Assume that \(PQRST\) is a convex pentagon such that \(PQ \parallel RT\), \(QR \parallel PS\), and \(PR \parallel ST\).  If \(\angle PQR = 120^\circ\), the length of \(PQ\) is \(3\), the length of \(QR\) is \(5\), and the length of \(ST\) is \(15\), calculate the sum of the numerator and denominator of the simplest form of the ratio of the area of triangle \(\triangle PQR\) to the area of triangle \(\triangle TQS\).

Assume that \(PQRST\) is a convex pentagon such that \(PQ \parallel RT\), \(QR \parallel PS\), and \(PR \parallel ST\). If \(\angle QRP = 120^\circ\), the length of \(PQ\) is \(3\), the length of \(QR\) is \(5\), and the length of \(ST\) is \(15\). Additionally, the area of the trapezoid \(PRTS\) is given to be \(100\). Calculate the sum of the numerators and denominators of the simplest form of the ratio of the area of triangle \(PQR\) to the area of triangle \(TQS\).

Let \(PR=a\) with \(QR=5\)\(PQ=3\) and \(\angle QRP=120^\circ\). By the law of cosines on \(\triangle PQR\), \(c^2=a^2+b^2-2ab\cos\gamma\), where \(c=PQ=3\), \(b=QR=5\), and \(\gamma=\angle QRP=120^\circ\). Plugging in,
 \[
 9 = a^2 + 25 - 2\cdot a\cdot 5\cdot \cos(120^\circ).
 \]
 Since \(\cos(120^\circ)=-\tfrac{1}{2}\), this becomes
 \[
 9 = a^2 + 25 + 5a \quad\Longrightarrow\quad a^2 + 5a + 16 = 0.
 \]
 The discriminant is \(\Delta = 5^2 - 4\cdot 1\cdot 16 = 25 - 64 = -39 < 0\), so there is no real solution for \(a=PR\). Therefore \(\triangle PQR\) cannot exist with the given side lengths and included angle.

Answer: 484


Consider the function \(g\), where \(g(y) = -2 \sin(\pi y)\). Identify the number of solutions for \(y\) within the range \(-2 \le y \le 2\) that satisfy the equation
 \[
 g(g(g(y))) = g(y).
 \]

Consider the function \(g\), where \(g(y) = y^2+3\). Solve for \(y\) within the range \(-2 \le y \le 2\) that satisfy the equation
 \[
 g(g(g(y))) = g(y).
 \]

Answer: 61


For some positive integers \(p, q, r, s\), the equation
 \[
 4 \cos 27^\circ = \sqrt{p + \sqrt{q}} + \sqrt{r - \sqrt{s}}
 \]
 is valid. Here, \(q\) isn't a multiple of any prime number's square, and likewise, \(s\) isn't a multiple of any prime number's square. Determine the value of \(p + q + r + s\).

For some positive integers \(p, q, r, s\), the equation (\(p+s > 20\))
 \[
 4 \cos 27^\circ = \sqrt{p + \sqrt{q}} + \sqrt{r - \sqrt{s}}
 \]
 is valid.

 Here, \(q\) isn't a multiple of any prime number's square, and likewise, \(s\) isn't a multiple of any prime number's square. Determine the value of \(p + q + r + s\).

Let
 \[
 X := 4\cos 27^\circ = \sqrt{p+\sqrt q} + \sqrt{r-\sqrt s},
 \qquad q,s \ \text{squarefree}.
 \]

 Squaring gives
 \[
 X^2 = p+r + (\sqrt q - \sqrt s) + 2\sqrt{\,pr - q + (r-p)\sqrt q\,}.
 \]

 On the other hand,
 \[
 X^2 = (4\cos 27^\circ)^2 = 8 + 2\sqrt{\,10 - 2\sqrt 5\,}.
 \]

 Since the right-hand side has no isolated radical of the form \(\sqrt{\cdot}\), it follows that
 \[
 \sqrt q - \sqrt s = 0 \quad \Rightarrow \quad q = s.
 \]

 Thus the nested radical is of the form \(\sqrt{A + B\sqrt q}\) with
 \[
 A = pr - q, \qquad B = r - p.
 \]

 Comparing with \(\sqrt{\,10 - 2\sqrt 5\,}\) forces the squarefree bases to match, hence
 \[
 q = s = 5.
 \]

 Now compare coefficients:
 \[
 p+r = 8, \qquad pr - q = 10 \;\Rightarrow\; pr = 15, \qquad r-p = -2.
 \]

 Solving gives \((p,r) = (5,3)\). Therefore,
 \[
 p+s = 5 + 5 = 10.
 \]

 Hence imposing \(p+s > 20\) contradicts the unique solution above.

Answer: 18


Given two real numbers, \(p\) and \(q\), satisfying the two conditions:\(\dfrac{\sin p}{\cos q} + \dfrac{\sin q}{\cos p} = 1\) and \(\dfrac{\cos p}{\sin q} + \dfrac{\cos q}{\sin p} = 6\). Find the value of:\(\dfrac{\tan p}{\tan q} + \dfrac{\tan q}{\tan p}\).

Given two real numbers, \(p\) and \(q\), satisfying the two conditions:\(\dfrac{\sin p}{\cos q} + \dfrac{\sin q}{\cos p} = -1\) and \(\dfrac{\cos p}{\sin q} + \dfrac{\cos q}{\sin p} = 6\). Find the value of: \(\dfrac{\tan p}{\tan q} + \dfrac{\tan q}{\tan p}\).

Given
 \[
 \frac{\sin p}{\cos q}+\frac{\sin q}{\cos p}=-1,\qquad
 \frac{\cos p}{\sin q}+\frac{\cos q}{\sin p}=6,
 \]
 set \(a:=\tan p,\; b:=\tan q\). Note
 \[
 \sin p\cos p=\frac{a}{1+a^2},\qquad
 \sin q\cos q=\frac{b}{1+b^2},
 \]
 and also
 \[
 \sin p\cos p+\sin q\cos q
 =\frac{\sin p}{\cos q}\cdot\cos p\cos q+\frac{\sin q}{\cos p}\cdot\cos p\cos q.
 \]
 Using the first equation we get
 \[
 \sin p\cos p+\sin q\cos q = -\cos p\cos q,
 \]
 and using the second,
 \[
 \sin p\cos p+\sin q\cos q = 6\sin p\sin q.
 \]
 Equating these two expressions and dividing by \(\cos p\cos q\neq0\) yields
 \[
 -1=6\tan p\tan q \quad\Longrightarrow\quad ab=-\tfrac{1}{6}.
 \]

 Next square the identity
 \[
 \frac{a}{1+a^2}+\frac{b}{1+b^2}=\sin p\cos p+\sin q\cos q,
 \]
 and use the relation coming from \(\bigl(\sin p\cos p+\sin q\cos q\bigr)^2=\cos^2p\cos^2q\) (which follows from the first given equation). After clearing denominators one obtains the polynomial condition
 \[
 396\,a^4+1632\,a^2+11=0,
 \]
 for \(a^2\). Let \(x=a^2\). Then
 \[
 396x^2+1632x+11=0
 \]
 has discriminant \(>0\) but both roots are negative:
 \[
 x\in\Big\{-\tfrac{68}{33}-\tfrac{35\sqrt{15}}{66},\;-\tfrac{68}{33}+\tfrac{35\sqrt{15}}{66}\Big\}
 \approx\{-4.1144,\,-0.006751\}.
 \]
 Hence \(a^2<0\), which is impossible for a real \(a\).

Answer: \frac{124}{13}


There exists a five-figure number \(Z = 14{,}9XY\) which is a multiple of 12. The digits \(X\) and \(Y\) are both nonzero.
 What is the least possible value that \(Z\) could be?

There exists a five-figure number \(Z = 14{,}9XY\) which is a multiple of 20. The digits \(X\) and \(Y\) are both nonzero.
 What is the least possible value that \(Z\) could be?

If a number is a multiple of \(20\) then it must be divisible by \(5\) and by \(4\). Divisibility by \(5\) forces the last digit \(Y\) to be \(0\) or \(5\). The problem states \(Y\) is nonzero, so \(Y=5\), which is not a multiple of 4.

Answer: 14916


Suppose \(x,\) \(y,\) \(z,\) and \(w\) are positive real numbers which, when added together, total 10. Determine the maximum possible value of the expression \(xw^2 z^3 y^4.\)

Suppose \(x,\) \(y,\) \(z,\) and \(w\) are real numbers which, when added together, total 10. Determine the maximum possible value of the expression \(xw^2 z^3 y^4.\)

Let \(t>0\) and choose
 \[
 w=1,\qquad x=9,\qquad z=t,\qquad y=-t.
 \]

 Then
 \[
 x+y+z+w = 9 + (-t) + t + 1 = 10,
 \]
 so the constraint is satisfied for every \(t>0\).

 The product becomes
 \[
 xw^2 z^3 y^4 = 9\cdot 1^2 \cdot t^3 \cdot (-t)^4 = 9\,t^7.
 \]

 As \(t\to\infty\), \(9t^7\to\infty\).

Answer: 27648


Let's introduce two quantities, \(a = \sum_{n = 1}^\infty \dfrac{1}{n^2} \quad \text{and} \quad b = \sum_{n = 1}^\infty \dfrac{1}{n^3}.\) Your task is to express the following summation \(\sum_{i = 1}^\infty \sum_{n = 1}^\infty \dfrac{1}{(i + n)^3}\) in terms of the variables \(a\) and \(b\).

Let's introduce two quantities, \(a = \sum_{n = 1}^\infty \dfrac{1}{n^2} \quad \text{and} \quad b = \sum_{n = 1}^\infty \dfrac{1}{n^3}.\) Your task is to express the following summation \(\sum_{i = 1}^\infty \sum_{n = 1}^\infty \dfrac{1}{(i + n)}\) in terms of the variables \(a\) and \(b\).

For any fixed \(i\),

 \(
 \sum_{n=1}^\infty\frac{1}{i+n}\;\ge\;\sum_{n=1}^\infty\frac{1}{n+1}=\sum_{m=2}^\infty\frac{1}{m}=\infty,
 \)

 so each inner sum diverges and therefore \(S=\infty\). No finite expression in terms of \(a\) and \(b\) exists for the series as written.

