(s\), \(t\), and \(u\) are positive whole numbers and \(u\) is not a multiple of the square of any prime number, calculate the sum \(s+t+u\).

Consider a square \(WXYZ\) with center \(M\). The length of side \(WZ\) is \(900\) units. Points \(P\) and \(Q\) are located on the line segment \(WZ\) such that \(WP < ZQ\), and point \(P\) is between point \(W\) and point \(Q\). The measure of angle \(\angle QMP\) is \(45^\circ\), and the distance between points \(P\) and \(Q\) is \(350\) units. If \(ZQ = s + t\sqrt{r}\), where \(s, t, r\) are positive integers and \(r\) is not a multiple of the square of any prime number, calculate the sum \(s + t + r\).

First, we fix a convenient coordinate system by placing \(W=(0,0)\), \(Z=(900,0)\), and hence the center of the square at \(M=(450,450)\). Any points \(P\) and \(Q\) on the side \(WZ\) with \(PQ=350\) can then be written as
 \[
 P=(x,0),\quad Q=(x+350,0),
 \]
 where \(0<x<275\) ensures that \(Q\) still lies between \(W\) and \(Z\).

 Next, we translate the requirement \(\angle QMP=45^\circ\) into an algebraic condition. The vectors from \(M\) to \(P\) and \(Q\) are
 \[
 \overrightarrow{MP}=(x-450,-450),\quad
 \overrightarrow{MQ}=(x-100,-450).
 \]
 Requiring their angle to be \(45^\circ\) means
 \[
 \frac{\overrightarrow{MP}\cdot\overrightarrow{MQ}}{\|\overrightarrow{MP}\|\;\|\overrightarrow{MQ}\|}
 =\cos45^\circ=\frac{\sqrt2}{2}.
 \]
 Squaring both sides and clearing denominators leads to the single equation
 \[
 \bigl((x-450)^2+450^2\bigr)\,\bigl((x-100)^2+450^2\bigr)
 =2\bigl(x^2-550x+247500\bigr)^2.
 \]

 Finally, expanding and collecting like terms gives a quartic equation in \(x\):
 \[
 -x^4 +1100x^3 -797500x^2 +272250000x -36450000000=0.
 \]
 A numerical check shows that all four roots of this polynomial are nonreal (they each have a nonzero imaginary part, with real part exactly \(275\)). Since none lie in the interval \((0,275)\), there is no admissible real \(x\) for which both \(PQ=350\) and \(\angle QMP=45^\circ\) hold simultaneously.

Answer: 307


On the circumference of the base of a cylinder, two points \(P\) and \(Q\) are chosen such that the central angle subtended by arc \(PQ\) is \(120^\circ\). The cylinder has radius \(6\) and height \(8\), and its entire outer surface is painted red. The cylinder is cut in half by a plane passing through \(P\), \(Q\), and the center of the cylinder, exposing an unpainted surface on each half. The area of one such unpainted surface can be expressed as \(x\pi + y\sqrt{z}\), where \(x,y,z\) are integers with \(z\) is not divisible by the square of any prime number. Determine the value of \(x+y+z\).

On the circumference of the base of a conical clay block, two points \(P\) and \(Q\) are chosen such that the central angle subtended by arc \(PQ\) is \(\alpha\). The clay block is cut in half by a plane passing through \(P\), \(Q\), and the midpoint \(M\) of the cone’s axis. The radius of the base is \(5\), and the height is \(18\). The volume of the larger portion can be expressed as \(x\pi + \sqrt{y}\). Given that \(x + y = 222067\), where \(x\) and \(y\) are positive integers, determine the value of \(\alpha\).

Given \(x + y = 222067\), where \(x, y\) are both positive integers, the minimum value of \(x \cdot \pi + \sqrt{y}\) is \(\sqrt{222067} > 150\pi\). Since \(150\pi\) represents the total volume of the cone, the condition cannot be satisfied.

Answer: 53


Calculate the sum of the sequence \( 1_6, 2_6, 3_6, ..., 45_6 \). The result should be in base 6.

Calculate the sum of the sequence \(1_6, 2_6, 3_6, \ldots, 45_6\). The result is \(\overline{abc}_6\). Find \( c \).

The sum is \(2003_6\) (a four-digit number), which cannot be represented as \(\overline{abc}_6\) (a three-digit number in base 6).

Answer: 2003


Consider a series of unique real numbers \(s_1, s_2, \dots, s_n\). This series can be sorted in increasing order using one or more "swap cycles". A swap cycle involves comparing the first and second terms and swapping them if the second is smaller, then comparing the second and third terms and swapping them if the third is smaller, and continuing in this way until the final term, \(s_n\), is compared with its immediate predecessor and swapped if the final term is smaller. Suppose the 40 terms are in a uniformly random order. Let \(\dfrac{p}{q}\) (in lowest terms) be the probability that the number which begins in position \(20\) ends up, after one swap cycle, exactly in position \(30\), where \(p/q\) is in the simplest fractional form (\(p>0\)). Find the value of \(p + q\).

Consider a sequence \(s_1, s_2, \ldots, s_{40}\) composed of \(40\) distinct real numbers. This sequence can be sorted in ascending order through one or more "exchange cycle" operations. The so-called "exchange cycle" operation is as follows: first, compare the first and the second terms, and if the second term is smaller, swap their positions; then compare the second and the third terms, and if the third term is smaller, swap their positions; and so on, until comparing the \((n - 1)\)-th and the \(n\)-th terms, and if the \(n\)-th term is smaller, swap their positions. Now assume that the terms in the initial sequence \(s_1, s_2, \ldots, s_{40}\) are randomly arranged. Let \(p/q\) represent the probability that the number initially located at the \(s_{20}\) position will eventually stay at the \(17^{\text{th}}\) position after one exchange cycle, where \(p/q\) is in the simplest fractional form (\(p>0\)). Find the value of \(p + q\).

In a single “exchange cycle” (one bubble-sort pass), each element can move left by at most one position (but may move right arbitrarily far). To go from position 20 to position 17 requires three leftward moves, so that event has probability 0—contradicting the requirement \(p>0\) and rendering the problem inconsistent.

Answer: 931


Determine the smallest positive integer \(P\) that satisfies the following condition: picking any one number from \(4\), \(9\), \(25\), and \(49\), at least one number in \(P\), \(P+1\), and \(P+2\) will be divisible by the number you picked.

For the integer \(P\) that satisfies the following condition: picking any one number from \(4\), \(25\), and \(49\), at least one number in \(P\), \(P+1\), and \(P+2\) will be divisible by the number you picked, if \( P \) is a 3-digit number \( P = \overline{abc} \), what is \( c \)?

When \(P = 48\), the condition is satisfied. \(48 = 2^2 \cdot 12,\; 49 = 7^2,\; 50 = 2 \cdot 5^2.\) Since 49 = 1 \cdot 7^2, and \(47, 48, 49\) do not satisfy the condition, 48 is already the smallest one. Therefore, the smallest \(P\) is not a three-digit number, and hence no hundreds digit.

Answer: 98


Consider the function \( F(m) \) which represents the sum of the digits of a given positive integer \( m \). For example, \( F(1507)=13 \). For a certain positive integer \(m\), (F(m) = 1274\). What is largest value of \(F(m+1)\) that is a multiple of 59?

Consider the function \( F(m) \) which represents the sum of the digits of a given positive integer \( m \). For example, \( F(1507)=13 \). For a certain 140 - digit positive integer \( m \), \( F(m)=1274 \) and \( F(m + 1)=6 \). What is the remainder when \( m \) is divided by 9?

\(F(m) = 1274\), and since \(1274 = 9 \times 141 + 5\), \(m\) must have at least 142 digits.

Answer: 1239


If \(a,b,c>0\) and \(a^2+b^2=c^2\), find the maximum real constant \(k\), such that
\[
\frac{a^3+b^3+c^3}{abc}\ge k.
\]

If \(a,b,c>0\) and \(a^2+b^2=c^2\), find the minimum real constant \(k\)，such that
\[
\frac{a^3+b^3+c^3}{abc}\le k.
\]

\(k\) should be the maximum of the formula in the left-hand side. From \(a^2+b^2=c^2\) and \(a,b,c>0\), we can make a trigonometric substitution \(a=c\cos\theta,\, b=c\sin\theta,\, \theta\in(0,\tfrac{\pi}{2})\). Thus
\[
\frac{a^3+b^3+c^3}{abc}
=\frac{\sin^3\theta+\cos^3\theta+1}{\sin\theta\cos\theta}.
\]
Let \(t=\sin\theta+\cos\theta\).
Since \(\theta\in(0,\tfrac{\pi}{2})\), we have \(t\in(1,\sqrt2]\). From \(\sin\theta \cos\theta = \frac{t^2-1}{2}\) and
\[
\sin^3\theta+\cos^3\theta
=(\sin\theta+\cos\theta)(\sin^2\theta - \sin\theta \cos\theta+\cos^2\theta)
= t\left(1-\frac{t^2-1}{2}\right),
\]
substituting back we get
\[
\frac{a^3+b^3+c^3}{abc} = \frac{t^2-t-2}{t-1}.
\]
Now let \(m=t-1\in(0,\sqrt2-1],\)
then the expression becomes \(\frac{2}{m}-m-1\). When \(m>0\), the expression is monotonically decreasing. As \(m\) approaches 0, the expression tends to positive infinity. Therefore, the expression has no maximum value. So we cannot find such a minimum constant \(k\).

2025 Southeast University (China) Strong Foundation Program

Answer: 2+\sqrt{2}


There are exactly three integer values of \(y\) that satisfy the inequality \[y^2 + dy + 2 \le 0.\] Determine the count of integer values of \(d\).

There are exactly three integer values of \(y\) that satisfy the inequality \[-y^2 + dy + 2 \le 0.\] Determine the value of \(d\).

First, note that for any real \( d \), the inequality
     \[
     -y^2 + dy + 2 \leq 0
     \]
     rewrites to
     \[
     y^2 - dy - 2 \geq 0,
     \]
     whose integer solution set is
     \[
     \{\,y \;|\; y \leq r_1\,\} \;\cup\; \{\,y \;|\; y \geq r_2\,\},
     \]
     where
     \[
     r_{1,2} = \frac{d \mp \sqrt{\,d^{2} + 8\,}}{2}.
     \]

 Both solution regions are infinite, so as stated, there is no real value of \( d \) for which the set of integer solutions is finite—let alone exactly three.

Answer: 2


On the coordinate plane, consider three points: \( P = (0,0) \), \( Q = (11,0) \), \( R = (18,0) \). Three lines are defined as follows: Line \( m_P \): passes through \( P \) with an initial slope of \(1\). Line \( m_Q \): passes through \( Q \) and is perpendicular to the \(x\)-axis. Line \( m_R \): passes through \( R \) with an initial slope of \(-1\). These three lines, \( m_P, m_Q, m_R \), start to rotate clockwise around their respective points \( P, Q, R \) at the same angular speed. At any given moment, the three lines form a triangle. What is the maximum possible area of this triangle?

On the coordinate plane, consider three points: \(P = (a, 0), \quad Q = (11, 0), \quad R = (18, 0).\)

- Line \(\ell_P\) is perpendicular to the \(x\)-axis and passes through point \(P\).
- Line \(\ell_Q\) has a slope of \(1\) and passes through point \(Q\).
- Line \(\ell_R\) passes through point \(R\) and is parallel to the \(x\)-axis.

These three lines \(\ell_P, \ell_Q, \ell_R\) begin rotating clockwise around their respective points \(P, Q, R\). The rotation speeds are as follows:
The rotation speed of \(\ell_Q\) is twice that of \(\ell_R\). The rotation speed of \(\ell_P\) is three times that of \(\ell_R\).

At any given moment, the following two triangles are formed:

1. The triangle bounded by \(\ell_P\), \(\ell_Q\), and the \(x\)-axis.
2. The triangle bounded by \(\ell_Q\), \(\ell_R\), and the \(x\)-axis.

If the maximum possible area of these triangles is 2000, find the value of \(a\).

We denote the rotation of \(\ell_R\) by
\(\theta_R(t) = -\,\omega t\)
since it starts horizontal and turns clockwise. The line \(\ell_Q\), initially at slope \(1\), follows
\(\theta_Q(t) = \tfrac\pi4 - 2\omega t\),
rotating twice as fast, while \(\ell_P\) begins vertical with
\(\theta_P(t) = \tfrac\pi2 - 3\omega t\),
at three times the speed.

Equating \(\theta_P\) and \(\theta_Q\) gives
\[
\tfrac\pi4 - \omega t = 0 \;\;\Longrightarrow\;\; t = \frac{\pi}{4\omega}.
\]

At that same moment \(\theta_Q - \theta_R = \tfrac\pi4 - \omega t = 0\), so all three lines become parallel together. As any two lines align, their intersection recedes without bound, sending the triangle’s height over the \(x\)-axis to infinity and thus making its area diverge to infinity.  Because the areas blow up at the instant of parallelism, there is no finite peak value, and hence no choice of \(a\) can make the maximum area equal to 2000.

Answer: 85


Find the value of \(10^{-1}\pmod{1001}\). Provide your answer as an integer between \(0\) and \(1000\), inclusive.

Find the value of \(10^{3}\pmod{1001}\). Provide your answer as an integer between \(1\) and \(999\), inclusive.

We note that
 \[
 10^3 = 1000,
 \]
 and dividing \(1000\) by \(1001\) gives remainder \(1000\). Hence
 \[
 10^3 \bmod 1001 = 1000.
 \]
 Because \(1000\) lies outside the specified range \(1\) through \(999\), no integer in that interval can satisfy the congruence.

Answer: 901


A convex pentagon has interior angle measures that are five consecutive integers (in degrees). Determine the measure of the largest angle in degrees.

One of the interior angles of a convex pentagon, whose interior angles are five consecutive integers (in degrees), measures \(120^\circ\). Determine the measure of the largest interior angle in degrees.

We set the five angles as consecutive integers \(n, n+1, n+2, n+3, n+4\), whose sum is
 \[
 5n + 10.
 \]
 Because any convex pentagon’s interior angles total \(540^\circ\), we solve
 \[
 5n + 10 = 540
 \quad\Longrightarrow\quad
 n = 106,
 \]
 giving the angles \(106, 107, 108, 109, 110\). Since none of these equals \(120^\circ\), the requirement that one angle measure \(120^\circ\) is inconsistent. Thus no such pentagon exists, and there is no largest angle to determine.

Answer: 110^\circ


Let \(q(y)\) be a nonzero quintic polynomial with leading coefficient \(1\), satisfying \(q(1)=1,\; q(2)=2,\; q(3)=3,\; q(4)=4,\; q(5)=5,\; q(6)=6.\) Determine one possible value of \(q(7)\).

Define
 \[
 r(y)=q(y)-y,
 \]
 so that \(r(1)=r(2)=\cdots=r(6)=0\). If \(r\) were not the zero polynomial then
 \[
 r(y)=A\,(y-1)(y-2)\cdots(y-6),
 \]
 which is degree 6 and would force \(q(y)=y+r(y)\) to be degree 6, contradicting the assumption that \(q\) is quintic. Hence \(A=0\) and \(q(y)=y\), but that is linear, not quintic. No degree-5 polynomial with leading coefficient 1 can meet the six conditions, so \(q(7)\) is not defined.

Answer: 727


Given that the function \(y = f(x)\) is the inverse function of \(y = 2^x\), and \(g(x)\) is an odd function. For \(x > 0\), we have  \( g(x) = f(x) - x.\) Find \(g(-8)\).

Given that the function \(y = f(x)\) is the inverse function of \(y = 2^x\), and \(g(x)\) is an odd function.  For \(x > 1\), we have  \( f(g(x)) + g(f(x)) = 2.\)  Find \(g(-8)\).

\(f(x)=\log_2 x\quad(x>0).\) For \(x>1) the equation becomes \(\log_2(g(x))\;+\;g(\log_2 x)\;=\;2\), and \(g(x) = 2^{2 - g(\log_{2} x)}\). Observing the equation above, we can get that the value of \(g\) at \(x\) depends on its value at \(\log_{2} x\), which forms a recurrence relation \(x_{n+1} = \log_2 x_n\). We can easily prove that for any \(x_0>1\), one can arrive a point \(x_n\) in \((0, 1]\) for a finite \(n\). Then since \(x_n\leq1\), the recursion does not apply, so the value of \(g(x)\) at \((0, 1]\) can be arbitrarily defined. Let us define \(\phi(x)=g(x)\) on \((0,1]\), where \(\phi(x)\) is an arbitrary function. As long as we can define \(\phi(x)\), then the value out of \((0,1]\) can be inferred by the definition of odd function and the recursion relation: \[ g(x) = \begin{cases} 2^{2 - g(\log_{2} x)}, & x > 1, \\[6pt] \phi(x), & 0 < x \leq 1, \\[6pt] -g(-x), & x < 0. \end{cases} \] Since we can arbitrarily define \(\phi(x)\), the value of \(g(-8)=-g(8)\) is also arbitrary. Thus, the given condition is not enough to determine \(g(-8)\). We now prove that for any \(x_0>1\), one can arrive a point \(x_n\) in \((0, 1]\) for a finite \(n\). First, notice that the sequence \({x_n}\) is decreasing, since \(\log_2x<x\) when \(x>0\). If the sequence has a lower bound 1, then it should converge to a certain number. Suppose the limit is \(L\). \(L = \lim_{n \to \infty} a_{n+1} = \lim_{n \to \infty} \log_{2} a_n = \log_{2} L\). Obviously no \(L>0\) satisfies \(L=\log_{2} L\), so the sequence is not bounded by 1. Then if the sequence does not arrive \((0, 1]\), and just exist an \(x_m \leq 0\). Then we have \(x_{m-1} \in (0, 1]\), leading to contradiction again. Thus, for any \(x_0>1\), one can arrive a point \(x_n\) in \((0, 1]\) for a finite \(n\).

2024 Guangdong Shenzhen Bao’an High School Monthly Exam

Answer: 5


A scalene triangle \(\triangle PQR\) has two sides measuring \(3\) centimeters and \(5\) centimeters. How many distinct whole-number lengths, in centimeters, could the third side have?

A scalene triangle \(\triangle PQR\) has two sides measuring \(2\) centimeters and \(5\) centimeters, with the cosine of the included angle between these sides equal to \(\frac{1}{5}\). Provide one possible whole-number length, in centimeters, for the third side.

We denote the unknown side opposite the given angle by \(x\). The Law of Cosines then gives
 \(x^2 = 2^2 + 5^2 - 2 \cdot 2 \cdot 5 \cdot \tfrac{1}{5} = 29 - 4 = 25,\)
 so \(x = 5\). However, this makes the side-lengths \(2, 5, 5\), which is isosceles rather than scalene.

Answer: 3


Given that \(f(2x)\) is symmetric about the origin, and satisfies  \( f(x+1) + f(3-x) = 0.\)
When \(x \in (2,4)\), we have  \( f(x) = -\log_{\tfrac{1}{2}}(x-2) + m. \) If \(\frac{f(2022) - 1}{2} = f(-1),\)  find the value of \(m\).

Given that \(f(2x)\) is symmetric about the origin, and satisfies  \( f(x+1) + f(3-x) = 0.\)
When \(x \in (2,4)\), we have  \( f(x) = -\log_{\tfrac{1}{2}}(x-2) + m. \)
where \(m\) is positive.  If
\[
\frac{f(2022+\varepsilon) - 1}{2} = f(-1), \quad \text{for some } \varepsilon \in (0,1),
\]
find the smallest value of \(m\).

Since \(f(2x)\) is symmetric about the origin, we can see that \(f\) is odd: \(f(-x)=-f(x)\). From \(f(x+1)+f(3-x)=0\) (put \(y=x+1\)) ⇒ \(f(y)+f(4-y)=0\). Using oddness gives \(f(x+4)=f(x)\), so \(f\) is 4-periodic. Hence for \(\varepsilon\in(0,1)\), \(f(2022+\varepsilon)=f(2+\varepsilon)\). For \(x\in(2,4)\),
  \[
  f(2+\varepsilon)=-\log_{1/2}\!\varepsilon + m=\log_2\varepsilon + m.
  \]
Also \(f(3)=m\), so \(f(1)=-m\) and \(f(-1)=m\). The equation \(\frac{f(2022+\varepsilon)-1}{2}=f(-1)\) becomes
  \[
  \frac{\log_2\varepsilon + m -1}{2}=m \quad\Longrightarrow\quad \log_2\varepsilon = m+1.
  \]
  But \(\log_2\varepsilon<0\) for \(\varepsilon\in(0,1)\), so \(m<-1\), contradicting \(m>0\).

2025 Hunan Normal University Affiliated High School Monthly Exam

Answer: -1/2


A rectangle, \(\mathrm{PQRS}\), and a semicircle with \(\mathrm{PQ}\) lie in the same plane and do not overlap. Let \(\mathcal{Z}\) represent the region enclosed by the rectangle and the semicircle. Line \(\gamma\) intersects the semicircular arc, the segment \(\mathrm{PQ}\), and the segment \(\mathrm{RS}\) at distinct points \(E\), \(F\), and \(G\), respectively. The line \(\gamma\) divides \(\mathcal{Z}\) into two parts whose areas are in the ratio \(1:2\). It is given that \(PF = 84\), \(PE = 126\), and \(FQ = 168\). Express \(SP\) in the form \(a\sqrt{b}\), where \(a\) and \(b\) are positive integers with \(b\) square‐free. Find \(a + b\).

A rectangle, \(\mathrm{PQRS}\), and a semicircle with \(\mathrm{PQ}\) lie in the same plane and do not overlap. Let \(\mathcal{Z}\) represent the region enclosed by the rectangle and the semicircle. Line \(\gamma\) intersects the semicircular arc, the segment \(\mathrm{PQ}\), and the segment \(\mathrm{RS}\) at distinct points \(E\), \(F\), and \(G\), respectively. The line \(\gamma\) divides \(\mathcal{Z}\) into two parts whose areas are in the ratio \(3:5\). It is given that \(PE=1\), \(PF=2\), and \(FQ=4\). Determine the length of \(PS\).

With \(E\) and \(F\) fixed, the slope of \(\gamma\) is \(m=-\frac{\sqrt{35}}{11}\).
  For \(G\) to be on the segment \(RS\), the rectangle’s height satisfies
  \[
  h \le h_{\max} = \frac{4\sqrt{35}}{11}.
  \]

  Let \(A_P(h)\) be the part of \(\mathcal{Z}\) on the \(P\)-side of \(\gamma\), and \(A_{\mathrm{tot}}(h)\) the total area.
  By extending \(PS\) to meet \(\gamma\) at \(O\), the \(P\)-side region is contained in triangle \(OSQ\), so
  \[
  A_P(h) \le \tfrac12 \cdot 6 \cdot \left(h + \tfrac{2\sqrt{35}}{11}\right)
  = 3\left(h + \tfrac{2\sqrt{35}}{11}\right).
  \]
  The total area is
  \[
  A_{\mathrm{tot}}(h) = 6h + \frac{9\pi}{2}.
  \]
  Thus the ratio satisfies
  \[
  R(h) \le \frac{3\left(h + \frac{2\sqrt{35}}{11}\right)}{6h + \frac{9\pi}{2}}.
  \]
  Since \(R(h)\) increases with \(h\), the largest possible ratio occurs at \(h_{\max}\).

  At \(h = \frac{4\sqrt{35}}{11}\),
  \[
  R_{\max} \le \frac{\frac{18\sqrt{35}}{11}}{\frac{24\sqrt{35}}{11} + \frac{9\pi}{2}}
  = \frac{36\sqrt{35}}{48\sqrt{35} + 99\pi}.
  \]
  To have \(R_{\max} \ge \frac{3}{8}\) would require \(144\sqrt{35} \ge 297\pi\).
  Dividing by 9 gives \(16\sqrt{35} \ge 33\pi\), but
  \[
  16\sqrt{35} < 16\sqrt{36} = 96 < 99 = 33\cdot 3 < 33\pi,
  \]
  so the inequality fails.

Answer: 69


The vertices of a isosceles trapezoid \(\mathrm{PQRS}\) all have integer coordinates, with \(P = (20, 100)\) and \(S = (21, 107)\). The trapezoid doesn't have any horizontal or vertical sides, with only \(\overline{PQ}\) and \(\overline{RS}\) being parallel to each other. If we express the sum of the absolute values of all possible slopes for line \(\overline{PQ}\) as a fraction \(a/b\), where \(a\) and \(b\) are coprime positive integers, determine the value of \(a + b\).

The vertices of a right trapezoid \(PQRS\) all have integer coordinates, with \(P = (20, 100)\) and \(R = (27, 101)\), and \(PQ\) parallel to \(RS\), \(\angle S\) is the right angle. If we express the sum of the absolute values of all possible slopes of \(QR\) as an irreducible fraction, what is the sum of the numerator and denominator of such a fraction?

Consider the point S = (21, 98). One can verify that PS is perpendicular to SR. For this point to be a vertex in a right trapezoid with right angles at P and S, the slope of PS is -2, and the slope of of PQ is 1/2. Since the base PQ passes through P=(20, 100) with a slope of 1/2, any integer point Q on this line can be written as: \(Q_k=(20+2k,100+k)\) for any non-zero integer \(k\). The slope of the leg QR for any such point \(Q_k\) is: \(k_{QR}=\frac{101−(100+k)}{27−(20+2k)}=\frac{1−k}{7−2k}=\frac{k-1}{2k-7}\). Now, let's examine the sum of the absolute values of these slopes. When \(k>4\), \(\frac{k-1}{2k-7}\) is larger than 1/2. There are infinitely many terms in the sum, so the sum of the absolute values of \(k_{QR}\) is infinite.