Answer: a - b


Consider a fifth-degree polynomial, denote it as \(q(y)\). It satisfies the equation \(q(m) = \dfrac{m}{m^2 - 1}\) for \(m = 2, 3, 4, \dots, 7\). What is the value of \(q(8)?\)

Consider a sixth-degree polynomial, denote it as \(q(y)\). It is known to satisfy the equation \(q(y) = \dfrac{y}{y^2 - 1}\) for \(y = 2, 3, 4, 5, 6, 7\). What is the value of \(q(8)?\)

A degree-6 polynomial has 7 degrees of freedom, so 7 data points are required for uniqueness, but only 6 values \(q(y)=\dfrac{y}{y^2-1}\) at \(y=2,3,4,5,6,7\) are given. Let \(P(x)\) be any polynomial of degree \(\le5\) interpolating those six points. Then for any constant \(c\), \(q(x)=P(x)+c\prod_{y=2}^7 (x-y)\) is a degree-6 polynomial that still fits all six conditions. Since \(\prod_{y=2}^7 (8-y)=6!=720\neq0\),varying \(c\) changes \(q(8)\). Therefore \(q(8)\) is undetermined.

Answer: \frac{3}{56}


Given a function \(g\) that satisfies the equation \(g(p) + g(q) = g(p + q) - pq - 1\) for each and every real number \(p\) and \(q\). If we know that \(g(1) = 1\), your task is to determine all the integer values \(m\) for which \(g(m) = m.\) Please list all such integers, separated by commas.

Given functions \(f, g\) that satisfy the equation \(g(p) + g(q) = f(p + q) - pq - 1\) for each and every real number \(q\). If we know that \(g(1) = 1\), your task is to determine the three integer values \(m\) for which \(g(m) = m.\)

From
 \[
 g(p)+g(q)=f(p+q)-pq-1\quad(\forall p,q),
 \]
 put \(p=0\) to get \(f(x)=g(0)+g(x)+1\). Substituting back yields
 \[
 g(p+q)=g(p)+g(q)-g(0)+pq\qquad(\forall p,q).
 \]
 Put
 \[
 G(x):=g(x)-\tfrac12x^2,
 \]
 then
 \[
 G(p+q)=G(p)+G(q)-g(0).
 \]
 Let \(H(x):=G(x)-g(0)\). Then \(H(p+q)=H(p)+H(q)\), so for integers \(n\) we have \(H(n)=nH(1)\). Using \(g(1)=1\) gives \(H(1)=\tfrac12-g(0)\). Hence for integer \(m\),
 \[
 g(m)=\tfrac12 m^2 + mH(1)+g(0)=\tfrac12 m^2 +\Big(\tfrac12-g(0)\Big)m + g(0).
 \]
 Solving \(g(m)=m\) gives the quadratic
 \[
 m^2-(1+2g(0))m+2g(0)=0,
 \]
 whose roots are \(m=1\) and \(m=2g(0)\). Thus at most two integers can satisfy \(g(m)=m\); three distinct integer solutions is impossible.

Answer: 1,-2


For all real numbers denoted by \(x\), what is the greatest possible value of the expression \(4(x + 7)(2 - x)\)?

The function \( f(x) = 4(x+7)(p-x) \) describes a parabolic arch with its base on the \(x\)-axis, where \(p > -7\). A rectangle is inscribed under this arch, with its base lying on the \(x\)-axis. It is known that the maximum possible area of this inscribed rectangle is \(\mathbf{432}\), and that the vertex height of the parabolic arch is \(\mathbf{81}\). What is the value of \(p\)?

The vertex is at the midpoint of the roots, i.e. at

 \[
 x_v = \frac{p-(-7)}{2} = \frac{p+7}{2}.
 \]

 The vertex height is

 \[
 f(x_v) = 4\left(\frac{p+7}{2}\right)^2 = (p+7)^2.
 \]

 Given the vertex height is \(81\), we get

 \[
 (p+7)^2 = 81 \;\;\Rightarrow\;\; p = 2 \quad (\text{since } p>-7).
 \]

 For \(p=2\), the parabola is

 \[
 f(x) = 4(x+7)(2-x) = 81 - 4\left(x+\tfrac{5}{2}\right)^2.
 \]

 At a rectangle of height \(h\), the half-width is

 \[
 \sqrt{81 - h}.
 \]

 So the rectangle’s area is

 \[
 A(h) = h\sqrt{81-h}, \quad 0\leq h\leq 81.
 \]

 Maximizing this gives):

 \[
 A_{\max} = \frac{2}{3}\cdot 81 \cdot \sqrt{\frac{81}{3}}
 = 162\sqrt{3}\approx 280.6.
 \]

 But the problem statement requires \(A_{\max} = 432\).

Answer: 81


How many real solutions does the equation \(\sqrt{120-\sqrt{y}} = n\), where \(n\) is an integer, have for the variable \(y\)?

Let \(N\) be a fixed positive integer. We define the set \(S\) to consist of all real numbers \(x\) such that the expression \(\sqrt{\,N - \sqrt{x}\,}\) evaluates to an integer. It is known that the set \(S\) satisfies the following two conditions: \(S\) is a finite set, and its number of elements (i.e., its cardinality) is 14. Also, the set \(S\) has a maximum element, and this maximum is 22500. Determine the value of the integer \(N\) that satisfies all of the above conditions.

We have: if \(\sqrt{\,N-\sqrt{x}\,}\) is an integer \(k\ge0\), then
 \[
 N-\sqrt{x}=k^2 \implies \sqrt{x}=N-k^2 \ge0 \implies x=(N-k^2)^2.
 \]
 So
 \[
 S=\{(N-k^2)^2:\; k=0,1,\dots,\lfloor\sqrt{N}\rfloor\},
 \]
 hence \(|S|\le \lfloor\sqrt{N}\rfloor+1\), and \(\max S=N^2\). Given \(\max S=22500\) we have \(N^2=22500\), therefore \(N=150\). But \(\lfloor\sqrt{150}\rfloor=12\) (since \(12^2=144\) and \(13^2=169\)), so
 \[
 |S|\le 12+1=13,
 \]
 which contradicts the given \(|S|=14\).

Answer: 11


Let \(y\) be a real number. Find all solutions to the equation (using commas to separate them)

  \[\sin \left( \tan^{-1} (y) + \cot^{-1} \left( \frac{1}{y} \right) \right) = \frac{1}{3}.\]

Let \(y\) be a real number. Find all solutions to the equation (using commas to separate them)

 \[
 \sin\!\left(\arcsin(y) + \arccos\!\left(\sqrt{1-y^2}\right)\right) = -\tfrac{1}{2}.
 \]

Let \(A=\arcsin(y)\). Then \(A\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\) and \(y=\sin A\). We have
 \[
 \sqrt{1-y^2}=\sqrt{1-\sin^2 A}=\sqrt{\cos^2 A}=|\cos A|.
 \]
 Since \(\cos A\ge 0\) for \(A\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\), it follows that \(\sqrt{1-y^2}=\cos A\). Hence
 \[
 \arccos\!\big(\sqrt{1-y^2}\big)=\arccos(\cos A)=|A|.
 \]
 The given equation becomes
 \[
 \sin\big(\arcsin(y)+\arccos(\sqrt{1-y^2})\big)=\sin(A+|A|)=-\tfrac{1}{2}.
 \]

 If \(A<0\), then \(|A|=-A\) and \(A+|A|=0\). Thus \(\sin(A+|A|)=\sin 0=0\neq -\tfrac{1}{2}\). If \(A\ge 0\), then \(|A|=A\) and \(A+|A|=2A\). Since \(A\in[0,\tfrac{\pi}{2}]\), we have \(2A\in[0,\pi]\) and therefore \(\sin(2A)\ge 0\), so \(\sin(2A)=-\tfrac{1}{2}\) is impossible. Therefore no real \(A\) (hence no real \(y\)) satisfies the equation.

Answer: 3 \pm 2\sqrt{2}


Define \(T\) as the collection of all ordered triplets \((p,q,r)\) of positive integers that can be the lengths of the sides of a triangle. Find the value of
  \[\sum_{(p,q,r) \in T} \frac{2^p}{3^q 5^r}.\]

Let \(T\) be the set of all triples \((a,b,c)\) of positive integers for which there exist triangles with side lengths \(a, b, c\). Define the function \(S(x)\) for real values of \(x\) as: \( S(x) = \sum_{(a,b,c) \in T} \frac{x^a}{3^b \, 5^c}.\) It is known that \(S(2)\) converges to a finite rational value. Given \(S(4)+B=2\) for some real value \(B\), determine the value of \(B\).

Take \(n\ge1\) and the triples \((a,b,c)=(2n,n,n+1)\in T\) since \(b+c=2n+1>a\). Their contributions to \(S(4)\) are
 \[
 \frac{4^{2n}}{3^n5^{\,n+1}}=\frac{1}{5}\Big(\frac{16}{15}\Big)^n.
 \]
 Thus the subseries \(\sum_{n\ge1}\dfrac{1}{5}\big(\dfrac{16}{15}\big)^n\) is geometric with ratio \(16/15>1\), so it diverges. Therefore \(S(4)=+\infty\), and no finite real \(B\) can make \(S(4)+B=2\).

Answer: \frac{17}{21}


Consider the function \(p(a,b)\) which defines a surface: \(p = (2a - b)^2 - 2b^2 - 3b.\) In a strategic challenge, Euler and Newton are participants. Euler goes first and selects a value for \(a.\) Next, Newton chooses a value for \(b.\) Euler's objective is to achieve the lowest possible value of \(p\), while Newton aims to maximize \(p.\) If we assume Newton will make the best possible choice, what value of \(a\) should Euler select?

Consider the function \(p(a,b)\) which defines a surface: \(p = (2a - b)^2 + 2b^2 - 3b. \) In a strategic challenge, Euler and Newton are participants. Euler goes first and selects a real number for \(a\). Next, Newton chooses a real number for \(b\). Euler's objective is to achieve the lowest possible value of \(p\), while Newton aims to maximize \(p\). If we assume Newton will make the best possible choice, what value of \(a\) should Euler select?

Note that for any fixed \(a\), the function becomes
 \[
 p(b) = (2a - b)^2 + 2b^2 - 3b,
 \]
 which is a quadratic in \(b\) with leading coefficient \(+2+1=3>0\). Hence \(p(b)\) is a parabola that opens upwards. Therefore, it cannot be maximized.