Answer: 131


Identify the range of the real number \(a\) for which the following equation \(y^3 - ay^2 - 2ay + a^2 - 1 = 0\) yields exactly one real solution for \(y\).

Identify the range of the real number \(a\) for which the following equation \( y^4 - 6y^3 + (13 + 2a)y^2 + (-12 - 6a)y + a^2 + 8a + 7 = 0 \) yields only two real roots (counting multiplicities).

Factoring the equation gives \((y^2 - 5y + (a + 7))(y^2 - y + (a + 1)) = 0\). The discriminant of each quadratic factor is \(-4a-3\). Therefore, counting multiplicities, each factor either contributes two real roots or none. Thus, the original equation has either four real roots or none.

Answer: \left( -\infty, \frac{3}{4} \right)


Suppose we have \(y = \sqrt{\frac{\sqrt{53}}{2} + \frac{3}{2}}\). We are looking for unique positive integers \(p,\) \(q,\) \(r\) that satisfy the equation
 \[
 y^{100} = 2y^{98} + 14y^{96} + 11y^{94} - y^{50} + py^{46} + qy^{44} + ry^{40}.
 \]
 Find the value of \(p + q + r\).

Suppose we have \( y = \sqrt{\frac{\sqrt{53} + 3}{2}} \). We are looking for unique positive integers \( p \), \( q \), and \( r \) such that
 \[
 y^{100} = 2y^{98} + 14y^{96} + 11y^{94} + py^{50} + qy^{46} + ry^{44} + 121y^{40}.
 \]
 Find the value of \( p + q + r \).

Answer: 157


In the complex plane, the three vertices of a non-degenerate square correspond to the complex numbers \( p \), \( p^2 \), and \( p^3 \). Please enumerate all potential areas that this square could have, separated by commas.

In the complex plane, the three vertices of an equilateral triangle whose center is not at the origin correspond to the complex numbers \( p \), \( p^2 \), and \( p^3 \). Find the modulus of \( p \).

Assume the three vertices are \(p,p^2,p^3\) with \(p\neq0,1\). A necessary and sufficient condition for three complex numbers \(z_1,z_2,z_3\) to form an equilateral triangle is

 \[
 (p-p^2)^2+(p^2-p^3)^2+(p^3-p)^2
 =2p^2(p-1)^2(p^2+p+1)=0.
 \]
 Excluding the degenerate cases \(p=0\) or \(p=1\), we obtain
 \[
 p^2+p+1=0\quad\Longrightarrow\quad p=e^{\pm 2\pi i/3},
 \]
 so \(|p|=1\). But for these values
 \[
 p+p^2+p^3=p+p^2+1=0,
 \]
 hence the triangle's centroid (and center) is the origin. This contradicts the hypothesis that the centre is not at the origin.

Answer: \frac{5}{8}, 2, 10


Given that for any value \(y \ge 1,\) we have a function \(h\) defined as:
 \[
 h(y) = \left\{
 \begin{array}{cl}
 \lfloor y \rfloor \left| y - \lfloor y \rfloor - \dfrac{1}{2 \lfloor y \rfloor} \right| & \text{when \(y < \lfloor y \rfloor + \dfrac{1}{\lfloor y \rfloor}\)}, \\
 h \left( y - \dfrac{1}{\lfloor y \rfloor} \right) & \text{otherwise}.
 \end{array}
 \right.
 \]
 Let \(g(y) = 2^{y - 2007}.\) Determine the number of points where \(h(y)\) and \(g(y)\) intersect.

Given that the domain of \(h(y)\) is \([1,+\infty)\), and \(k(y)\) is the floor function which equals the greatest integer not exceeding \(y\).

When \(y < k(y) + \dfrac{1}{k(y)}\),
\[
h(y) = k(y)\,\left|\,y - k(y) - \frac{1}{2k(y)}\,\right|;
\]
otherwise
\[
h(y) = \biggl(y - \frac{1}{k(y)}\biggr)\times
\left|\,y - \frac{1}{k(y)} - k\!\Bigl(y - \frac{1}{k(y)}\Bigr) - \frac{1}{2\,k\!\Bigl(y - \frac{1}{k(y)}\Bigr)}\,\right|.
\]

Let \(g(y) = 2^{\sqrt{(y-2)^2}} = 2^{|y-2|}.\) Determine the points where \(h(y)\) and \(g(y)\) intersect.

If \(y<n+\dfrac{1}{n}\) then
\[
h(y)=n\bigl|y-\bigl(n+\tfrac{1}{2n}\bigr)\bigr|.
\]
The center \(n+\tfrac{1}{2n}\) lies inside the strip, so
\[
0\le h(y)\le n\cdot\frac{1}{2n}=\tfrac12.
\]

If \(y\in\bigl[n+\tfrac{1}{n},\,n+1\bigr)\) then \(\lfloor y-\tfrac{1}{n}\rfloor=n\) and
\[
h(y)=(y-\tfrac{1}{n})\bigl|y-\bigl(n+\tfrac{1}{n}+\tfrac{1}{2n}\bigr)\bigr|.
\]
Numerically this expression on the subinterval is maximized near the right endpoint; denote that maximum by \(M_n\). For example
\[
M_2=\tfrac58,\quad M_3=\tfrac{11}{6}\approx1.833,\quad M_4\approx2.968,\ \ldots
\]

The other function is \(g(y)=2^{\lvert y-2\rvert}\). On the strip \([n,n+1)\) its minimum (infimum) is
\[
\inf_{y\in[n,n+1)} g(y)=
\begin{cases}
1, & n=1\ \text{(infimum as }y\to2^-),\\[4pt]
1, & n=2\ \text{(attained at }y=2),\\[4pt]
2^{\,n-2}, & n\ge3\ \text{(attained at }y=n).
\end{cases}
\]

Compare the two: - On \([1,2)\): \(h(y)\le\tfrac12<1\le g(y)\) (so \(h<g\) on \([1,2)\)). - On \([2,3)\): \(h(y)\le\tfrac58<1=g(2)\) (so \(h<g\) on \([2,3)\)). - For \(n\ge3\): \(M_n\) grows only polynomially in \(n\), while \(2^{\,n-2}\) is exponential, and in fact \(M_n<2^{\,n-2}\) for every \(n\ge3\) (check \(M_3=1.833<2,\ M_4\approx2.968<4,\) etc.). Hence \(h(y)<g(y)\) on every strip \([n,n+1)\), \(n\ge3\). Hence for every \(y\ge1\) we have \(h(y)<g(y)\). Therefore there are no intersection points.

Answer: 4022030


\( g: \mathbb{R} \rightarrow \mathbb{R} \) satisfies
 \[
 g(p^2) + g(q^2) = g(p + q)^2 - 2pq.
 \]
 for any real numbers \( p \) and \( q \). Define
 \[
 T = \sum_{m = -2019}^{2019} g(m).
 \]
What is the count of feasible values for \(T\)?

\( g: \mathbb{R} \rightarrow \mathbb{R} \) satisfies
 \[
 g(p + q) = 2g(p) + 2g(q) + 3pq
 \]
 for any real numbers \( p \) and \( q \). If \( 2g(1)g(-1) \geq \frac{9}{8} \). What is value of \( g\!\left(\frac{3}{4}\right) \).

By taking \(p = q = 0\), we get \(g(0) = 4g(0)\), which solves to \(g(0) = 0\).
 By taking \(p = 0\) and \(q \neq 0\), we get \(g(q) = 2g(q)\), which solves to \(g(q) = 0\) for any real number \(q\).

Answer: 2039191

The insolvable problem is from Preliminary Contest of the National High School Mathematics League (Xinjiang Division). The problem itself is already problematic.


Let \(P(x,y)\)  denote the least common multiple of positive integers \(x\) and \(y\). Determine all ordered triples \((m,n,p)\) of positive integers such that \(P(m,n) = 1000\), \(P(n,p) = 2000\), and \(P(p,m) = 2000\).

Let \(P(x,y)\) denote the least common multiple of positive integers \(x\) and \(y\). Determine all ordered triples \((m,n,p)\) of positive integers such that \(P(m,n) = 291600000\), \(P(n,p) = 29160000\), and \(P(p,m) = 69984000\).

The prime factorizations of \(m, n, p\) are \(2^7 \times 3^6 \times 5^5\), \(2^6 \times 3^6 \times 5^4\), and \(2^8 \times 3^7 \times 5^3\), respectively.
 Let the exponents of 5 in \(m, n, p\) be \(m_1, n_1, p_1\). Then we obtain
 \[\max(m_1, n_1) = 5,\quad \max(n_1, p_1) = 4,\quad \max(m_1, p_1) = 3.\]
 The first condition requires that either \(m_1\) or \(n_1\) equals 5. However, either choice contradicts the latter two conditions.

Answer: 70


Suppose that \(a,\) \(b,\) and \(c\) are positive real number values which satisfy
 \[
 \frac{1}{a^4} + \frac{1}{b^4} + \frac{1}{c^4} = 1.
 \]
 Determine the minimum of
 \[
 \frac{a^4 b^4 + a^4 c^4 + b^4 c^4}{a^3 b^2 c^3}.
 \]

Suppose that \(a,\) \(b,\) and \(c\) are positive rational numbers which satisfy
 \[
 a^{-4} + b^{-4} + c^{-4} = 1.
 \]
 Determine the minimum of
 \[
 \frac{a^4 b^4 + a^4 c^4 + b^4 c^4}{a^3 b^2 c^3}.
 \]

Over the reals, the unique minimizer of the problem occurs when
 \[
 \left(\tfrac{1}{a^4},\tfrac{1}{b^4},\tfrac{1}{c^4}\right)=\left(\tfrac14,\tfrac12,\tfrac14\right),
 \]
 giving the value \(2\sqrt{2}\).

 Translating back,
 \[
 a=\tfrac{1}{\sqrt{2}},\quad b=\tfrac{1}{2^{1/4}},\quad c=\tfrac{1}{\sqrt{2}},
 \]
 which are not all rational. Because the real minimizer is unique, any rational triple strictly increases the value. However, rationals are dense, so we can find rational triples arbitrarily close to \(\left(\tfrac14,\tfrac12,\tfrac14\right)\), making the expression arbitrarily close to \(2\sqrt{2}\). Thus over rationals, the infimum is \(2\sqrt{2}\) but it is never attained — there is no minimum.

Answer: 2 \sqrt{2}


Consider a triangle with sides measuring 7, 8, and 9 units. There exist two lines that divide both the triangle's perimeter and area in half at the same time. Let the acute angle formed by the intersection of these two lines be represented by \(\psi\). Determine the value of \(\tan \psi.\)

Consider a triangle with sides measuring 5, 8, and 11 units. There exist two distinct lines that simultaneously divide both the perimeter and the area of the triangle into equal halves. Let the angle formed at their intersection be denoted by \(\psi\). Determine all possible values of this angle.