Answer: -\frac{3}{8}


Given that the sum of \(\arccos y\), \(\arccos 2y\), and \(\arccos 3y\) is equal to \(\pi\), we know that \(y\) is a root of a cubic equation in the form of\(py^3 + qy^2 + ry + s = 0,\)where \(p, q, r,\) and \(s\) are all integers, and \(p \neq 0.\) Determine the smallest possible sum of the absolute values of \(p, q, r,\) and \(s.\)

Given that the sum of \(\arccos y\), \(\arccos 2y\), and \(\arccos 3y\) is equal to \(0\), we know that \(y\) is a root of a cubic equation in the form of \(py^3 + qy^2 + ry + s = 0,\) where \(p, q, r,\) and \(s\) are all integers, and \(p \neq 0\). Determine the smallest possible sum of the absolute values of \(p, q, r,\) and \(s\).

If \(\arccos y+\arccos(2y)+\arccos(3y)=0\) then each \(\arccos(\cdot)=0\), so \(y=1\), \(2y=1\), \(3y=1\) simultaneously, which is impossible.

Answer: 27


Consider a right-angled triangle \(\triangle ABC\), with the right angle at vertex \(C\). The length of the hypotenuse, \(AB\), is \(29 \,\text{cm}\). The radius of the circle inscribed within the triangle (the inradius) is \(6 \,\text{cm}\). Calculate the area of the triangle.

Consider a right-angled triangle \(\triangle ABC\), with the right angle at vertex \(C\). The length of the hypotenuse, \(AB\), is \(26 \,\text{cm}\). The radius of the circle inscribed within the triangle (the inradius) is \(6 \,\text{cm}\). Calculate the area of the triangle.

For a right triangle with legs \(a,b\) and hypotenuse \(c\),
 \[
 r=\frac{a+b-c}{2}\quad\Longrightarrow\quad a+b=2r+c.
 \]
 With \(c=26\) and \(r=6\) this gives \(a+b=38\).

 But
 \[
 (a+b)^2 = a^2+b^2+2ab = c^2 + 2ab.
 \]
 Since \((a-b)^2\ge0\) we have \(a^2+b^2\ge 2ab\), so \(2ab\le a^2+b^2=c^2=676\). Therefore
 \[
 (a+b)^2 \le c^2 + c^2 = 2c^2 = 2\cdot 676 = 1352.
 \]
 However \(38^2 = 1444 > 1352\). This contradiction shows no such \(a,b\) exist. \

Answer: 210\text{ cm}^2


Determine the quadratic polynomial with real coefficients such that \(3+i\) is a root, and the coefficient of the \(z^2\) term is 2.

Given that \(2z^2 - 12z - c = 0\) for positive \(c\), and the modulus of its imaginary roots is \(\sqrt{10}\), determine the value of \(c\).

The discriminant is: \(\Delta = (-12)^2 - 4(2)(-c) = 144 + 8c.\) For nonreal roots we need \(\Delta<0\): \(144+8c<0 \Rightarrow c<-18.\) Thus no positive \(c\) works for \(2z^2-12z-c=0\).

Answer: 2z^2 - 12z + 20


Let \(P=(0,1)\), \(Q=(2,5)\), \(R=(5,2)\), and \(S=(7,0)\). A quadrilateral is formed by joining \(P\) to \(Q\), \(Q\) to \(R\), \(R\) to \(S\), and \(S\) back to \(P\). The perimeter of \(PQRS\) can be written as \(d\sqrt{2}+e\sqrt{5}\), where \(d\) and \(e\) are integers. Find the value of \(d+e\).

Determine the sum of the interior angles of the quadrilateral formed by the points \(P = (0, 7)\), \(Q = (2, 5)\), \(R = (5, 2)\), and \(S = (7, 0)\).

(\displaystyle \text{slope}_{PQ}=\frac{5-7}{2-0}=-1,\;
 \text{slope}_{QR}=\frac{2-5}{5-2}=-1,\;
 \text{slope}_{RS}=\frac{0-2}{7-5}=-1,\;
 \text{slope}_{SP}=\frac{7-0}{0-7}=-1.\)

 All four slopes are \(-1\), so the points lie on the line \(y=-x+7\) and are collinear. The points do not form a nondegenerate quadrilateral.

Answer: 12


Consider all polynomials in a complex variable given by
 \(Q(w) = 4w^4 + ew^3 + fw^2 + gw + h\),
 where \(e,f,g,\) and \(h\) are integers satisfying \(0 \le h \le g \le f \le e \le 4\). Suppose the polynomial has a root \(w_0\) such that \(|w_0|=1\). Determine the sum of all possible values of \(Q(1)\) over every polynomial with these conditions.

Consider all polynomials in a complex variable given by
 \(Q(w) = 0.4w^4 + ew^3 + fw^2 + gw + h\),
 where \(e,f,g,\) and \(h\) are integers satisfying \(0 \le h \le g \le f \le e \le 0.15\). Suppose the polynomial has a root \(w_0\) such that \(|w_0|=1\). Determine the sum of all possible values of \(Q(1)\) over every polynomial with these conditions.

Integers with \(0\le h\le g\le f\le e\le 0.15\) imply \(e=f=g=h=0\). Thus \(Q(w)=0.4w^4\), whose only root is \(w=0\) (so \(|w|=0\neq1\)). No polynomial meets the "has a root with \(|w_0|=1\)" condition.

Answer: 92


Define a sequence \(\{s_m\}\) with \(s_1=1\). For \(m>1\), the rule is: if \(m\) is even, then \(s_m = 1+s_{m/2}\); if \(m\) is odd, then \(s_m = \tfrac{1}{s_{m-1}}\). Suppose that \(s_m = \tfrac{19}{87}\). Determine the value of \(m\).

Define a sequence \(\{s_m\}\) with \(s_1 = 1\). For \(m > 1\), the rule is: If \(m\) is even, then \(s_m = 10 \times s_{m/2}\);  If \(m\) is odd, then \(s_m = 1 + s_{m-1}\). Suppose \(s_m = 1200111011\). Determine the value of \(m\).

We can prove that for any \(s_m\) its digits only have 0 and 1 by induction. If \(m\) is even, then \(s_m\) is obtained by multiplying \(s_{m/2}\) by 10. So as long as \(s_{m/2}\) contains only 0s and 1s, \(s_m\) will also contain only 0s and 1s. In particular, all even-indexed terms will end in 0. If \(m\) is odd, then \(s_m\) is equal to \(s_{m-1} + 1\). Since \(s_{m-1}\) ends in 0, adding 1 will still result in a number with only digits 0 and 1. The base case is \(s_1 = 1\), so every element in \(s_m\) only contains 1s and 0s. 1200111011 is not a valid element.

Answer: 1905


Find how many solutions satisfy the equation
 \(2\sin^3 y - 5\sin^2 y + 2\sin y = 0\)
 for \(0 \leq y \leq 2\pi\).

Suppose \(\sin z = \sin(2\pi - z)\). Determine the average of the solutions to the equation

 \[
 8w^9 - 60w^8 + 174w^7 - 257w^6 + 234w^5 - 159w^4 + 74w^3 - 27w^2 + 6w - 1 = 0
 \]

 within the interval \(0 \leq z \leq 2\pi\), where \(w = \sin z\).

Using the identity
 \[
 \sin(2\pi-z)=\sin 2\pi\cos z-\cos 2\pi\sin z = -\sin z,
 \]
 the equation \(\sin z=\sin(2\pi-z)\) becomes \(\sin z=-\sin z\), so
 \(\sin z=0\). Thus
 \[
 z\in\{0,\pi,2\pi\}\quad (0\le z\le2\pi),
 \]
 and hence \(w=\sin z=0\) for each candidate \(z\).

 Plug \(w=0\) into the polynomial:
 \[
 8\cdot0^9-60\cdot0^8+\cdots+6\cdot0-1 = -1 \ne 0, so \(w=0\) is not a root. Therefore there are no \(z\in[0,2\pi]\) satisfying both conditions; the solution set is empty and its average is undefined.

Answer: 5


Let \(h(y) = 3y^2 - 7\). Suppose \(k(h(4)) = 9\). Determine the value of \(k(h(-4))\).

Let \(h(y) = 3y^2 - 7\). Suppose there exists a real number \(m\) such that \(h(m) = -9\). Determine the value of \(h(-m)\).

We have \(h(y)=3y^2-7\). If \(h(m)=-9\) then
 \[
 3m^2-7=-9 \implies 3m^2=-2 \implies m^2=-\tfrac{2}{3},
 \]
 which has no real solution. Hence no real \(m\) exists and the real-case hypothesis is impossible.

Answer: 9


Four prime numbers \(p, q, r,\) and \(s\) form a strictly increasing arithmetic progression with common difference 6, and it is given that \(p > 5\). What is the units digit of \(p\)?

Four prime numbers \(p, q, r,\) and \(s\) form a strictly increasing arithmetic progression with common difference 6, and it is given that \(250 > p > 100\). What is the units digit of \(p\)?

From the problem we require four primes \(p,\;p+6,\;p+12,\;p+18\). Since \(6\equiv1\pmod5\), these are congruent to \(p,\;p+1,\;p+2,\;p+3\) (mod \(5\)). To avoid a multiple of \(5\) (and hence a composite term, because \(p>100\) so none can be \(5\)), we must have \(p\equiv1\pmod5\). A direct search finds starting values \(p=5,11,41,61,251,601,\dots\). The largest working \(p\) below \(100\) is \(61\); the next is \(251\). Since the condition requires \(100<p<250\) and there is no working \(p\) in that open interval.

Answer: 1


An ellipse is tangent to both the coordinate axes, and its foci are located at \((2,\,-3+\sqrt{5})\) and \((2,\,-3-\sqrt{5})\). Determine the length of its major axis.

An ellipse is tangent to both the coordinate axes, and its foci are located at \((2,\,3+\sqrt{5})\) and \((2,\,-3-\sqrt{5})\). Determine the length of its major axis.

The foci are \( (2,\,3+\sqrt5) \) and \( (2,\,-3-\sqrt5) \). The center is their midpoint:
  \( \bigl(2,\tfrac{(3+\sqrt5)+(-3-\sqrt5)}{2}\bigr)=(2,0) \).
  So the major axis is vertical and we can write
  \( \frac{(x-2)^2}{b^2}+\frac{y^2}{a^2}=1 \) with \( c=3+\sqrt5 \) and \( c^2=a^2-b^2 \). For tangency to the \(y\)-axis \( (x=0) \) plug \( x=0 \): \( \frac{4}{b^2}+\frac{y^2}{a^2}=1 \). Tangency means a single \(y\)-solution, so \( 1-\tfrac{4}{b^2}=0 \Rightarrow b^2=4 \) (hence \(b=2\)). The tangency point is \( (0,0) \). For the \(x\)-axis \( (y=0) \) plug \( y=0 \):
  \( \frac{(x-2)^2}{b^2}=1 \Rightarrow x=2\pm b \).
  With \(b=2\) the ellipse meets the \(x\)-axis at \(x=0\) and \(x=4\), two distinct points — not a tangency. Thus the ellipse cannot be tangent to both axes simultaneously given the specified foci.