 Assume the two lines partition the triangle into four regions with areas (cyclically) \(a_1,a_2,a_3,a_4>0\) and they cut the boundary into four consecutive arc-lengths \(p_1,p_2,p_3,p_4\). Because \(L_1\) halves area, then
\[
a_1+a_2 = a_3+a_4 = \tfrac{S}{2},
\]
and because \(L_2\) halves area,
\[
a_2+a_3 = a_4+a_1 = \tfrac{S}{2}.
\]
Subtracting gives \(a_1=a_3\) and \(a_2=a_4\). Hence \(a_1=a_2=a_3=a_4=\tfrac{S}{4}\). Let \(p_i\) be the boundary length between successive intersection points. We can also derive that \(p_1=p_3\) and \(p_2=p_4\). So far, the four interior pieces have equal area \(S/4\), and opposite boundary arcs are equal in length. Cut the three sides by the four boundary intersection points and introduce nonnegative segment variables. Let the splits be
\[
AB:\; x,\;y,\;z \quad (x+y+z=11),\qquad
BC:\; u,\;v \quad (u+v=5),\qquad
CA:\; m,\;n \quad (m+n=8).
\]
The four consecutive boundary arcs are
\[
\,p_1=y,\quad p_2=z+u,\quad p_3=v+m,\quad p_4=n+x\,.
\tag{A}
\]
Perimeter equalities \(p_1=p_3,\ p_2=p_4\) give two linear relations
\[
\,y = v + m\,,\qquad z + u = n + x\,
\tag{P}
\]

Let \(h_a,h_b,h_c\) denote the perpendicular distances from \(X\) to sides \(BC,CA,AB\) respectively. Each region area equals half (base × corresponding height) of the boundary pieces that belong to it:
\[
\begin{aligned}
a_1 &= \tfrac12\,y\,h_c,\\
a_2 &= \tfrac12\bigl(z\,h_c + u\,h_a\bigr),\\
a_3 &= \tfrac12\bigl(v\,h_a + m\,h_b\bigr),\\
a_4 &= \tfrac12\bigl(n\,h_b + x\,h_c\bigr).
\end{aligned}
\]
Using \(a_1=a_3\) and \(a_2=a_4\) and clearing the \(\tfrac12\) factor yields
\[
\boxed{\,y\,h_c = v\,h_a + m\,h_b\,,\qquad z\,h_c + u\,h_a = n\,h_b + x\,h_c\,.}
\tag{A}
\]

Use \(v=y-m\) from the first of (P) in the first area relation to get
\[
y h_c = (y-m)h_a + m h_b \quad\Longrightarrow\quad y(h_c-h_a)=m(h_b-h_a).
\tag{1}
\]
Use \(u=n+x-z\) from the second of (P) in the second area relation to get
\[
z h_c + (n+x-z)h_a = n h_b + x h_c
\]
which rearranges to
\[
z(h_c-h_b)=x(h_a-h_b).
\tag{2}
\]

Equations (1) and (2) are two independent linear proportionality relations:
\[
m \;=\; y\cdot\frac{h_c-h_a}{\,h_b-h_a\,},\qquad
x \;=\; z\cdot\frac{h_c-h_b}{\,h_a-h_b\,},
\]
We still have the three side-sum constraints
\[
x+y+z=11,\qquad u+v=5,\qquad m+n=8,
\]
together with the two perimeter relations (P). Substituting the expressions for \(m\) and \(x\) from (1),(2) into these side-sum/perimeter relations produces (after routine elimination) two independent linear relations among the three total side lengths \(AB,BC,CA\). Concretely, the algebra reduces to a system of the form
\[
\alpha_1\cdot AB + \beta_1\cdot BC + \gamma_1\cdot CA = 0,\qquad
\alpha_2\cdot AB + \beta_2\cdot BC + \gamma_2\cdot CA = 0,
\]
with coefficients determined by the height differences \((h_c-h_a),(h_b-h_a),(h_c-h_b),(h_a-h_b)\). The only way to hold for the given positive numbers \(AB,BC,CA\) is that two of the side-lengths are equal. But our triangle has pairwise distinct sides \(5,8,11\). No choice of interior point \(X\) or directions of the two lines can make the exceptional proportionality conditions hold for these three fixed, distinct side–lengths.

Answer: 3 \sqrt{5} + 2 \sqrt{10}


Consider a square with side length \(2\). Define the set \(\mathcal{T}\) as the collection of all line segments of length \(2\) whose endpoints lie on two neighboring sides of the square. The midpoints of the line segments in \(\mathcal{T}\) form a closed region. If the area of this region, rounded to the nearest hundredth, is \(m\), determine the value of \(100m\).

Consider a square with side length \(2\). Let \(\mathcal{T}\) be the set of all line segments of length \(\sqrt{3}\) whose endpoints lie on two adjacent sides of the square. The midpoints of the line segments in \(\mathcal{T}\) form a region. Determine the area of this region.

Take the square with vertices \((0,0),(2,0),(2,2),(0,2)\).
 For the two adjacent sides at the origin let the endpoints be
 \[
 A=(u,0),\qquad B=(0,v),\qquad u^2+v^2=3,\quad 0\le u,v\le2.
 \]
 The midpoint is \(M=(\tfrac{u}{2},\tfrac{v}{2})\), hence
 \[
 \Big(\frac{u}{2}\Big)^2+\Big(\frac{v}{2}\Big)^2=\frac{3}{4},
 \]
 so the midpoints from this corner lie on the quarter-circle
 \[
 x^2+y^2=\frac{3}{4},\qquad x\ge0,\ y\ge0,
 \]
 i.e. radius \(r=\dfrac{\sqrt3}{2}\). Repeating for all four corners gives the union of four quarter-circles centered at
 \((0,0),(2,0),(2,2),(0,2)\) each of radius \(\dfrac{\sqrt3}{2}\).
 Along an edge the two neighbouring quarter-arcs meet at points \(x=r\) and \(x=2-r\),
 and the gap between them is
 \[
 (2-r)-r=2-2r=2-\sqrt3>0.
 \]
 Thus the four arcs are disconnected and do not form a single closed curve; no region is enclosed.

Answer: 86


Consider \(H\) to be the set of polynomials
 \[
 Q(w) = w^m + d_{m-1}w^{m-1} + \cdots + d_2w^2 + d_1w + 50,
 \]
 where \(d_1, d_2, \dots, d_{m-1}\) are integer coefficients and \(Q(w)\) possesses unique roots which can be expressed as \(p + iq\) where both \(p\) and \(q\) are integers. How many polynomials does the set \(H\) contain?

Consider \(H\) to be the set of polynomials
 \[
 Q(w) = w^m + d_{m-1}w^{m-1} + \cdots + d_1 w + 50,
 \]
 where \(d_1, d_2, \dots, d_{m-1}\) are integers. Assume that every root of \(Q(w) = 0\) is a complex number whose real and imaginary parts are integers. What is the number of elements in the set \(H\)?

There is no restriction on the multiplicity of the roots. For any element in this set, multiplying it by \(z+1\) or +(z^2+1\) also leads to a valid element in this set. Thus, there are infinitely many elements in this set.

Answer: 528


Consider the function
\[
g(x) = \frac{1}{e^{x-1}+1} + \ln\!\left(\frac{2+x}{4-x}\right) + a(x-1).
\]
Determine the range of the parameter \(a\) such that the condition
\[
g(x) > 0.5 \quad \text{if and only if} \quad 1 < x < 4
\]
is satisfied.

Consider the function
 \[
 g(x) = \frac{1}{e^{x-1}+1} + \ln\!\left(\frac{2+x}{4-x}\right) + a(x-1).
 \]
 Determine the range of the parameter \(a\) such that the inequality
 \[
 g(x) > 0.49
 \]
 holds true if and only if \(1 < x < 4\).

At \(x = 1\),
\[
g(1) = \frac{1}{e^{0} + 1} + \ln\!\left(\frac{2 + 1}{4 - 1}\right) + a(1 - 1)
= \frac{1}{2} + \ln 1 + 0
= 0.5.
\]

The condition says \(g(x) > 0.49\) if and only if \(1 < x < 4\).  This means that for \(x = 1\) (which is not in the open interval), we must have \(g(1) \le 0.49\).  But \(g(1) = 0.5 > 0.49\), and this value does not depend on \(a\). Therefore, no choice of \(a\) can satisfy the stated condition.

2024 New Chinese College Entrance Exam, Paper 1, Problem 18(3)

Answer: a \ge -\frac{5}{12}


In the rectangular coordinate system \(xOy\), consider the parabola \(\Gamma: y^2 = 8x\). Let \(P(-1,0)\) be a point, and draw a line \(AB\) through \(P\) that intersects the parabola \(\Gamma\) at two distinct points \(A\) and \(B\). Let \(Q\) be the midpoint of \(AB\), and let \(F\) be the focus of the parabola. If \(QF\) is perpendicular to the \(x\)-axis, find the slope of \(AB\).

In the rectangular coordinate system \(xOy\), consider the parabola \(\Gamma: y^2 = 8x\). Let \(P(-2, 0)\) be a point, and draw a line \(AB\) through \(P\) that intersects the parabola \(\Gamma\) at two distinct points \(A\) and \(B\). Let \(Q\) be the midpoint of \(AB\), and let \(F\) be the focus of \(\Gamma\). If \(QF\) is perpendicular to the \(x\)-axis, find the slope of line \(AB\).

Let the line through \(P(-2,0)\) be \(y = m(x+2)\). Intersections with \(\Gamma: y^2=8x\) satisfy
 \[
 m^2(x+2)^2 = 8x
 \;\Longrightarrow\;
 m^2 x^2 + (4m^2 - 8)x + 4m^2 = 0.
 \]
 If the intersection points are \(A(x_1,\,\cdot)\) and \(B(x_2,\,\cdot)\), the midpoint \(Q\) has
 \[
 x_Q = \frac{x_1+x_2}{2} = \frac{-\,(4m^2-8)/m^2}{2} = -2 + \frac{4}{m^2}.
 \]
 The focus of \(y^2=8x\) is \(F=(2,0)\). The condition “\(QF\) is perpendicular to the \(x\)-axis” means \(QF\) is vertical, so \(x_Q=2\). Hence
 \[
 -2 + \frac{4}{m^2} = 2 \;\Longrightarrow\; m^2 = 1.
 \]
 But for two distinct intersections we need the quadratic’s discriminant
 \[
 \Delta = (4m^2-8)^2 - 16m^4 = 64(1 - m^2)
 \]
 to be positive. When \(m^2=1\), \(\Delta=0\), so the line is tangent (one repeated intersection), contradicting the requirement of two distinct points.

2013 Zhejiang College Entrance Exam – Mathematics

Answer: \pm\frac{2\sqrt{3}}{3}


Let \( f(x) \) be a polynomial of degree five. If \( f(n) = \frac{n}{n+1} \) holds for all \( n \in \{ 0,1,2,3,4,5\} \), find the value of \( f(6) \).

Let \( f(x) \) be a polynomial of degree five. If \( f(n) = n \) holds for all \( n \in \{ 1, 3, 4, 5, 6, 7 \} \), find the value of \( f(2) \).

If \(f(x)\) is a quintic polynomial, then \(f(n) = n\) means
 \[
 f(x) - x = 0.
 \]

 Here, \(f(x) - x\) is also a degree-five polynomial, which can have at most five real roots. However, the condition \(f(n) = n\) for \(n \in \{1, 3, 4, 5, 6, 7\}\) gives six distinct real roots.

2023 Fudan University Strong Foundation Program

Answer: 1


The strictly increasing function \( f(x) \) defined on the set of positive integers satisfies \( f(f(x))=3x \) for any positive integer \( x \). Determine the value of \( f(6) \).

The strictly increasing function \( f(x) \) defined on the set of positive integers satisfies \( f(f(x))=x^2 \) for any positive integer \( x \). Determine the exact value of \( f(6)+ f(7)+f(8)+f(9)+f(10) \).

Write \(a_n := f(n)\). Monotonicity and the relation \(f(f(x))=x^2\) force
 \[
 a_1=1,\; a_2=3,\; a_3=4,\; a_4=9,\; a_9=16,
 \]
 and \(a_5,a_6,a_7,a_8\) must be four increasing integers chosen from \(\{10,11,12,13,14,15\}\).
 For each \(x\in\{1,\dots,10\}\) the relation also forces \(f(a_x)=x^2\), so once the vector
 \((a_1,\dots,a_{10})\) is chosen the indices \(a_x\) carry fixed values \(x^2\).
 Feasibility of a choice reduces to the usual spacing condition for strictly increasing integer sequences: if some indices \(i_1<i_2<\dots<i_m\) are preassigned values \(v_1< v_2<\dots<v_m\), there must be enough integers between each \(v_k\) and \(v_{k+1}\) to fill the intervening positions \(i_k+1,\dots,i_{k+1}-1\). Checking all possible choices for
 \((a_5,a_6,a_7,a_8)\) and all admissible \(a_{10}>16\) and applying this spacing test yields the feasible sums \(S\) and their extremes. The minimum \(S_{\min}=72\) occurs for example with
  \[
  (a_5,a_6,a_7,a_8,a_9,a_{10})=(11,12,13,14,16,17),
  \]
 The maximum \(S_{\max}=83\) occurs for example with
 \[
 (a_5,a_6,a_7,a_8,a_9,a_{10})=(10,13,14,15,16,25),
 \]
 giving \(S=13+14+15+16+25=83\). Every integer between 72 and 83 occurs for at least one admissible choice.

Answer: 9


Given that \(x\) and \(y\) are positive real numbers and satisfy \(2x + 3y = 5\), find the minimum value of \( x^2 + y^2\).

Given that \(x\) and \(y\) are positive rational numbers and satisfy \(2x + 3y = 5\), find the minimum value of \(\frac{2}{x} + \frac{5}{y}\).

Let \(F(x,y)=\frac{2}{x}+\frac{5}{y}\) with \(2x+3y=5\) and \(x,y>0\). Using Lagrange multipliers:
 \[
 -\frac{2}{x^2}=2\lambda,\quad -\frac{5}{y^2}=3\lambda
 \]
 Hence
 \[
 \frac{1}{x^2}=\frac{5}{3y^2}\Rightarrow y=\sqrt{\frac{5}{3}}x.
 \]
 Substituting into \(2x+3y=5\) gives
 \[
 x=\frac{5}{2+3\sqrt{5/3}},\quad y=\sqrt{\frac{5}{3}}x,
 \]
 which are irrational. Therefore no pair of positive rationals attains the minimum.

Answer: \frac{25}{13}


Given the curve \(\Gamma: x^3 + y^3 = 8xy\) and the point \(P(4, 4)\), draw a line passing through point \(P\) and intersecting \(\Gamma\) at another point \(A\). Determine the one slope that is impossible.

Given the curve \(\Gamma: x^3 + y^3 = 8xy\) and the point \(P(4, 4)\), draw a line passing through point \(P\) and intersecting \(\Gamma\) at only one another point \(A\), such that the line does not pass through the origin. Find the range of possible slopes of such lines.

Given \(\Gamma: x^3+y^3=8xy\) and \(P(4,4)\). A line through \(P\) with slope \(m\) is \(y=mx+4-4m\). Substituting into the curve gives \[ (x-4)\bigl((m^3+1)x^2+(-8m^3+12m^2-8m+4)x+16m^3-48m^2+48m-16\bigr)=0. \] Let the quadratic factor be \(Ax^2+Bx+C\). Its discriminant is \[ \Delta=B^2-4AC=16\,(m-1)^2(5m^2-6m+5). \] Since \(5m^2-6m+5>0\) for all real \(m\), one has \(\Delta=0\) exactly when \(m=1\). When \(m=1\) the line is \(y=x\) and it passes through the origin, so it is excluded. For every \(m\neq1\) the quadratic has two distinct real roots, so the line meets \(\Gamma\) in three distinct real points unless one of those roots equals \(x=4\). The condition \(x=4\) in the quadratic forces \(m=-1\), but for \(m=-1\) the intersection polynomial reduces to \[ 32(x-4)^2=0, \] so the only real intersection is \(P\) itself and there is no other distinct point. Therefore there is no slope \(m\) (with the line not passing through the origin) for which the line through \(P\) meets \(\Gamma\) at exactly one other distinct point.

Answer: -1


If \( f(x) = (x - 0.25)^4 + (x - a)^2 \), where \( x \in \mathbb{R} \), and the graph of \( f(x) \) is symmetric about an axis, find the value of \( a \).

If \( f(x) = (x - 0.25)^4 + (x - a)^2 \), where \( x \in \mathbb{Z} \), and the graph of \( f(x) \) is symmetric about an axis, find the value of \( a \).

Let \(f(x)=(x-\tfrac14)^4+(x-a)^2,\;x\in\mathbb Z.\) Assume the graph is symmetric about the vertical axis \(x=c\). Then for every integer \(x\) we have \(f(x)=f(2c-x)\), hence \(2c\in\mathbb Z\) and the polynomial identity \(f(c+u)=f(c-u)\) holds for all real \(u\). Put \(A=c-\tfrac14,\;B=c-a\). Then
 \[
 f(c+u)=(A+u)^4+(B+u)^2
 =A^4+4A^3u+6A^2u^2+4Au^3+u^4+B^2+2Bu+u^2.
 \]
 For this to be even in \(u\) the odd coefficients must vanish:
 \[
 4A=0,\qquad 4A^3+2B=0.
 \]
 Thus \(A=0\) and then \(B=0\), so \(c=\tfrac14\) and \(a=\tfrac14\).

 But \(2c=\tfrac12\notin\mathbb Z\), contradicting the necessary condition \(2c\in\mathbb Z\).

Answer: \frac{1}{4}


In the Cartesian plane \(xOy\), consider the ellipse \(\Gamma: \tfrac{x^2}{a^2} + \tfrac{y^2}{1} = 1\). Suppose there exists a circle with center on the \(x\)-axis and radius \(1\) that intersects \(\Gamma\) at four points \(A, B, C, D\). Assume there also exists a point \(E\) in the plane such that the pentagon \(ABCDE\) is a regular pentagon. Find all possible \(x\)-coordinates of the center of this circle. The point \(E\) need not lie on \(\Gamma\); it only needs to exist somewhere in the plane satisfying the above condition.

In the Cartesian plane \(xOy\), let the ellipse \(\Gamma: \tfrac{x^2}{6} + \tfrac{y^2}{1} = 1\). Suppose there exists a circle with center on the \(x\)-axis and radius \(1\) that intersects \(\Gamma\) in four points \(A,B,C,D\), and that there exists another point \(E\) in the plane such that the pentagon \(ABCDE\) is a regular pentagon. Find the \(x\)-coordinate of the center of this circle. The point \(E\) need not lie on \(\Gamma\); it only needs to exist somewhere in the plane satisfying the above condition.

Ellipse \(\dfrac{x^2}{6}+y^2=1\). Circle center \((h,0)\), radius \(1\): points \((h+\cos t,\ \sin t)\). Intersection condition: \[ \frac{(h+\cos t)^2}{6}+\sin^2 t=1 \quad\Rightarrow\quad 5\cos^2 t-2h\cos t-h^2=0. \] Let the roots be \(u_1,u_2\). By Vieta, \[ u_1+u_2=\frac{2h}{5},\qquad u_1u_2=-\frac{h^2}{5}. \] For a regular pentagon on the unit circle to yield only two cosine values among four vertices, the configuration must be symmetric about the \(x\)-axis. Up to rotation the four angles are \(\pm\frac{2\pi}{5},\ \pm\frac{4\pi}{5}\) or \(\pm\frac{\pi}{5},\ \pm\frac{3\pi}{5}\). Hence \[ \{u_1,u_2\}=\{\cos\tfrac{2\pi}{5},\ \cos\tfrac{4\pi}{5}\} \quad\text{or}\quad \{u_1,u_2\}=\{\cos\tfrac{\pi}{5},\ \cos\tfrac{3\pi}{5}\}. \] Using \(\cos\tfrac{\pi}{5}=\frac{1+\sqrt5}{4},\ \cos\tfrac{2\pi}{5}=\frac{\sqrt5-1}{4}\) and \(\cos\tfrac{3\pi}{5}=-\cos\tfrac{2\pi}{5},\ \cos\tfrac{4\pi}{5}=-\cos\tfrac{\pi}{5}\), one gets \[ u_1+u_2=\pm\frac12,\qquad u_1u_2=-\frac14. \] Equating with Vieta: \[ \frac{2h}{5}=\pm\frac12,\qquad -\frac{h^2}{5}=-\frac14 \ \Rightarrow\ h=\pm\frac{5}{4},\quad h^2=\frac54, \] which is impossible since \(\left(\pm\frac54\right)^2=\frac{25}{16}\ne\frac54\).

Answer: -1,1


What is the lowest degree of polynomial function \(f(x)\) such that it has both a local maximum and a local minimum, and that the value of the local maximum is less than the value of the local minimum?

The quartic polynomial function \(f(x)\) satisfies the condition that there exists a local maximum and a local minimum such that the local maximum is less than the local minimum. Find the analytical expression of \(f(x)\).

Let quartic with both a local maximum and a local minimum \(\Rightarrow f'(x)\) is a cubic with three real simple roots \(r_1<r_2<r_3\).
 For \(a>0\) (leading coefficient of \(f\)): sign \(f'\) on intervals \((-\infty,r_1),(r_1,r_2),(r_2,r_3),(r_3,\infty)\) is \(-, +, -, +\). Thus \(r_1\) is a local minimum, \(r_2\) a local maximum, \(r_3\) a local minimum, and \(f\) increases on \((r_1,r_2)\), decreases on \((r_2,r_3)\), so \(f(r_2)>f(r_1)\) and \(f(r_2)>f(r_3)\).
 For \(a<0\): sign \(f'\) is \(+, -, +, -\). Thus \(r_1\) is a local maximum, \(r_2\) a local minimum, \(r_3\) a local maximum, and \(f(r_1)>f(r_2)\) and \(f(r_3)>f(r_2)\).
 In both cases local maximum value exceeds every local minimum value.

Answer: 5


In the Cartesian plane with axes \(xOy\), let the hyperbola \(\Gamma\) be defined by \(\Gamma: \tfrac{x^2}{b} - y^2 = 1.\) A function \(f(x)\) has the property that for every point \(A\) on \(\Gamma\), there exists a point \(B\) on the graph of \(y=f(x)\) such that \(\lvert AB \rvert = \lvert OA \rvert = \lvert OB \rvert.\) Find the value of \(b\).

In the Cartesian plane with axes \(xOy\), let the hyperbola \(\Gamma\) be defined by
 \(\Gamma:\;\tfrac{x^2}{2}-y^2=1.\) A function \(f(x)\) has the property that for every point \(A\) on \(\Gamma\), there exists another point \(B\) on the graph of \(y=f(x)\) such that
 \(\lvert AB\rvert = \lvert OA\rvert = \lvert OB\rvert.\) Find the explicit expression of \(f(x)\) and determine its domain.

If \(|AB|=|OA|=|OB|\) then \(B\) is obtained from \(A=(x,y)\) by a rotation about the origin through \(\pm 60^\circ\).
 Writing \(B=(X,Y)=R_{\pi/3}(x,y)\) and inverting gives
 \[
 x = X\cos\frac{\pi}{3}+Y\sin\frac{\pi}{3},\qquad
 y = -X\sin\frac{\pi}{3}+Y\cos\frac{\pi}{3}.
 \]
 Substituting into \(\dfrac{x^2}{2}-y^2=1\) yields the rotated conic
 \[
 5X^2-6\sqrt{3}\,X Y - Y^2 + 8 = 0.
 \]
 Viewed as a quadratic in \(Y\),
 \[
 Y^2 + 6\sqrt{3}\,X Y - 5X^2 - 8 = 0,
 \]
 whose discriminant is
 \[
 \Delta = (6\sqrt{3}\,X)^2 - 4\cdot 1\cdot(-5X^2-8)
 =128X^2+32 = 32(4X^2+1) > 0\quad\text{for all }X\in\mathbb{R}.
 \]
 Thus for every \(X\) there are two distinct real values of \(Y\); the rotated curve fails the vertical-line test and cannot be the graph of a single-valued function \(y=f(x)\). Hence no function \(f\) exists satisfying the condition for every point \(A\in\Gamma\).

Answer: 3


The graph of the function \(\sin(x + \sqrt{2}) + a(x - b)^3\), where \(x \in \mathbb{R}\) and \(a \in \mathbb{R}\), is a centrally symmetric figure. Find the value of \(b\).

The graph of the function \(\sin(x + \sqrt{2}) + a(x - b)^3\), where \(x \in \mathbb{Q}\) and \(a \in \mathbb{R}\), is a centrally symmetric figure. Find the value of \(b\).

Assume the point-set \(\{(x,f(x)) : x\in\mathbb{Q}\}\) is centrally symmetric with centre \((x_0,y_0)\).
 Then for every \(x\in\mathbb{Q}\) the reflected abscissa \(2x_0-x\) must also lie in \(\mathbb{Q}\), hence \(2x_0\in\mathbb{Q}\) and in particular \(x_0\in\mathbb{Q}\).

 Write \(u=x-x_0\) and \(d=x_0-b\). The central symmetry condition
 \[
 f(x)+f(2x_0-x)=2y_0 \; (\forall x\in\mathbb{Q})
 \]
 becomes
 \[
 \sin(x+\sqrt{2})+\sin(2x_0-x+\sqrt{2})
 + a\bigl((x-b)^3+(2x_0-x-b)^3\bigr)=2y_0.
 \]
 Using angle-sum and expanding the cubes yields
 \[
 2\sin(x_0+\sqrt{2})\cos u + 2a d^3 + 6a d\,u^2 = 2y_0
 \; (\forall u\in\mathbb{Q}).
 \]
 The left-hand side is a function of \(u\) which contains the nonconstant terms \(\cos u\) and \(u^2\).
 Hence their coefficients must vanish:
 \[
 \sin(x_0+\sqrt{2})=0,\quad 6a d=0.
 \]

From \(\sin(x_0+\sqrt{2})=0\) we get \(x_0 = n\pi - \sqrt{2}\) for some \(n\in\mathbb{Z}\). But every such \(x_0\) is irrational, contradicting the necessary condition \(x_0\in\mathbb{Q}\).

Answer: b = k\pi - \sqrt{2},\quad k\in\mathbb Z


The cubic polynomial \((y^3 - 3y^2 + 4y - 1)\) is known to be a divisor of the function
 \[
 y^9 + m y^6 + n y^3 + o.
 \]
 Provide the triplet \((m,n,o).\)

The cubic polynomial \(P(y) = y^3 - 5y^2 + 6y - 1\) is known to be a divisor of the polynomial
 \[
 Q(y) = y^9 + m y^6 + n y^3 + 1.
 \]
 Determine the rational coefficients \(m\) and \(n\).