Answer: 6


In a right triangle, the longer leg measures one foot less than twice the shorter leg. If the area of the triangle is 60 square feet, determine the length of the hypotenuse in feet.

In a right triangle, the longer leg measures one foot less than twice the shorter leg. The area of the triangle is 60 square feet, and one of its angles measures \(32^\circ\). Determine the length of the hypotenuse in feet.

Let the shorter leg be \(S\) and the longer leg \(L\). Given \(L=2S-1\). Area \(=\tfrac12 S L = 60\) so \(S L = 120\). Substitute: \(S(2S-1)=120 \Rightarrow 2S^2 - S -120 = 0\).
 The discriminant is \(D=1+960=961=31^2\). Hence \( S=\dfrac{1+31}{4}=8\). Then \(L=2\cdot8-1=15\). The hypotenuse is \( h=\sqrt{8^2+15^2}=\sqrt{289}=17.\) But the acute angles are \( \arcsin(8/17)\approx 28.07^\circ \) and \( \arcsin(15/17)\approx 61.93^\circ\), so neither is \(32^\circ\). Therefore “one angle is \(32^\circ\)” contradicts the other given information.

Answer: 17


Suppose the real number \(y\) satisfies \(y - \tfrac{1}{y} = 4\). Determine the value of \(y^3 - \tfrac{1}{y^3}\).

Suppose the real number \(y\) satisfies \(y + \tfrac{1}{y} = 1\). Determine the value of \(y^3 + \tfrac{1}{y^3}\).

No real \(y\) satisfies \(y+\tfrac{1}{y}=1\) (discriminant \(1-4=-3<0\)).

Answer: 76


Consider three squares \(\,WXYZ\), \(\,WXAB\), and \(\,WZCB\) which are consecutive faces of a cube, where \(\,WC = 8\) units. Let \(D\) be the cube's remaining vertex. Suppose that \(L\), \(M\), and \(N\) are points on \(WX\), \(WZ\), and \(WC\) correspondingly, such that \(WL = WM = WN = 2\) units. We obtain a 3D structure \(R\) by boring a passageway through the cube. The walls of this passageway are the planes parallel to \(DW\) and containing the edges \(LM\), \(MN\), and \(NL\). The total surface area of the 3D structure \(R\), including the passage walls, is expressed as \(x + y\sqrt{z}\), where \(x\), \(y\), and \(z\) are positive integers and \(z\) is not a multiple of any prime's square. Determine the sum of \(x\), \(y\), and \(z\).

Consider three square faces \(\,WXYZ\), \(\,WXAB\), and \(\,WZCB\) of a cube that share a common vertex \(W\), where the edge \(WC\) has length 1 unit. Let \(D\) denote the remaining vertex of the cube. Suppose points \(L\), \(M\), and \(N\) lie on edges \(WX\), \(WZ\), and \(WC\) respectively, such that \(WL = WM = WN = 2\) units. A three-dimensional structure \(R\) is formed by removing a tunnel from the cube, where the tunnel walls consist of planes parallel to \(DW\) and passing through the edges \(LM\), \(MN\), and \(NL\). The total surface area of \(R\) (including the tunnel walls) is given by \(x + y\sqrt{z}\), with \(x\), \(y\), \(z\) positive integers and \(z\) square-free. Find the sum \(x + y + z\).

The cube edge length is \(WC = 1\). By definition, all cube edges from \(W\) have length \(1\). But \(WL = WM = WN = 2\) exceeds the edge length.

Answer: 417


Consider a regular hexagon with its center at the origin on the complex plane. The hexagon has opposite sides at a distance of one unit apart, with one pair of sides running parallel to the imaginary axis. Define \(Q\) as the region that exists outside this hexagon, and let \(T = \left\lbrace \tfrac{1}{y} \ \middle|\ y \in Q \right\rbrace\). What is the area of \(T\)?

Consider a regular hexagon centered at the origin on the complex plane. The distance between its opposite sides is 1 unit, and one pair of its sides is parallel to the imaginary axis. Let \( Q \) be the region inside the hexagon, and define \( T= \left\lbrace\frac{1}{y} \mid y \in Q\right\rbrace \). What is the area of the region \( T \)?

The hexagon has circumradius
 \[
 R = \frac{1}{\sqrt{3}},
 \]
 so it contains the origin.

 Define
 \[
 T = \left\{ \frac{1}{y} : y \in Q \right\}.
 \]
 Since \(0 \in Q\), the mapping \(y \mapsto \tfrac{1}{y}\) sends the origin to infinity. Thus \(T\) is unbounded, and its area is
 \[
 \infty.
 \]

Answer: 3 \sqrt{3} + 2 \pi


Larry chose an acute angle \(y\) (strictly between 0 and 90 degrees) and computed the values of \(\sin y\), \(\cos y\), and \(\tan y\). He then wrote these values on three separate pieces of paper. He distributed these papers to three learners, Anthony, Bethany, and Charlotte, each receiving one paper. He then challenged them to deduce which trigonometric function (sin, cos, or tan) corresponded to the value on their paper. Despite discussing the values on their papers amongst themselves, only Anthony was able to definitively ascertain which function resulted in the value on his paper. Calculate the sum of all possible values that Larry wrote on Anthony's paper.

Larry chose an acute angle \(y\) (strictly between 0 and 90 degrees) and computed the values of \(\sin y\), \(\cos y\), and \(\tan y\). He then wrote these values on three separate pieces of paper. He distributed these papers to three learners, Anthony, Bethany, and Charlotte, each receiving one paper. He then challenged them to deduce which trigonometric function (sin, cos, or tan) corresponded to the value on their paper. Despite discussing the values on their papers amongst themselves, only Anthony was able to definitively ascertain which function resulted in the value on his paper.  Calculate the third smallest possible value written on Anthony's paper.

There exists a maximal \(y_0\approx 0.0005848034309229532\) radians (\(\approx 0.0335067684^\circ\)) such that for every \(y\in(0,y_0)\) only the \(\cos\)-holder has a unique consistent identity. Numerically \(\cos(y_0)\approx 0.9999998290024784\). There is one further isolated case \(y=\tfrac{\pi}{4}\) giving the multiset \(\{\tfrac{\sqrt2}{2},\tfrac{\sqrt2}{2},1\}\), where only the \(\tan\)-holder (value \(1\)) identifies.

 \[
 \text{Possible values}=\{\cos y:0<y<y_0\}\ \cup\ \{1\},
 \]
 i.e. the open interval \((\cos(y_0),1)\) together with the point \(1\). The admissible Anthony values form a continuum (an interval arbitrarily close to \(1\)) plus the point \(1\); therefore there is no well-defined “smallest possible value” .

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


Given \(f(x)\) is defined on \(\mathbb{R}\), and satisfies \(f(x+y) + f(x-y) = f(x) f(y).\) and \(f(1) = 1\), compute  \(\sum_{k=1}^{22} f(k).\)

Let \(f:\mathbb{Z}\to\mathbb{R}\) satisfy \[ f(x+y)+f(x-y)=f(x)\,f(y+3)\quad\text{for all even integers x,y}\] and \(f(0)=0\), and also \(f\) is not constant on the even integers. Compute \[ \sum_{k=1}^{22} f(k). \]

Put \(x=2m,\;y=2n\) and define \(g(m):=f(2m),\;p(n):=f(2n+3)\). Then the identity (valid for all \(m,n\in\mathbb Z\)) becomes
\[
g(m+n)+g(m-n)=g(m)\,p(n).
\]
In particular, since \(p(0)=f(3)\), setting \(n=0\) gives
\[
g(m+0)+g(m-0)=g(m)\,p(0),
\]
that is,
\[
2g(m)=g(m)\,p(0).
\]
Since \(f\) is not constant on the even integers, we must have \(p(0)=2\). Now use the identity
\[
\sin((m+n)\theta)+\sin((m-n)\theta)=2\sin(m\theta)\cos(n\theta).
\]
It follows that the choice
\[
g(m)=C\sin(m\theta), \qquad p(n)=2\cos(n\theta)
\]
satisfies the functional equation for all \(m,n\). The pair \((g,p)\) determines \(f\) on all even and odd integers respectively:
\(g(m)=f(2m)\) (evens) and \(p(n)=f(2n+3)\) (odds). Hence,
\[
\sum_{k=1}^{22} f(k)
= C\sum_{m=1}^{11}\sin(m\theta) \;+\; 2\sum_{n=-1}^{9}\cos(n\theta).
\]
Since \(C,\theta\) can take arbitrary real values, the solutions are uncountably many.

Answer: -3


Consider \(g(y)\) to be a real-coefficient polynomial satisfying \(g(0) = 1\), \(g(2)+g(3)=125\) , and for all real values of \(y\),\(g(y)\cdot g(2y^2) = g(2y^3 + y)\). Determine the value of \(g(5)\).

Consider \( g(y) \) to be a real-coefficient polynomial satisfying \( g(0) = 1 \), \( g(2) + g(3) = 91 \), and for all real numbers \( y \), the equation \( g(y) \cdot g(y^2) = g(2y^3 + y) \) holds. Determine the value of \( g(5) \).

Suppose \(\deg g = d \geq 1,\)
 \[
 g(y) = a_d y^d + a_{d-1} y^{d-1} + \cdots + a_1 y + a_0,\quad a_0 = g(0) = 1.
 \]

 Compare leading terms in
 \[
 g(y) g(y^2) = g(2y^3 + y).
 \]

 The LHS leading term is \(a_d^2 y^{3d}\); the RHS leading term is \(a_d (2y^3)^d = a_d \, 2^d y^{3d}\). Hence
 \[
 a_d^2 = a_d \, 2^d \quad \Rightarrow \quad a_d = 2^d.
 \]

 Now compare the coefficient of \(y^{3d-1}\). On the LHS the only contribution is \(a_d \cdot a_{d-1}\). On the RHS there is no \(y^{3d-1}\) term, since \((2y^3 + y)^d\) expands to powers \(y^{3d}, y^{3d-2}, y^{3d-4}, \dots\).
 Thus
 \[
 a_d a_{d-1} = 0 \;\;\Rightarrow\;\; a_{d-1} = 0.
 \]

 Repeating this top–down coefficient comparison  (the RHS produces only exponents \(3d, 3d-2, 3d-4, \dots\))
 forces
 \[
 a_{d-2} = a_{d-3} = \cdots = a_1 = 0.
 \]

 Therefore \(g\) is constant, and with \(g(0)=1\) we get \(g \equiv 1\). But \(g \equiv 1\) gives \(g(2) + g(3) = 2\), contradicting the condition \(g(2)+g(3)=91\). Hence no real-coefficient polynomial \(g\) satisfies all the given conditions.

Answer: 676


For triangle \(XYZ\), there exists a single point \(Q\) such that the following equation holds:
 \[
 XZ^2 + XQ^2 + ZQ^2 = XY^2 + XQ^2 + YQ^2 = YZ^2 + ZQ^2 + YQ^2.
 \]
 Can you express the common value above in terms of the length of the sides \(p, q, r\), and the circumradius \(S\) of triangle \(XYZ\)?

For triangle \(XYZ\), there exist two distinct points \(P\) and \(Q\) such that the following equations hold:
 \[
 XZ^2 + XQ^2 + ZQ^2 = XY^2 + XQ^2 + YQ^2 = YZ^2 + ZQ^2 + YQ^2,
 \]
 \[
 XZ^2 + XP^2 + ZP^2 = XY^2 + XP^2 + YP^2 = YZ^2 + ZP^2 + YP^2.
 \]
 Calculate the distance between points \(P\) and \(Q\).

Let \(X,Y,Z\) be the position vectors of the triangle's vertices and let \(R\) be any point
 satisfying
 \[
 XZ^2 + XR^2 + ZR^2 \;=\; XY^2 + XR^2 + YR^2 \;=\; YZ^2 + ZR^2 + YR^2.
 \]

 Subtracting the equal sums yields
 \[
 \|R-X\|^2-\|R-Y\|^2 = YZ^2 - XZ^2,
 \qquad
 \|R-Y\|^2-\|R-Z\|^2 = XZ^2 - XY^2.
 \]

 But for vectors these differences are linear in \(R\):
 \[
 \|R-X\|^2-\|R-Y\|^2 = (\|X\|^2-\|Y\|^2) - 2R\cdot (X-Y),
 \]
 so the two equalities become the linear system
 \[
 R\cdot (X-Y) = C_1,\qquad R\cdot (Y-Z) = C_2,
 \]
 for constants \(C_1,C_2\) depending only on \(X,Y,Z\).

 If \(X,Y,Z\) are not collinear then \((X-Y)\) and \((Y-Z)\) are linearly independent, so the two linear equations have a single solution \(R\). Hence at most one point \(R\) can satisfy the original equalities. Therefore the existence of two distinct points \(P\) and \(Q\) is impossible for a genuine triangle.

Answer: 32S^2 - 3(p^2 + q^2 + r^2)


Consider an acute, non-equilateral triangle \(\triangle PQR\) with circumcircle \(\Theta\). The tangents to \(\Theta\) at points \(Q\) and \(R\) cross at a point \(V\). Let \(Z\) and \(W\) be the projections of \(V\) onto lines \(PQ\) and \(PR\), respectively. Given that \(VQ = VR = 16\), \(QR = 22\), and \(VZ^2 + VW^2 + ZW^2 = 1143\), compute the value of \(ZW^2\).

Consider an acute triangle \(\triangle PQR\) with circumcircle \(\Theta\). The tangents to \(\Theta\) at points \(Q\) and \(R\) intersect at point \(V\). Let \(Z\) and \(W\) be the feet of the perpendiculars from \(V\) to lines \(PQ\) and \(PR\), respectively. Given \(VQ = VR = 10\), \(QR = 4\), \(PQ=PR), and \(VZ^2 + VW^2 + ZW^2 = 300\), find the value of \(VZ^2 + VP^2\).

Place the circumcenter at the origin \(O=(0,0)\).  Let the symmetry axis be the \(x\)-axis.
Set
\[
Q=(x_0,2), \quad R=(x_0,-2) \quad (QR=4),
\]
let the circle radius be \(r\), the tangent intersection be \(V=(v,0)\), and let \(P\) lie on the circle. Since \(Q\) lies on the circle,
\[
x_0^2+2^2=r^2 \;\;\Longrightarrow\;\; x_0^2+4=r^2. \tag{1}
\]

From similar triangles \(\triangle VQO\sim\triangle VMQ\),
\[
\frac{OQ}{QM}=\frac{VO}{VQ}.
\]
With \(OQ=r,\; QM=2,\; VQ=10\),
\[
\frac{r}{2}=\frac{v}{10} \;\;\Longrightarrow\;\; r=\frac{v}{5}. \tag{2}
\]

From tangent–secant relation,
\[
VQ^2=VO^2-OQ^2 \;\;\Longrightarrow\;\; 100=v^2-r^2. \tag{3}
\]
Substitute (2):
\[
100=v^2-\Big(\tfrac{v}{5}\Big)^2
=\tfrac{24}{25}v^2
\;\;\Longrightarrow\;\;
v^2=\tfrac{625}{6},\;\; v=\tfrac{25}{\sqrt6}.
\]
Hence
\[
r=\tfrac{5}{\sqrt6},\quad x_0^2=r^2-4=\tfrac{1}{6},\quad x_0=\tfrac{1}{\sqrt6}.
\]

So
\[
O=(0,0),\;
Q=\Big(\tfrac{1}{\sqrt6},2\Big),\;
R=\Big(\tfrac{1}{\sqrt6},-2\Big),\;
V=\Big(\tfrac{25}{\sqrt6},0\Big),\;
r=\tfrac{5}{\sqrt6}.
\]

Since \(PQ=PR\) and \(\triangle PQR\) is acute, then \(P\) lies on the perpendicular bisector of \(QR\), so
\[
P=(-r,0)=\Big(-\tfrac{5}{\sqrt6},0\Big).
\]

Foot \(Z\) from \(V\) onto \(PQ\):
\[
Q-P=(\sqrt6,2),\;\;|Q-P|^2=10,\;\;V-P=(5\sqrt6,0).
\]
Projection parameter:
\[
t=\frac{(V-P)\cdot(Q-P)}{|Q-P|^2}
=\frac{(5\sqrt6)(\sqrt6)}{10}=3.
\]
So
\[
Z=P+3(Q-P)=\Big(\tfrac{13}{\sqrt6},6\Big),\qquad
W=\Big(\tfrac{13}{\sqrt6},-6\Big).
\]
Thus \(ZW=12.\)

Then
\[
VZ=\Big(\tfrac{12}{\sqrt6},-6\Big)\;\;\Longrightarrow\;\;VZ^2=60,\quad VW^2=60.
\]
So
\[
VZ^2+VW^2=120,\quad VZ^2+VW^2+ZW^2=264 \neq 300
\]

Answer: 717


Consider a triangle, \(\triangle XYZ\). The median drawn from vertex \(X\) intersects at a right angle with the median drawn from vertex \(Y\). Provided that the lengths of \(YZ\) and \(XZ\) are 7 and 6 units respectively, determine the length of \(XY\).

Consider a triangle \(\triangle XYZ\). The medians XM from vertex X and YN from vertex Y are perpendicular to each other. Given that the lengths of sides YZ and XZ are 7 units and 6 units respectively, and point P is a point on the line segment MN. Find the distance from the orthocenter of triangle XYP to point Z.

Place \(Z=(0,0)\), \(X=(6,0)\). From the perpendicular-medians condition
 \(Y=\big(\tfrac{17}{3},\pm\tfrac{2\sqrt{38}}{3}\big)\). Midpoints: \(M=(\tfrac{17}{6},\tfrac{\sqrt{38}}{3})\), \(N=(3,0)\). Parametrize \(P\in MN\) by \(P=M+t(N-M)\), \(t\in[0,1]\):
 \[
 P=\Big(\tfrac{17}{6}+\tfrac{t}{6},\;\tfrac{\sqrt{38}}{3}(1-t)\Big).
 \]
 Let \(H=(h_x,h_y)\) be the orthocenter of \(\triangle XYP\). Solving
 \((H-X)\cdot(Y-P)=0\) and \((H-Y)\cdot(X-P)=0\) yields
 \[
 h_x=\frac{17}{12}t^2-\frac{1}{6}t+\frac{53}{12},\qquad
 h_y=\sqrt{38}\Big(\frac{17}{912}t^2-\frac{77}{228}t+\frac{17}{48}\Big).
 \]
 Hence
 \[
 HZ(t)=\sqrt{\frac{4913\,t^4-2312\,t^3+42262\,t^2-25688\,t+59033}{2432}}.
 \]
 which is dependent on \(t\). So the distance from the orthocenter of triangle XYP to point Z is underdetermined.

Answer: \sqrt{17}


Points \(M,N,P,Q\) are collinear, with \(P\) and \(Q\) on the ray \(MN\) beyond \(N\) and \(MN<MP<MQ\), \(MN=4\). Let \(O\) be a point not on line \(MN\) such that \(OM=6\), and \(ON=5\). Let \(A\) be the second intersection point (other than \(O\)) of the circumcircles of \(\triangle MOP\) and \(\triangle NOQ\). Suppose \(PA=2\) and \(QA=7\). Then \(NQ=\frac{m+n\sqrt{p}}{q},\) where \(m,n,p,q\) are positive integers, \(p\) is squarefree, and \(\gcd(m,q)=1\). Find \(m+n+p+q\).

Points \(M,N,P,Q\) are collinear, with \(P\) and \(Q\) on the ray \(MN\) beyond \(N\) and \(MN<MP<MQ\), \(MN=4\). Let \(O\) be a point not on line \(MN\) such that \(OM=6\), and \(ON=5\). Let \(A\) be the second intersection point (other than \(O\)) of the circumcircles of \(\triangle MOP\) and \(\triangle NOQ\). Suppose \(PA=2\) and \(QA=7\). There is another point \(B\) sliding on the segment \(AM\), and the distance between the center of the circumcircle of \(BPQ\) and the center of the circumcircle of \(APQ\) is a constant. Compute such a constant.

We first show that \(P, Q, A\) can be uniquely determined. Put the line \(MN\) on the \(x\)-axis with \(M=(0,0),\qquad N=(4,0).\). WLOG assume that \(O) is above line \MN\), then \(O\) is determined, since the lengths of three sides of triangle \(MNO\) is known. Then we set \(P=(p,0),\quad Q=(q,0),\quad\text{with } 4<p<q.\), and let \(A=(x,y)\) with \(y\neq 0\). Now we have four constraints, \(PA=2\), \(QA=7\), \(M,O,P,A \text{ concyclic},\text{and} N,O,Q, A \text{ concyclic}.\) Then it would be enough to determine the coordinate of \(P, Q, A\). So the center of the circumcircle of triangle \(APQ\) is fixed. However, when \(B\) is sliding on the segment \(AM\), triangle \(BPQ\)'s height is changing, and obviously the center of its circumcircle is also changing. Then the distance between the centers of two circumcircles is also changing, so it is underdetermined.