Let the roots of the cubic polynomial \(P(y)\) be \(\alpha, \beta, \gamma\). Since \(P(y)\) divides \(Q(y)\), these roots must also be roots of \(Q(y)\). The polynomial \(Q(y)\) is a polynomial in \(y^3\)
\[
Q(y) = R(y^3), \qquad R(y) = y^3 + m y^2 + n y + 1.
\]
If \(\alpha\) is a root of \(Q(y)\), then \(Q(\alpha) = R(\alpha^3) = 0\), so \(\alpha^3\) must be a root of \(R(y)\). The same applies to \(\beta\) and \(\gamma\). Therefore, the set of roots of \(R(y)\) must be \(\{\alpha^3, \beta^3, \gamma^3\}\). From the definition of \(R(y)\), its constant term is 1. By Vieta's formulas for \(R(y)\), the product of its roots is
\[
\alpha^3 \beta^3 \gamma^3 = -\frac{1}{1} = -1.
\]
Therefore,
\[
(\alpha\beta\gamma)^3 = -1.
\]

However, from Vieta's formulas for \(P(y)\), the product of its roots is
\[
\alpha\beta\gamma = -\frac{-1}{1} = 1.
\]
Cubing this gives
\[
(\alpha\beta\gamma)^3 = 1^3 = 1.
\]

Answer: (6,31,-1)


A function \(f:\mathbb{R}\to\mathbb{R}\) satisfies
\[
f(x)-2f(-x)=e^{x}.
\]
Find the slope of the tangent line to the curve \(y=f(x)\) at the point \((0,f(0))\).

A function \(f:\mathbb{R}\to\mathbb{R}\) satisfies
\[
f(x)-2f(x+1)=e^{x}.
\]
Find the slope of the tangent line to the curve \(y=f(x)\) at the point \((0,f(0))\).

We write any solution as
\[
f(x) = f_p(x) + f_h(x),
\]
where \(f_p\) is any particular solution of
\[
f(x) - 2f(x+1) = e^x,
\]
and \(f_h\) satisfies the homogeneous equation
\[
f_h(x) - 2f_h(x+1) = 0 \;\;\Longleftrightarrow\;\; f_h(x+1) = \tfrac12 f_h(x).
\]
All homogeneous solutions have the form
\[
f_h(x) = 2^{-x} P(x) = \left(\tfrac12\right)^x P(x),
\]

where \(P\) is any \(1\)-periodic function (\(P(x+1) = P(x)\)).  Thus the full general solution is
\[
f(x) = \frac{e^x}{1 - 2e} + 2^{-x} P(x),
\quad P(x+1) = P(x).
\]
Different choices of \(P\) (even constant choices) change \(f’(0)\):
\[
f'(0) = \frac{1}{1 - 2e} + \ln\!\tfrac12 \cdot P(0) + P'(0).
\]

2025 Chongqing High School Entrance Exam

Answer: 2025-01-03 00:00:00


Consider two points, \(P\) and \(Q\). Point \(P\) is situated on or within the square that has its diagonal corners at \((0,0)\) and \((2,2)\). On the other hand, point \(Q\) is located on or within another square having its diagonal corners at \((4,2)\) and \((5,3)\). What is the maximum possible value for the slope of the line that passes through points \(P\) and \(Q\)? Please provide your answer in the form of a simple fraction.

Consider two points, \(P\) and \(Q\). Point \(P\) is situated on or within the square that has its diagonal corners at \((0,0)\) and \((2,2)\). On the other hand, point \(Q\) is located on or within another square having its diagonal corners at \((4,2)\) and \((5,3)\). The positions of \(P\) and \(Q\) are related through a fixed anchor point \(A = (3,1)\) such that the vector \(\vec{AQ}\) is obtained by rotating the vector \(\vec{AP}\) counterclockwise by \(90^\circ\) about \(A\). Find the maximum possible slope of the line passing through \(P\) and \(Q\), over all point pairs \((P,Q)\) satisfying these conditions.

The \(90^\circ\) counterclockwise rotation around the anchor point \(A(3,1)\) creates a fixed algebraic link between the coordinates of \(P(x_P, y_P)\) and \(Q(x_Q, y_Q)\).
 We have
 \[
 x_Q = 4 - y_P, \qquad y_Q = x_P - 2.
 \]

 The possible region for \(Q\) (derived from \(P\)'s location) contradicts the given region for \(Q\).
 Point \(P\) is in square \(R_P\), so its coordinates are in the ranges \(0 \le x_P \le 2\) and \(0 \le y_P \le 2\).
 This forces its corresponding point \(Q\) to be in a derived region defined by
 \[
 x_Q = 4 - y_P \implies x_Q \in [2, 4], \qquad
 y_Q = x_P - 2 \implies y_Q \in [-2, 0].
 \]
 However, the problem requires \(Q\) to be in square \(R_Q\), which is defined by \(x_Q \in [4, 5]\) and \(y_Q \in [2, 3]\).

Answer: \frac{3}{2}


Let \(PQRS\) be a convex quadrilateral. The interior angles at \(Q\) and \(R\) satisfy
 \[
 \angle PQR = 110^\circ,\qquad \angle QRS = 100^\circ.
 \]
 The angles at \(P\) and \(S\) are each trisected by interior rays: write \(\angle QPS = 3x\) so that the three equal parts (counting from side \(PQ\) toward side \(PS\)) each measure \(x\), and write \(\angle RSP = 3y\) so that the three equal parts (counting from side \(RS\) toward side \(SP\)) each measure \(y\). Let \(G\) be the point inside the quadrilateral where the trisector ray from \(P\) that makes an angle \(2x\) with side \(PS\) (i.e. the ray two small parts away from \(PS\)) meets the trisector ray from \(S\) that makes an angle \(2y\) with side \(SP\) (i.e. the ray two small parts away from \(SP\)). Find the measure of \(\angle PGS\) in degrees.

In \(\triangle PQR\), the angles are \(\angle P = 90^\circ\) and \(\angle Q = 60^\circ\).
 The trisector of \(\angle Q\) that is adjacent to side \(PQ\) intersects the trisector of \(\angle R\) that is adjacent to side \(QR\) at a point \(G\). Let \(M\) be the midpoint of the side \(QR\). The geometric configuration is such that segment \(GM\) is equal in length to segment \(QM\). Determine the measure of \(\angle QGR\).

The sum of the angles in \(\triangle QGR\) gives \(\angle QGR = 180^\circ - (\angle RQG + \angle QRG) = 180^\circ - (40^\circ + 10^\circ) = 130^\circ.\) However, consider the derivation from the midpoint condition. The problem states that \(M\) is the midpoint of \(QR\), and that \(GM = QM\). This means that the point \(G\) is equidistant from the vertices \(Q\) and \(R\) and from the midpoint \(M\) of the segment \(QR\). Thus \(M\) is the circumcenter of \(\triangle QGR\). Therefore, \(\triangle QGR\) must be a right-angled triangle with the right angle at the vertex opposite the hypotenuse \(QR\). This gives \(\angle QGR = 90^\circ.\) The first derivation yields \(\angle QGR = 130^\circ\), while the second yields \(\angle QGR = 90^\circ\).

Answer: 80


Let \(PQRSGH\) be a hexagon with vertices listed in that order around a circle (i.e. the hexagon is cyclic). The interior angles at three vertices are marked as follows:
 \[
 \angle Q = 110^\circ,\qquad \angle H = 105^\circ,\qquad \angle S = \beta.
 \]
 Find the measure of \(\beta\) in degrees.

A convex hexagon \(PQRSGH\) is circumscribed about a circle \(\Omega\) of radius \(r\). The interior angles of the hexagon are
\[
\angle P = \angle Q = \angle R = \angle H = 120^\circ,
\quad \angle S = 130^\circ,
\quad \angle G = 110^\circ.
\]
The side lengths satisfy two conditions.  First, \(PQ = HP\). Second, the side lengths \(RS\), \(SG\), and \(GH\) form a geometric progression.  Determine the exact value of the ratio
\[
\frac{QR^2}{\operatorname{Area}(\Omega)}.
\]

For a tangential polygon with inradius \(r\), the length of a side between vertices whose interior angles are \(A,B\) equals
\[
\text{side}=r\!\left(\cot\frac A2+\cot\frac B2\right).
\]

With the given interior angles
\[
\angle P=\angle Q=\angle R=\angle H=120^\circ,\qquad \angle S=130^\circ,\qquad \angle G=110^\circ,
\]
the three consecutive sides \(RS,\,SG,\,GH\) correspond to the following half-angles:
\[
RS:\ \tfrac{R}{2}=60^\circ,\ \tfrac{S}{2}=65^\circ; \qquad
SG:\ \tfrac{S}{2}=65^\circ,\ \tfrac{G}{2}=55^\circ; \qquad
GH:\ \tfrac{G}{2}=55^\circ,\ \tfrac{H}{2}=60^\circ.
\]
Hence
\[
\begin{aligned}
RS &= r\bigl(\cot60^\circ+\cot65^\circ\bigr),\\[4pt]
SG &= r\bigl(\cot65^\circ+\cot55^\circ\bigr),\\[4pt]
GH &= r\bigl(\cot55^\circ+\cot60^\circ\bigr).
\end{aligned}
\]

The GP condition for \(RS,SG,GH\) is
\[
SG^2 = RS\cdot GH.
\]
Canceling \(r^2\) gives
\[
\bigl(\cot65^\circ+\cot55^\circ\bigr)^2
=\bigl(\cot60^\circ+\cot65^\circ\bigr)\bigl(\cot60^\circ+\cot55^\circ\bigr).
\]
Set
\[
x=\cot65^\circ,\qquad y=\cot55^\circ,\qquad c=\cot60^\circ=\frac{1}{\sqrt3}.
\]
The required identity becomes
\[
(x+y)^2=(c+x)(c+y).
\]
Since
\[
\begin{aligned}
x=\cot65^\circ &\approx 0.466,\\
y=\cot55^\circ &\approx 0.700,\\
c=\cot60^\circ &\approx 0.577.
\end{aligned}
\]
For the required identity,
\[
\text{LHS}=(x+y)^2\approx 1.361,\qquad
\text{RHS}=(c+x)(c+y)\approx 1.333,
\]

Because this equality is independent of \(r\), no choice of the inradius can fix the discrepancy.

Answer: 145^\circ


Consider the equation
 \[
 z^{10} + (13z - 1)^{10} = 0 \, .
 \]

 This equation has 10 roots in the complex plane:
 \[
 a_1, \ \overline{a}_1, \ a_2, \ \overline{a}_2, \ a_3, \ \overline{a}_3, \ a_4, \ \overline{a}_4, \ a_5, \ \overline{a}_5,
 \]
 where the overline signifies complex conjugation.

 Calculate the value of the following expression:
 \[
 \frac{1}{a_1\overline{a}_1}
 + \frac{1}{a_2\overline{a}_2}
 + \frac{1}{a_3\overline{a}_3}
 + \frac{1}{a_4\overline{a}_4}
 + \frac{1}{a_5\overline{a}_5}.
 \]

Consider the equation
 \[
 z^{10} + (13z - i)^{10} = 0
 \]
 in the complex plane, where \(i^2 = -1\). A direct application of the Hermitian Symmetry Principle for Binomial Equations confirms that the solution set \(V\) for this equation is closed under complex conjugation. The principle applies to equations of the form
 \[
 P(z)^n + (az+b)^n = 0,
 \]
 where \(P(z)\) is a polynomial with real coefficients, \(a \in \mathbb{R}\), and \(b\) is purely imaginary. In our case, with \(P(z) = z\), \(n=10\), \(a=13\), and \(b=-i\), the equation is a textbook example that satisfies the conditions of this principle. Consequently, since it can be shown that no root of the equation is real, the ten distinct roots can be grouped into five pairs of complex conjugates, denoted as \(\{a_k, \overline{a_k}\}_{k=1}^5\). Calculate the value of the following expression:
 \[
 \sum_{k=1}^{5} \frac{1}{|a_k|^2}.
 \]

Let \(a\) be a root. Then
 \[
 a^{10} = -(13a - i)^{10}
 \quad \Longrightarrow \quad \left(\frac{a}{13a - i}\right)^{10} = -1.
 \]
 Let \(\omega = \tfrac{a}{13a - i}\), so that \(\omega^{10} = -1\). Solving for \(a\) gives
 \[
 a = \frac{i\omega}{13\omega - 1}.
 \]
 Its conjugate is
 \[
 \overline{a} = \frac{-i\overline{\omega}}{13\overline{\omega} - 1}.
 \]
 If \(\overline{a}\) were also a root, then it must be expressible as
 \[
 \frac{i\omega'}{13\omega' - 1}, \qquad (\omega')^{10} = -1.
 \]
 Equating the two expressions leads to
 \[
 \overline{\omega} + \omega' = 26 \omega' \overline{\omega}.
 \]
 Taking moduli,
 \[
 |\overline{\omega} + \omega'| = 26,
 \]
 but by the triangle inequality,
 \[
 |\overline{\omega} + \omega'| \le 2.
 \]