Answer: 32


Consider a square named \(WXYZ\) with each side measuring 13. Inside this square, there is a point \(P\) such that the distance from \(P\) to \(W\) is 5 and the distance from \(P\) to \(Y\) is 12. Can you determine the distance from point \(P\) to side \(WZ\)?

Consider a square \(WXYZ\) with side length 13. Point \(P\) lies inside the square, with \(PW = 5\) and \(PY = 12\). Point \(Q\) is on the line segment \(PX\) such that \(QX = 3\), and \(QZ = 13\). Find the distance between points \(P\) and \(Q\).

Place coordinates \(W=(0,0),\; X=(13,0),\; Y=(13,13),\; Z=(0,13)\). Then the distance between opposite vertices is \(WY=\sqrt{(13-0)^2+(13-0)^2}=13\sqrt2.\) The triangle-inequality constraint for any point \(P\) gives \(|PY-PW|\le WY \le PY+PW.\) Plugging the given values: \(|12-5|=7 \le WY \le 12+5=17.\ But \(WY=13\sqrt2 \approx 18.384 > 17.\)

Answer: 1\frac{12}{13}


The parabola with equation \(y = x^2 - 4\) is rotated \(60^\circ\) counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has \(y\)-coordinate \(\tfrac{a - \sqrt{b}}{c}\), where \(a\), \(b\), and \(c\) are positive integers, and \(a\) and \(c\) are relatively prime. Find \(a + b + c\).

The equation of the parabola is \( y = x^2 - 4 \). After rotating it counterclockwise by \( 60^\circ \) about the origin, a new image is obtained. The original parabola and its rotated image form a region \( T \) that contains the point \( (1, 1) \). Let \( S \) be the image of \( T \) under inversion with respect to the origin. Find the area of region \( S \).

The area enclosed by the original and rotated curves covers the origin, so its inversion is infinitely large.

Answer: 62


Let \(A_1A_2\dots A_{11}\) be a non-convex \(11\)-gon such that

 - The area of \(A_iA_1A_{i+1}\) is \(1\) for each \(2 \le i \le 10\),
 - \(\cos(\angle A_iA_1A_{i+1}) = \tfrac{12}{13}\) for each \(2 \le i \le 10\),
 - The perimeter of \(A_1A_2\dots A_{11}\) is \(20\).

 If \(A_1A_2 + A_1A_{11}\) can be expressed as \(\tfrac{m\sqrt{n} - p}{q}\) for positive integers \(m,n,p,q\) with \(n\) squarefree and \(\gcd(m,p,q) = 1\), find \(m+n+p+q\).

Let \(A_1A_2\dots A_{11}\) be a non-convex \(11\)-gon such that

 - The area of \(A_iA_1A_{i+1}\) is \(1\) for each \(2 \le i \le 10\),
 - \(\cos(\angle A_iA_1A_{i+1}) = \tfrac{12}{13}\) for each \(2 \le i \le 10\),
 - The perimeter of \(A_1A_2\dots A_{11}\) is \(10\).

 If \(A_1A_2 + A_1A_{11}\) can be expressed as \(\tfrac{m\sqrt{n} - p}{q}\) for positive integers \(m,n,p,q\) with \(n\) squarefree and \(\gcd(m,p,q) = 1\), find \(m+n+p+q\).

Set \(A_1A_2 = x\) and \(A_1A_3 = y\). By the first condition, we have \(\tfrac{1}{2}xy \sin \theta = 1\), where \(\theta = \angle A_2A_1A_3\). Since \(\cos \theta = \tfrac{12}{13}\), we have \(\sin \theta = \tfrac{5}{13}\), so \(xy = \tfrac{26}{5}\). Repeating this process for \(\triangle A_iA_1A_{i+1}\), we get \(A_1A_2 = A_1A_4 = \dots = A_1A_{10} = x\) and \(A_1A_3 = A_1A_5 = \dots A_1A_{11} = y\). Since the included angle of these 9 triangles is \(\theta\), the square of the third side is
 \[
 x^2+y^2-2xy\cos\theta = x^2+y^2-\tfrac{52}{5}\cdot\tfrac{12}{13}
 = x^2+y^2-\tfrac{48}{5}=(x+y)^2-20.
 \]
 Thus the third side has length \(\sqrt{(x+y)^2-20}\). The perimeter is constructed from 9 of these lengths, plus \(A_{11}A_1 + A_1A_2 = x+y\), so \(9\sqrt{(x+y)^2-20}+x+y=10.\) Let \(x+y=a\), we have \(9\sqrt{a^2-20}+a=10\). Moving \(a\) to the right-hand side and squaring both sides, we get \(4a^2+a-86=0.\), and solving it yields \(a=\frac{-1 + 9\sqrt{17}}{8}\), since the side length should be positive. We can verify \(a^2-20 > 0\), so this is the solution of the equation \(9\sqrt{a^2-20}+a=10\). However, according to AM-GM Inequality, \(x+y \geq 2\sqrt{xy} = 2\sqrt{\frac{26}{5}}\). One can verify that \(2\sqrt{\frac{26}{5}} > \frac{-1 + 9\sqrt{17}}{8}\). Thus, there is no real solution for \(x\) and \(y\).

Answer: 19

The annotation is modified from the solution in: https://artofproblemsolving.com/wiki/index.php/2025_AIME_II_Problems/Problem_12


 Consider the vectors \(\mathbf{u} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},\; \mathbf{v} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix},\; \mathbf{w} = \begin{pmatrix} 4 \\ 1 \\ -5 \end{pmatrix}\), which are pairwise orthogonal.
 There exist real numbers \(d,\; e,\; f\) such that

 \[
 \begin{pmatrix} -4 \\ 7 \\ 3 \end{pmatrix} = d\mathbf{u} + e\mathbf{v} + f\mathbf{w}.
 \]

 Find the ordered triple \((d,e,f)\).

Consider the vectors \(\mathbf{u} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\),
 \(\mathbf{v}\), and \(\mathbf{w} = \begin{pmatrix} 4 \\ 1 \\ -5 \end{pmatrix}\),
 which are pairwise orthogonal.

 There exist real numbers \(d,\; e,\; f\) such that

 \[
 \begin{pmatrix} -4 \\ 7 \\ 3 \end{pmatrix} = d\mathbf{u} + e\mathbf{v} + f\mathbf{w}.
 \]

 Determine the ordered triple \((d,e,f)\) and the vector \(\mathbf{v}\).

We need \(\mathbf{u}=(1,1,1)\), \(\mathbf{v}\), \(\mathbf{w}=(4,1,-5)\) to be pairwise orthogonal.
 This forces \(\mathbf{v}=(2k,-3k,k)\).

 Now expand
 \[
 (-4,7,3)=d\mathbf{u}+e\mathbf{v}+f\mathbf{w}.
 \]

 From orthogonality we get
 \[
 d=2,\quad f=-\tfrac{4}{7},\quad e=\frac{\mathbf{v}\cdot(-4,7,3)}{\mathbf{v}\cdot \mathbf{v}}
  =-\frac{13}{7k}.
 \]

 \(e\) depends on the arbitrary scale factor \(k\).
 Since scaling \(\mathbf{v}\) should not change the decomposition, the coefficient \(e\) is not well-defined.

Answer: (2, -\frac{13}{7}, -\frac{4}{7})


If  \(\frac{a_{n+1}-a_n}{a_n} =\frac{a_{n+2}-a_{n+1}}{a_{n+2}}  )\),  for \(n\in\mathbb{N}^*\) and \(a_1a_2 = 1\), what is the simplest fraction, simply expressed in terms of \(n\) (without any other free variable), for \(S_n = a_2a_3 + \cdots + a_n a_{n+1}\)

Put \(b_n:=1/a_n\). The recurrence is equivalent to the linear second-order difference
\[
b_{n+2}-2b_{n+1}+b_n=0,
\]
so
\[
b_n = x + (n-1)d
\]
for constants \(x=b_1\) and \(d=b_2-b_1\). The condition \(a_1a_2=1\) becomes
\[
\frac{1}{b_1 b_2}=1 \quad\Longrightarrow\quad x(x+d)=1.
\]
That is a single equation for the two unknowns \(x,d\). Therefore one degree of freedom remains — \(x\) (or \(d\)) is not fixed by the given information. By telescoping
\[
S_n=\sum_{k=2}^n \frac{1}{b_k b_{k+1}}
=\frac{1}{d}\Big(\frac{1}{b_2}-\frac{1}{b_{n+1}}\Big)
=\frac{1}{d}\Big(\frac{1}{x+d}-\frac{1}{x+nd}\Big).
\]
Because \(x\) and \(d\) are linked only by \(x(x+d)=1\), the right-hand side still depends on the free choice of \(x\) (or \(d\)). Hence \(S_n\) is not determined uniquely by the data and thus cannot be simply expressed in terms of \(n\).

Answer: \frac{2n}{n+2}


Define the function

 \[
 q(y,z) =
  \begin{cases}
  y + z & \text{if } y \ge 0 \text{ and } z \ge 0, \\[6pt]
  y - 2z & \text{if } y < 0 \text{ and } z < 0, \\[6pt]
  3y + z & \text{otherwise}.
  \end{cases}
 \]

 Determine the value of \(q(q(1,-1),\, q(-5,-2))\).

Define the function

 \[
 q(y,z) =
  \begin{cases}
  y + z & \text{if } y \ge 0 \text{ and } z \ge 0, \\[6pt]
  y - 2z & \text{if } y \ge 0 \text{ and } z < 0, \\[6pt]
  q(y,-2z) & \text{otherwise}.
  \end{cases}
 \]

 Suppose that \(q(-1,1) = 1\). Find the value of \(q(q(1,-1),\, q(-5,-2))\).

\[
  q(-5,-2)=q(-5,4)=q(-5,-8)=q(-5,16)=\cdots
  \]
  The recursion never changes the sign of \(y\) (it stays \(-5\)), so no terminating case is ever reached. Hence \(q(-5,-2)\) is undefined.

Answer: 5


A circle is inscribed in quadrilateral \(WXYZ\), tangent to \(WX\) and \(YZ\) at points \(E\) and \(F\), respectively. If \(WE = 19\), \(EX = 26\), \(YF = 37\), and \(FZ = 23\), what is the value of square of radius of the inscribed circle?

A circle is inscribed in quadrilateral \(WXYZ\), tangent to \(WX\) and \(YZ\) at points \(E\) and \(F\), respectively. It is given that \(WE = 24\), \(EX = 16\), \(YF = 32\), and \(FZ = 12\). Find all the possible integer radius (radii) of the inscribed circle.

Let the tangent lengths at vertices \(W,X,Y,Z\) be
 \[
 a=24,\quad b=16,\quad c=32,\quad d=12,
 \]
 so the circle is tangent to the four sides with those contact-segment lengths. For a tangential quadrilateral with inradius \(r>0\) the center-to-tangent geometry gives the angle relation
 \[
 \arctan\!\frac{a}{r}+\arctan\!\frac{b}{r}+\arctan\!\frac{c}{r}+\arctan\!\frac{d}{r}=\pi.
 \]
 Take the tangent of both sides; since \(\tan\pi=0\) we get
 \[
 \tan\Big(\arctan\frac{a}{r}+\arctan\frac{b}{r}+\arctan\frac{c}{r}+\arctan\frac{d}{r}\Big)=0.
 \]
 Using repeated tangent-addition formulas and simplifying the numerator (the denominator cannot be zero for positive \(r\)) yields the polynomial factor
 \[
 84\,r\,(r^2-384)=0.
 \]
 Discarding \(r=0\) (not a valid radius), we obtain
 \[
 r^2=384 \quad\Rightarrow\quad r=8\sqrt6.
 \]
 Thus the unique positive radius is \(8\sqrt6\), which is not an integer.

Answer: 647


Consider the interval \(\pi \leq \alpha < 2\pi\). We define two series as follows:

 \[
 A = \tfrac12 \cos \alpha - \tfrac14 \sin 2\alpha - \tfrac18 \cos 3\alpha + \tfrac1{16} \sin 4\alpha + \tfrac1{32} \cos 5\alpha - \tfrac1{64} \sin 6\alpha - \tfrac1{128} \cos 7\alpha + \dotsb
 \]

 and

 \[
 B = 1 - \tfrac12 \sin \alpha - \tfrac14 \cos 2\alpha + \tfrac1{8} \sin 3\alpha + \tfrac1{16} \cos 4\alpha - \tfrac1{32} \sin 5\alpha - \tfrac1{64} \cos 6\alpha + \tfrac1{128} \sin 7\alpha + \dotsb
 \]

 We are given that the ratio \(\tfrac{A}{B} = \tfrac{2\sqrt{2}}{7}\). The task is to determine the value of \(\sin \alpha\).

Consider the interval \(\tfrac{\pi}{2} \leq \alpha < \pi\). We define two series as follows:

 \[
 A = \tfrac12 \cos \alpha - \tfrac14 \sin 2\alpha - \tfrac18 \cos 3\alpha + \tfrac1{16} \sin 4\alpha + \tfrac1{32} \cos 5\alpha - \tfrac1{64} \sin 6\alpha - \tfrac1{128} \cos 7\alpha + \dotsb
 \]

 and

 \[
 B = 1 - \tfrac12 \sin \alpha - \tfrac14 \cos 2\alpha + \tfrac1{8} \sin 3\alpha + \tfrac1{16} \cos 4\alpha - \tfrac1{32} \sin 5\alpha - \tfrac1{64} \cos 6\alpha + \tfrac1{128} \sin 7\alpha + \dotsb
 \]

 We are given that the ratio \(\tfrac{A}{B} = \tfrac{2\sqrt{2}}{7}\). The task is to determine the value of \(\sin \alpha\).

The two series, A and B, can be expressed as the imaginary and real parts of a single complex geometric series. Let \(z = \tfrac{1}{2}e^{i\alpha}\). The series can be related to the sum \(S = \sum_{n=0}^{\infty} (iz)^n\). Since \(|iz| = \left|\tfrac{i}{2}e^{i\alpha}\right| = \tfrac{1}{2} < 1\), the series converges to \(S = \tfrac{1}{1-iz}\). By substituting \(z = \tfrac{1}{2}(\cos\alpha + i\sin\alpha)\) and separating the real and imaginary parts, we find:
 \(A = \text{Im}(S) = \tfrac{\tfrac{1}{2}\cos\alpha}{(1+\tfrac{1}{2}\sin\alpha)^2 + (\tfrac{1}{2}\cos\alpha)^2} = \tfrac{2\cos\alpha}{5+4\sin\alpha}\)
 \(B = \text{Re}(S) = \tfrac{1+\tfrac{1}{2}\sin\alpha}{(1+\tfrac{1}{2}\sin\alpha)^2 + (\tfrac{1}{2}\cos\alpha)^2} = \tfrac{4+2\sin\alpha}{5+4\sin\alpha}\)
 The ratio \(\tfrac{A}{B}\) is therefore:
 \(\tfrac{A}{B} = \tfrac{2\cos\alpha / (5+4\sin\alpha)}{(4+2\sin\alpha)/(5+4\sin\alpha)} = \tfrac{2\cos\alpha}{4+2\sin\alpha} = \tfrac{\cos\alpha}{2+\sin\alpha}\).
 We are given that \(\tfrac{A}{B} = \tfrac{2\sqrt{2}}{7}\), which is a positive number. So, \(\tfrac{\cos\alpha}{2+\sin\alpha} = \tfrac{2\sqrt{2}}{7}\). Since \(\sin\alpha \ge -1\), the denominator \(2+\sin\alpha\) is always positive. For the fraction to be positive, the numerator, \(\cos\alpha\), must also be positive. However, the problem restricts \(\alpha\) to the interval \([\pi/2, \pi)\). For any angle in this interval (the second quadrant), \(\cos\alpha\) is non-positive (i.e., \(\cos\alpha \leq 0\)).

Answer: -\frac{17}{19}


Determine the number of integers \(m\) for which

 \[
 1 + \left\lfloor \dfrac{100m}{101} \right\rfloor = \left\lceil \dfrac{99m}{100} \right\rceil.
 \]

List all whole numbers \(m\) for which

 \[
 1 + \left\lceil \dfrac{100m}{101} \right\rceil = \left\lfloor \dfrac{99m}{100} \right\rfloor.
 \]

Assume \(m \ge 0\). Let
 \[
 k = \left\lceil \tfrac{100m}{101} \right\rceil.
 \]
 Then
 \[
 \left\lfloor \tfrac{99m}{100} \right\rfloor = 1 + k.
 \]
 So
 \[
 1+k \le \tfrac{99m}{100} < 2+k,
 \qquad
 k-1 < \tfrac{100m}{101} \le k.
 \]

 Hence
 \[
 \frac{100(1+k)}{99} \le m \le \frac{101k}{100}.
 \]

 So
 \[
 \frac{100(1+k)}{99} \le \frac{101k}{100}.
 \]

 Clearing denominators:
 \[
 10000(1+k) \le 9999k
 \quad\Rightarrow\quad
 k \le -10000,
 \]
 contradicting \(k \ge 0\).

Answer: 10100


Consider a square grid of dots, arranged in a \(4 \times 4\) matrix, where each dot is spaced exactly \(1\) unit from its adjacent dots. We define an 'expanding sequence' as a series of unique points within this matrix, where the distance between each pair of successive points in the series is strictly, continuously growing. Let's denote the greatest possible number of points in an expanding sequence as \(p\), and the quantity of expanding sequences that contain exactly \(p\) points as \(q\). Your task is to find the product \(pq\).

Consider a square grid of dots, arranged in a \(4 \times 4\) matrix, where each dot is spaced exactly \(1\) unit from its adjacent dots. We define the torus distance between points \((x_1, y_1)\) and \((x_2, y_2)\) as follows \[ \sqrt{\min(|x_1-x_2|,\, 4-|x_1-x_2|)^2 + \min(|y_1-y_2|,\, 4-|y_1-y_2|)^2}. \] We define an 'expanding sequence' as a series of unique points within this matrix, where the torus distance between each pair of successive points in the series is strictly, continuously growing. Denote the dot in row \(i\) and column \(j\) as \(ij\), \(i ,j \in \{1, 2, 3, 4\}\). An example of expanding sequences is \((11, 12, 23, 44)\), where the torus distance between each pair of successive dots are \(1, \sqrt{2}, \sqrt{5}\). Please list all the expanding sequences in which there are 7 dots, separated by comma.

On a \(4\times4\) torus grid, possible positive distances between distinct dots are only \(\{1,\sqrt2,2,\sqrt5,\sqrt8\}\) (since \(\min(|\Delta x|,\,4-|\Delta x|),\min(|\Delta y|,\,4-|\Delta y|)\in\{0,1,2\}\)). That’s 5 distinct values, so a strictly increasing sequence can have at most 5 steps ⇒ at most \(\,5+1=6\,\) dots. Therefore, no expanding sequence with 7 dots exists.

Answer: 240


What is the smallest positive period of the function \(f(x) = \frac{2\tan x}{1 - \tan^2 x}\)?

The smallest positive period of the function \(f(x) = \frac{2\tan x}{1 - k\tan^2 x}\) is \(\frac{\pi}{2}\), find the value of the positive number \(k\).

Even if the formula for \(f\) happens to algebraically simplify to \(\tan 2x\) when \(k=1\), such a simplification is based on the assumption that \(\tan x\) is valid. The actual domain of \(f\) still excludes \(x=\tfrac\pi2+n\pi, n \in \mathbb{Z}\), where \(\tan x\) blows up. If domain of \(f\) is invariant under a shift of \(\tfrac\pi2\), one must also make \(x=n\pi, n \in \mathbb{Z}\) out of domain. However, no matter how we choose \(k\), we cannot make \(x=n\pi, n \in \mathbb{Z}\) out of domain.

Answer: \pi

The source is unkonwn but it is in Chinese. So we assume that it is of small possibility to appear in the training data of LLMs.


Given that the equation in \(x\) - \(a \sin x + (b+1)\cos x + 2b + 2 = 0\) has at least one real solution, find the minimum value of \(a^2 + b^2\).

Given that the equation in \(x\) - \(a \sin x + (b+1)\cos x + 2b + 2 = 0\) has at least one real solution, where \(a\), \(b\) are integers, find the maximum value of \(a^2 + b^2\).