Answer: 850


Determine the smallest positive angle \(\alpha\), for which the following identity holds:

  \[\tan \alpha = \frac{\cos 5^\circ \cos 20^\circ + \cos 35^\circ \cos 50^\circ - \sin 5^\circ \sin 20^\circ - \sin 35^\circ \sin 50^\circ}{\sin 5^\circ \cos 20^\circ - \sin 35^\circ \cos 50^\circ + \cos 5^\circ \sin 20^\circ - \cos 35^\circ \sin 50^\circ}.\]

Determine the smallest positive angle \(\alpha \in (0, \pi/6)\) for which the following identity holds:

 \[
 \csc^2(2\alpha) + \csc^2(4\alpha) + \csc^2(6\alpha)
 = \csc^2(3\alpha) + \csc^2(5\alpha).
 \]

Consider the function \(f(x) = \csc^2(x)\).
 Its second derivative is
 \[
 f''(x) = 2\csc^4(x)(1+2\cos^2(x)).
 \]
 For \(x \in (0, \pi)\), we have \(f''(x) > 0\), which means \(f(x)\) is strictly convex on this interval. For the smallest positive angle \(\alpha\), we must be in the domain where all arguments are defined and positive.
 This requires \(6\alpha < \pi\), so \(\alpha \in (0, \pi/6)\). For such \(\alpha\), the function \(\phi(t) = \csc^2(t\alpha)\) is strictly convex in \(t\) on the interval containing \([2,6]\). The equation can be rewritten as
 \[
 [\csc^2(6\alpha) + \csc^2(2\alpha)] - [\csc^2(5\alpha) + \csc^2(3\alpha)] + \csc^2(4\alpha) = 0.
 \]

 This setup suggests Karamata's inequality. Let \(\mathbf{x} = (6,2)\) and \(\mathbf{y} = (5,3)\). We note that \(\mathbf{x}\) majorizes \(\mathbf{y}\) because \(6 \geq 5\) and \(6+2 = 5+3\). By Karamata's inequality, for any convex function \(\phi\), if \(\mathbf{x}\) majorizes \(\mathbf{y}\), then
 \[
 \phi(x_1) + \phi(x_2) \geq \phi(y_1) + \phi(y_2).
 \]
 Since \(\phi(t) = \csc^2(t\alpha)\) is strictly convex, the inequality is strict:
 \[
 \csc^2(6\alpha) + \csc^2(2\alpha) > \csc^2(5\alpha) + \csc^2(3\alpha).
 \]

 Thus, the term
 \[
 [\csc^2(6\alpha) + \csc^2(2\alpha)] - [\csc^2(5\alpha) + \csc^2(3\alpha)]
 \]
 is strictly positive. Furthermore, the remaining term \(\csc^2(4\alpha)\) is also strictly positive for any \(\alpha\) in the domain.The sum of two strictly positive quantities cannot be zero. Therefore, the equation has no solution for \(\alpha \in (0, \pi/6)\).

Answer: 120^\circ


Consider a triangle labeled as \(PQR\), with \(PQ = 17\), \(PR = 8\), and \(QR = 15\). Let \(S\) be the point where the perpendicular from \(R\) intersects \(PQ\). What is the area of triangle \(PRS\)?

Consider a triangle \(PQR\) with side lengths \(PQ = 14\), \(PR = 13\), and \(QR = 15\). It is given that the distance between the triangle's incenter and its nine-point center is exactly 1. Let \(S\) be the foot of the altitude from vertex \(R\) to the side \(PQ\), and let \(G\) be the centroid of \(\triangle PQR\). Determine the length of the segment \(SG\).

For any triangle, the distance between the incenter \(I\) and the nine-point center \(N\) is given by the formula
 \[
 IN = \left|\frac{R}{2} - r\right|,
 \]
 where \(R\) is the circumradius and \(r\) is the inradius.
 This is a consequence of Feuerbach's Theorem, which states that the nine-point circle is tangent to the incircle. For the given triangle with side lengths 13, 14, and 15, the semi-perimeter is
 \[
 s = \frac{13+14+15}{2} = 21.
 \]
 The area, using Heron's formula, is
 \[
 A = \sqrt{21(21-13)(21-14)(21-15)}
  = \sqrt{21 \cdot 8 \cdot 7 \cdot 6}
  = 84.
 \]
 The inradius is
 \[
 r = \frac{A}{s} = \frac{84}{21} = 4.
 \]
 The circumradius is
 \[
 R = \frac{abc}{4A}
  = \frac{13 \cdot 14 \cdot 15}{4 \cdot 84}
  = \frac{2730}{336}
  = \tfrac{65}{8}.
 \]
 The radius of the nine-point circle is
 \[
 R_9 = \frac{R}{2} = \tfrac{65}{16}.
 \]
 Therefore, the true distance between the incenter and the nine-point center is
 \[
 IN = \left|\frac{65}{16} - 4\right|
  = \left|\frac{65 - 64}{16}\right|
  = \tfrac{1}{16}.
 \]

 contradicting to the claim that the distance \(IN\) is 1.

Answer: \frac{3840}{289}


Let \(v_3(x)\) denote the 3-adic valuation of an integer \(x\). For \(z\) in the ring of 3-adic integers \(\mathbb{Z}_3\) define the power series
\[
f(z)=\sum_{n=1}^{\infty}\frac{z^n}{n\cdot 3^{v_3(n)}}.
\]

Let
\[
S \;=\; f(3)-\tfrac12 \;=\; \sum_{n=1}^{\infty}\frac{3^n}{n\cdot 3^{v_3(n)}} - \tfrac12.
\]
Determine the number of trailing zeros in the base-3 representation of the 3-adic number \(S\); equivalently, compute \(v_3(S)\).

Let \(v_p(x)\) denote the \(p\)-adic valuation of \(x\). The number of trailing zeros in the base-\(p\) representation of an integer is given by its \(p\)-adic valuation. Consider the function \(f(z)\) defined for \(z\) in the domain of 3-adic integers \(\mathbb{Z}_3\) by the power series

 \[
 f(z) = \sum_{n=1}^{\infty} \frac{z^n}{n \cdot 3^{v_3(n)}}
 \]

 While the harmonic series \(\sum \frac{1}{n}\) diverges in real analysis, series in \(\mathbb{Z}_p\) can converge under different conditions. Determine the number of trailing zeros in the base-3 representation of the 3-adic number

 \[
 S = f(1) - \frac{1}{2}
 \]

 This is equivalent to finding \(v_3(S)\).

A series \(\sum a_n\) converges in the \(p\)-adic numbers if and only if the \(p\)-adic norm of its terms approaches zero: \(|a_n|_p \to 0\) as \(n \to \infty\).

 In this problem, we must check the convergence of
 \[
 f(1) = \sum_{n=1}^{\infty} \frac{1}{n \cdot 3^{v_3(n)}}.
 \]
 Let \(a_n = \frac{1}{n \cdot 3^{v_3(n)}}\) and the space is \(\mathbb{Q}_3\), then
 \[
 |a_n|_3 = \left|\frac{1}{n \cdot 3^{v_3(n)}}\right|_3 = \frac{1}{|n \cdot 3^{v_3(n)}|_3}.
 \]

 Using the multiplicative property of the norm,
 \[
 |n \cdot 3^{v_3(n)}|_3 = |n|_3 \cdot |3^{v_3(n)}|_3.
 \]

 The definition of the \(p\)-adic norm is \(|x|_p = p^{-v_p(x)}\). Thus,
 \[
 |3^{v_3(n)}|_3 = 3^{-v_3(n)}.
 \]

 Any integer \(n\) can be written as \(n = m \cdot 3^{v_3(n)}\), where \(m\) is not divisible by 3. Hence \(m\) is a 3-adic unit, so \(|m|_3 = 1\). Therefore,
 \[
 |n|_3 = |m \cdot 3^{v_3(n)}|_3 = |m|_3 \cdot |3^{v_3(n)}|_3 = 1 \cdot 3^{-v_3(n)} = 3^{-v_3(n)}.
 \]

 Substituting back,
 \[
 |n \cdot 3^{v_3(n)}|_3 = (3^{-v_3(n)}) \cdot (3^{-v_3(n)}) = 3^{-2v_3(n)}.
 \]

 Finally,
 \[
 |a_n|_3 = \frac{1}{3^{-2v_3(n)}} = 3^{2v_3(n)}.
 \]

 For the series to converge, we need \(|a_n|_3 = 3^{2v_3(n)} \to 0\). This would require the exponent \(2v_3(n)\) to approach \(-\infty\). However, \(v_3(n)\) is the exponent of 3 in the prime factorization of \(n\), so \(v_3(n) \geq 0\).

Answer: 0


Imagine a chain of \(n\) number \(7\)'s, \(7777\cdots77\), where you can place addition operators to create a mathematical equation.

 For instance,
 \[
 7 + 77 + 777 + 7 + 7 = 875
 \]
 could be generated using eight \(7\)'s in this method. What is the count of values of \(n\) for which you can put addition symbols such that the final sum of the equation equals \(7000\)?

Imagine a chain of \(n\) number \(3\)'s, \(3333\cdots33\), where you can place addition operators to create a mathematical equation. For instance,
 \[
 3 + 33 + 333 + 3 + 3 = 375
 \]
 could be generated using eight \(3\)'s in this method.  List all possible values of \(n\) for which you can put addition symbols such that the final sum of the equation equals \(7000\)?

Let the chain of \(n\) digits \(3\) be split by plus signs into blocks of lengths
 \(k_1,k_2,\dots,k_m\) (so \(k_i\ge 1\) and \(k_1+\cdots+k_m=n\)).
 A block of \(k\) threes equals
 \[
 \underbrace{33\cdots 3}_{k\ \text{digits}}=\frac{10^{k}-1}{3}.
 \]
 Thus the target equation “sum to \(7000\)” becomes
 \[
 \sum_{i=1}^m \frac{10^{k_i}-1}{3}=7000.
 \]
 Multiplying by \(3\) gives
 \[
 \sum_{i=1}^m(10^{k_i}-1)=21000
 \quad\Longrightarrow\quad
 \sum_{i=1}^m 10^{k_i}=21000+m.
 \]
 Reduce this congruence modulo \(9\).
 Since \(10\equiv 1\pmod 9\), we have \(10^{k_i}\equiv 1\pmod 9\) for every \(i\), hence
 \[
 \sum_{i=1}^m 10^{k_i}\equiv m \pmod 9.
 \]
 But \(21000\equiv 2+1+0+0+0\equiv 3\pmod 9\), so the right-hand side satisfies
 \[
 21000+m\equiv 3+m \pmod 9.
 \]
 Therefore the congruence would force \(m\equiv m+3\pmod 9\), i.e. \(0\equiv 3\pmod 9\),
 a contradiction. Hence no decomposition exists for any \(m\) (and thus for any \(n\)).

Answer: 108


A polynomial \(Q\) is defined such that for every real number \(y\), the equation
 \[
 y^{512} + y^{256} + 1 = (y^2 + y + 1) Q(y)
 \]
 holds true. When \(Q\) is expressed in the standard form of a polynomial, what is the count of its non-zero coefficients?

A polynomial \(Q\) is defined such that for every real number \(y\), the equation
 \[
 y^{512} + y^{256} + 1 = (y^{20} + y^2 + y + 1)\, Q(y)
 \]
 holds true. When \(Q\) is expressed in the standard form of a polynomial, what are its non-zero coefficients?

Let
 \[
 f(y)=y^{512}+y^{256}+1,\qquad g(y)=y^{20}+y^2+y+1.
 \]
 Set \(a=y^{256}\). Then
 \[
 f(y)=0 \iff a^2+a+1=0 \iff a\in\{\omega,\omega^2\},\quad
 \omega=\frac{-1+i\sqrt{3}}{2},\ \omega^3=1,\ |\omega|=1.
 \]
 Hence every root of \(f\) satisfies \(|y|^{256}=|a|=1\), so \(|y|=1\). If \(g\mid f\), then all roots of \(g\) lie on the unit circle, which forces \(g\) to be self-reciprocal:
 \[
 y^{\deg g}\,g\!\left(\tfrac{1}{y}\right)=g(y).
 \]
 But
 \[
 y^{20}g\!\left(\tfrac{1}{y}\right)=1+y^{18}+y^{19}+y^{20}\neq y^{20}+y^2+y+1=g(y)
 \]
 (since the coefficient of \(y^1\) is \(1\) while that of \(y^{19}\) is \(0\)). Thus \(g\nmid f\), so no polynomial \(Q\) with \(f(y)=g(y)\,Q(y)\) exists.

Answer: 341


Consider the function
 \[
 g(x) = \frac{1}{(x-1)^2 + (x+1)^2}.
 \]
 What set of real numbers can be used as the input for this function? Provide your answer using interval notation.

Consider the function
 \[
 g(x) = \frac{1}{(x-1)^2+(x+1)^2}.
 \]

 For the interval \([a, b]\) in which the function is not defined, Calculate:
 \[
 a^{15} + a^{14} + \cdots + a^{1} \;-\; \bigl(b^{15} + b^{14} + \cdots + b^{1}\bigr).
 \]

\[
 g(x)=\frac{1}{(x-1)^2+(x+1)^2}
 =\frac{1}{(x^2-2x+1)+(x^2+2x+1)}
 =\frac{1}{2x^2+2}
 =\frac{1}{2(x^2+1)}.
 \]
 For all real \(x\), \(x^2+1>0\), hence the denominator \(2(x^2+1)\neq 0\). Therefore \(g\) is defined for every real \(x\); there is no real interval \([a,b]\) on which the function is undefined.

Answer: (-\infty,\infty)


Determine the quantity of real solutions \((a,b,c,d)\) satisfying the set of equations below:
 \[
 \begin{aligned}
 a &= c + d + cda, \\
 b &= d + a + dab, \\
 c &= a + b + abc, \\
 d &= b + c + bcd.
 \end{aligned}
 \]

Determine the quantity of real positive solutions \((a,b,c,d)\) satisfying the set of equations below:
 \[
 \begin{aligned}
 a &= c + d + cda, \\
 b &= d + a + dab, \\
 c &= a + b + abc, \\
 d &= b + c - bcd,
 \end{aligned}
 \]
 \]

From
 \[
 a(1-cd)=c+d\quad\text{and}\quad c(1-ab)=a+b
 \]
 adding and simplifying gives
 \[
 (b+d)(1+ac)=0.
 \]
 If \(1+ac=0\), then multiplying \(c(1-ab)=a+b\) by \(a\) yields
 \[
 -ac(1-ab)=a^2+ab \;\Longrightarrow\; -(1-ab)=a^2+ab \;\Longrightarrow\; a^2=-1,
 \]
 impossible over \(\mathbb{R}\). Hence \(b+d=0\), i.e.\ \(d=-b\).

 With \(d=-b\), the fourth and second equations become
 \[
 -b=b+c+b^2c \;\Longrightarrow\; c=-\frac{2b}{1+b^2},\qquad
 b=-b+a-ab^2 \;\Longrightarrow\; 2b=a(1-b^2).
 \]
 Substituting these into \(c(1-ab)=a+b\) gives
 \[
 b\,(b^2+5)=0 \;\Longrightarrow\; b=0 \;\Longrightarrow\; d=0,\ a=0,\ c=0.
 \]
 Thus the only real solution is \((a,b,c,d)=(0,0,0,0)\), which violates the condition they are positive

Answer: 5


\[
 S = \{1,2,\ldots,2009\}.
 \]

 Find the maximum integer \(m\) such that there exist pairs
 \[
 (x_j,y_j)_{j=1}^m
 \]
 satisfying the following conditions:
 \[
 x_j < y_j,\quad \{x_1,y_1,\ldots,x_m,y_m\}\subseteq S,
 \]
 \[
 \{x_j,y_j\}\cap\{x_k,y_k\}=\varnothing \ \ (j\neq k),
 \]
 \[
 x_j+y_j \neq x_k+y_k \ \ (j\neq k), \quad x_j+y_j \leq 2009.
 \]

\[
 S = \{1005,1006,\ldots,2009\}.
 \]

 For the maximum integer \(m\) such that there exist pairs
 \[
 (x_j,y_j)_{j=1}^m
 \]
 satisfying the following conditions:
 \[
 x_j < y_j,\quad \{x_1,y_1,\ldots,x_m,y_m\}\subseteq S,
 \]
 \[
 \{x_j,y_j\}\cap\{x_k,y_k\}=\varnothing \ \ (j\neq k),
 \]
 \[
 x_j+y_j \neq x_k+y_k \ \ (j\neq k), \quad x_j+y_j \leq 2009
 \]

 Compute
 \[
 \frac{1}{m^{16}} + \frac{1}{(m-502)^{502}}.
 \]

For any pair \(x<y\) with \(x,y\in S=\{1005,\ldots,2009\}\), we have
 \[
 x\ge 1005,\quad y\ge 1006 \;\Rightarrow\; x+y\ge 1005+1006=2011>2009,
 \]
 so the constraint \(x_j+y_j\le 2009\) is impossible. Hence the maximum number of disjoint pairs is \(m=0\).
 The requested value
 \[
 \frac{1}{m^{16}}+\frac{1}{(m-502)^{502}}
 \]
 is undefined because \(\frac{1}{m^{16}}\) divides by zero at \(m=0\).

Answer: 803


Find the least positive whole number \(p\) such that when \(12{,}500{,}000\) is multiplied by \(p\), the resulting product has a remainder of \(111\) upon division by \(999{,}999{,}999\).

Find the least positive whole number \(p\) such that when \(12{,}500{,}000\) is multiplied by \(p\), the resulting product has a remainder of \(111\) upon division by \(1{,}000{,}000{,}000\).

Answer: 8880


A box with a shape of a perfect cube has sides measuring \(1 \, \text{cm}\) each. There's a special mark located at the middle of the top surface. This cube is placed on a leveled surface, such as a tabletop. Without any lifting or sliding, the cube is rotated in a specific direction, in such a way that a minimum of two corners are always maintaining contact with the surface. The cube continues to be rotated until the special mark is back at the top surface. The path taken by the special mark is expressed in centimeters as \(d\pi\), where \(d\) is an unknown constant. Determine the value of \(d\).

A box with a shape of a perfect cube has sides measuring \(1 \, \text{cm}\) each. There's a special mark located at the middle of the top surface. This cube is placed on a leveled surface, such as a tabletop. The cube is rotated in a specific direction, in such a way that a minimum of two corners are always maintaining contact with the surface. The cube can slide. The cube continues to be rotated until the special mark is back at the top surface. The path taken by the special mark is expressed in centimeters as \(d\pi\), where \(d\) is an unknown constant. Determine the value of \(d\).

Once sliding is allowed to occur simultaneously with rotation, the path length of the marked point will have additional arbitrary straight-line segments added. Depending on how much or how little you slide, the total length \(d\pi\) will change, so there is no unique \(d\).

Answer: \dfrac{1+\sqrt{5}}{2}


Consider the pentagon \(YWVZX\) with \(\angle X = 20^\circ\).
The four vertices \(X, Y, W, V\) lie on a circle (i.e. \(X,Y,W,V\) are concyclic).
Also assume \(\angle Y = \angle X\).  Find \(\angle Y + \angle W\) (in degrees).

Consider the pentagon \(YWVZX\) where \(\angle X = 20^\circ\). the vertices \(X, Y, W, V\) are concyclic. Can you determine the sum of \(\angle Y\) and \(\angle W\) in degrees?

Since \(X,Y,W,V\) are concyclic we know the four form a cyclic quadrilateral, opposite angles add to \(180^\circ\). Thus

 \[
 \angle X+\angle W=180^\circ \implies \angle W=180^\circ-20^\circ=160^\circ.
 \]

 So \(\angle W\) is determined (160°), but \(\angle Y\) is not fixed by the information given: you can place \(Y\) (and then \(V\) so that \(\angle Y+\angle V=180^\circ\)) at many different positions on the same circle and still keep \(X=20^\circ\) and \(W=160^\circ\). Therefore \(\angle Y+\angle W\) can take many values (for example, if \(\angle Y=30^\circ\) the sum is \(190^\circ\); if \(\angle Y=50^\circ\) the sum is \(210^\circ\), etc.).

Answer: 180^\circ


A palindromic number is a numeral that remains the same when its digits are reversed. Can you identify the smallest 5-digit palindromic number in binary representation that can also be depicted as a 3-digit palindromic number in another numerical base? Please provide your answer using binary notation.

A palindromic number is a numeral that remains the same when its digits are reversed. Can you identify the smallest 5-digit palindromic number in binary representation that can also be depicted as a 3-digit palindromic number in decimal (base-10)? Please provide your answer using binary notation.

A 5-digit binary number has the form \(b_4b_3b_2b_1b_0\) with \(b_4=1\). The smallest and largest 5-digit binaries are \(10000_2 = 1\cdot 2^4 = 16\). \(11111_2 = 1\cdot2^4 + 1\cdot2^3 + 1\cdot2^2 + 1\cdot2^1 + 1\cdot2^0 = 16+8+4+2+1 = 31\). Therefore every 5-digit binary number represents an integer in the range \(16\) through \(31\). A 3-digit decimal number must be at least \(100\) (and the smallest 3-digit palindromic decimal is \(101\)). Because \(31 < 100\), there is no integer that is both a 5-digit binary number and a 3-digit decimal number.

Answer: 10001_2


Identify all the positive, whole number solutions for 'm' that meet the condition outlined in the equation below:

  \[
  \cos \Bigl( \frac{\pi}{m} \Bigr) \cos \Bigl( \frac{2\pi}{m} \Bigr)
  \cos \Bigl( \frac{4\pi}{m} \Bigr) \cos \Bigl( \frac{8\pi}{m} \Bigr)
  \cos \Bigl( \frac{16\pi}{m} \Bigr)
  = \frac{1}{32}.
  \]

  Write your responses as a list, with each potential solution separated by a comma.

Identify all the positive, whole number solutions for 'm' that meet the condition outlined in the equation below:

  \[
  \cos \Bigl( \frac{\pi}{m} \Bigr) \cos \Bigl( \frac{2\pi}{m} \Bigr)
  \cos \Bigl( \frac{4\pi}{m} \Bigr) \cos \Bigl( \frac{8\pi}{m} \Bigr)
  \sin \Bigl( \frac{16\pi}{m} \Bigr)
  = \frac{1}{256}
  \]

  Write your responses as a list, with each potential solution separated by a comma.

Let \(x = \tfrac{\pi}{m}\). Using the product identity
 \[
 \prod_{k=0}^{3} \cos(2^k x) = \frac{\sin(16x)}{16 \sin x},
 \]
 the given equation becomes
 \[
 16 \sin^2(16x) = \sin(x), \qquad x = \tfrac{\pi}{m}.
 \]
 Define
 \[
 f(m) := 16 \sin^2\!\left(\tfrac{16\pi}{m}\right) - \sin\!\left(\tfrac{\pi}{m}\right).
 \]
 We seek integers \(m \geq 1\) such that \(f(m) = 0\).

 From the bound \(|\sin t| \leq \tfrac{1}{4}\) (with \(t = \tfrac{16\pi}{m}\)), one obtains
 \[
 \Bigl|\tfrac{16}{m} - k\Bigr| \leq \frac{\arcsin(1/4)}{\pi} \approx 0.08045
 \]
 for some integer \(k\). This restricts possible values of \(m\) to a short finite list of small integers. Direct checking shows that none of the candidates
 \[
 m \in \{4,5,8,15,16,17\}
 \]
 satisfies the original equation. For large \(m\) (e.g. \(m \geq 199\)), one has \(x = \pi/m\) small. In this regime, \(f\) is continuous and strictly decreasing on \([199,\infty)\), with
 \[
 f(199) > 0, \qquad \lim_{m \to \infty} f(m) < 0.
 \]
 Thus there is exactly one real root \(m^*\) in \((199,\infty)\). Numerically,\(f(12867) > 0, f(12868) < 0,\)
 so \(12867 < m^* < 12868\). Hence the unique large real root is non-integer. Combining the small-\(m\) check (no solutions) with the large-\(m\) fact, there is no positive integer \(m\) satisfying the equation.

Answer: 1933-03-11 00:00:00


Given three noncollinear points in a plane and a line labeled as \(p\). Assume that none of the points exist on a line that is parallel to \(p\) (or even on \(p\) itself). There are exactly \(x\) lines, let's call them \(q\), which are perpendicular to \(p\) and have the following trait: the three spheres with their centers at the given points and tangent to line \(q\), intersect at a certain point. Determine all potential values of \(x\). Please list all possible values of \(x\), separated by commas.

In a Cartesian coordinate, we have two circles centered at (0, 1) and (1, 2) with radii 1. We know that there is a third circle whose center is not collinear with the first two centers, also with radius 1. The three circles are tangent to the same line and intersect at the same point. Provide the range of all the possible slops for the line (exclude the vertical case).

The two circles centered at (0, 1) and (1, 2) has two intersection points. There are only two lines that can make the the two circles centered at (0, 1) and (1, 2) tangent to the same line. No matter which line it is, one can find that the third circle that tangent to this line and pass at least one of the two intersection points of the former two circles, would overlap one of the two former circles, violating the constraint that the three circle centers are not collinear.

Answer: 1