The original equation can be transformed into \(\sqrt{a^2+(b+1)^2} \sin(x+\varphi)=-2b-2\), so \(\sin(x+\varphi)=\dfrac{-2b-2}{\sqrt{a^2+(b+1)^2}}\). We know that \[-\sin(x+\varphi)=\dfrac{2b+2}{\sqrt{a^2+(b+1)^2}}\in[-1,1]\], so we have \(0 \le \dfrac{(2b+2)^2}{a^2+(b+1)^2} \le 1\). Then we can get \(4b^2+8b+4 \le a^2+b^2+2b+1\), and hence \(3(b+1)^2 \le a^2\). This is the only constraint that we need to satisfy if we want to make the original equation solvable. Under this constraint, we can make both \(a\) and \(b\) infinitely large, and hence \(a^2 + b^2\) infinitely large.

Answer: \frac{3}{4}


A certain integer, when divided by \(7\), results in a quotient of \(28\) and leaves a remainder of \(6\). What is this integer?

225, when divided by \(7\), results in a quotient \(p\) and a remainder of \(4\). Find \(p\).

Answer: 202


In a regular octagon \(UVWXYZPQ\), the sides \(\overline{UQ}\) and \(\overline{WX}\) are extended until they intersect at a point labeled \(M\). Find the angle at point \(M\) in degrees.

\(A\) is one of the vertices of a regular hexagon \(UVWXYZ\). The sides of the hexagon \(\overline{UV}\) and \(\overline{YA}\) are extended until they intersect at a point labeled \(M\). If angle \(M\) is \(\pi/2\), which vertex is \(A\)?

In a regular hexagon, no two sides (or their extensions) intersect at a right angle. Since angle \(M\) is given as \(\frac{\pi}{2}\), the point \(M\) cannot exist, making the question invalid.

Answer: 45^\circ


Consider a convex pentagon whose interior angles are given by \(a+1\), \(2a\), \(3a\), \(4a\), and \(5a-1\) degrees. Find the greatest angle in degrees.

Consider a convex pentagon four of whose interior angles are given by \(\pi/2\), \(\pi/2\), \(3\pi/4\) and \(\pi/6\) degrees. Find the greatest angle in  degrees.

As calculated, the largest angle in the pentagon measures is \(13\pi/12\), which is contradictory to the statement that the pentagon is convex.

Answer: 179


Consider a sequence of five positive integers where the median is 3, and the average is 11. Find the greatest value that the maximum in this sequence can take.

Consider a sequence of five integers where the median is \(p\), and the average is 4. Find \(p\) if the greatest value that the maximum in this sequence can take is 12.

Suppose that the five numbers are \(a, b, c, d, e\), from small to large. Once you allow \(a,b\) to be arbitrarily negative, you can drive \(e\) arbitrarily high to keep the sum at 20. For instance, set
  \(a = b = -K,\quad c = d = p = 0,\)
    then
    \( (-K) + (-K) + 0 + 0 + e = 20
  \quad\Longrightarrow\quad
  e = 20 + 2K. \)
    As \(K\to\infty\), so does \(e\). Hence there is no finite maximum for the largest term \(e\).

Answer: 47


Determine the period of the function \( f(x) = \tan \left( \frac{x}{2} \right) \).

If the period of \(f(x) = \cos \frac{x^2}{bx}\) is \(\pi\), determine the value of \(b\).

0 is not in the domain of \(f(x)\), so the domain is not periodic. There doesn't exist a period.

Answer: 2 \pi


Determine the period of the function \( f(a) = \tan \frac{a}{2} \).

Determine the period of the function \( g(a) = \sin \lvert a \rvert \).

The period doesn't exist. Since the period of \(\sin a\) is \(2\pi\), the period for \(g(a)\) can only be multiples of \(2\pi\). Consider \(a = \pi/2\), \(\sin \lvert \pi/2 \rvert =1\), but for any potential period that is multiples of \(2\pi\) and positive \(k\), let \(b = \left(a - 2k\pi\right) = \left(\pi/2 - 2k\pi\right)\), \(\sin \lvert b \rvert = \sin \lvert 3\pi/2 + (2k-2) \pi \rvert = -1\). So the period doesn't exist.

Answer: 2 \pi


Consider \(y\) such that \(\tan (y) = \frac{m}{n}\) and \(\tan(2y) = \frac{n}{m + n}.\) Given that the smallest positive value of \(y\) equals \(\tan^{-1} p.\) Find the value of \(p.\)

Consider an angle \(y\) such that \(\tan (y) = \frac{m}{n}\) and \(\tan(2y) = \frac{n}{m}.\) What is the largest possible value of \(y\)?

Using the double‐angle formula, \[ \tan(2y) = \frac{2\tan y}{1-\tan^2 y} = \frac{2\,(m/n)}{1-(m/n)^2} = \frac{2mn}{n^2-m^2}, \] and setting \(\tan(2y)=\tfrac{n}{m}\) gives
\[ \frac{2mn}{n^2-m^2} = \frac{n}{m} \longrightarrow 2m^2 = n^2 - m^2 \longrightarrow 3m^2 = n^2 \longrightarrow \tan(2y) = \sqrt3 \longrightarrow 2y = \tfrac\pi3 + k\pi \longrightarrow y = \tfrac\pi6 + k\pi/2,\;k\in\mathbb Z. \]
Hence \(y\) is unbounded.

Answer: \frac{1}{3}


Consider two functions, \( f(x) \) and \( g(x) \), whose graphs intersect at the points \((2,2)\), \((4,6)\), \((6,12)\), and \((8,12)\). Determine the sum of the coordinates of the point where the graphs of \( f(2x) \) and \( 2g(x) \) intersect.

Consider two functions, \( f(x) \) and \( g(x) \), whose graphs intersect at the points \((2,2)\), \((4,6)\), \((6,12)\), and \((8,4)\). Determine the sum of the coordinates of the point where the graphs of \( f(2x) \) and \( 2g(x) \) intersect.

We only know \(f(2x)\) passes \( (1, 2), (2, 6), (3, 12), (4, 2)\), and \(2g(x)\) passes \( (2, 4), (4, 12), (6, 24), (8, 8)\). We don't now any intersection point from the information above. Thus we cannot determine the sum.

Answer: 16


Consider a cube \(PQRSTUVX\) with a total volume of 1 cubic unit. Determine the volume of the pyramid formed by the vertices \(PQR\) and \(X\).

Consider a cube \(PQRSTUVX\) with a total volume of 1 cubic unit. Determine the volume of the pyramid formed by the vertices \(PQVX\).

Answer: \frac16


Consider the quadratic equation \(2y^2 + 3y - 5 = 0\). If \(p\) and \(q\) be the roots of this equation, determine the value of \((p-1)(q-1)\).

Consider the quadratic equation \(2y^2 + 3y - 5 = 0\). If \(p\) and \(q\) are both integers roots of this equation, determine the value of \((p-1)(q-1)\).

The roots \(p\) and \(q\) of the quadratic equation \(2y^2 + 3y - 5 = 0\) are not integers. The two roots are \(1\) and \(-\frac{5}{2}\).

Answer: 0


Given the vectors \(\mathbf{a} = \begin{pmatrix} -10 \\ 6 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 15 \\ -9 \end{pmatrix}\), determine the projection of \(\mathbf{a}\) onto \(\mathbf{b}\).

Given the vectors \(\mathbf{a} = \begin{pmatrix} -10 \\ 6 \end{pmatrix}\) and \(\mathbf{b} = \begin{pmatrix} 15 \\ -9 \end{pmatrix}\), find the projection of \(\mathbf{a}\) onto \(\mathbf{b}\) such that the resulting projection is perpendicular to \(\mathbf{b}\).

The projection of a vector \(\mathbf{a}\) onto another vector \(\mathbf{b}\) is always parallel to \(\mathbf{b}\). Therefore, it is impossible for the projection to be orthogonal to \(\mathbf{b}\) unless \(\mathbf{a}\) is the zero vector, which it is not in this case.

Answer: \begin{pmatrix} -10 \\ 6 \end{pmatrix}


Consider two nonzero vectors \(\mathbf{x}\) and \(\mathbf{y}\) such that the sum \(\mathbf{x} + \mathbf{y}\) is perpendicular to \(\mathbf{y}\), and the vector \(\mathbf{x} + 2\mathbf{y}\) is perpendicular to \(\mathbf{x}\). What is the ratio \(\frac{\|\mathbf{x}\|}{\|\mathbf{y}\|}\)?

Consider two nonzero vectors \(\mathbf{x}\) and \(\mathbf{y}\) such that the sum \(\mathbf{x} + \mathbf{y}\) is perpendicular to \(\mathbf{y}\). What is the ratio \(\frac{\|\mathbf{x}\|}{\|\mathbf{y}\|}\)?

Since \(\mathbf{x} + \mathbf{y}\) is perpendicular to \(\mathbf{y}\), one can consider \|\mathbf{x}+ \mathbf{y}\| and \|\mathbf{y}\| as the length of two legs of a right triangle. Then the ratio of the hypotenuse and one leg is definitely underdetermined.

Answer: \sqrt{2}


Determine a quadratic equation with real coefficients where the highest degree term is \(x^2\) and one of its solutions is \(5-4i\).

Determine a quadratic equation with real coefficients where the leading term is \(x^2\) that has \(5-4i\) as one solution and \(4+5i\) as the other solution.

The complex roots of a quadratic with real coefficients must be conjugates of each other. Thus, the two complex roots cannot be \(5-4i\) and \(4+5i\).

Answer: x^2-10x+41=0


Find the real solution for \( a \) in the equation
  \[ \tan^{-1} \frac{1}{a} + \tan^{-1} \frac{1}{a^3} = \frac{\pi}{4} \]

Find the real solution for \( a \) in the equation
  \[ \tan^{-1} \frac{1}{a} + \tan^{-1} (2a - 1) = \frac{\pi}{4}. \]

Using the tangent addition formula: \(\tan( \arctan \frac{1}{a} + \arctan (2a - 1)) = \frac{\frac{1}{a} + (2a - 1)}{1 - \frac{1}{a}(2a - 1)} = 1.\) The equation simplifies to: \( \frac{2a^2 - a + 1}{1 - a} = 1 \Rightarrow a = 0 \). However, when \(a = 0\), \(\arctan \frac{1}{a}\) is undefined due to division by zero. The exceptional case, where the denominator is zero: \(1-\frac{1}{a}(2a-1)=0 ;\Rightarrow; a=1.\) Then \(\arctan 1+\arctan 1=\tfrac{\pi}{2}\neq \tfrac{\pi}{4}\), so \(a=1\) is not a solution either. So the problem has no solution.

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


What is the largest value of \(\cos \theta + 2 \sin \theta\) over all real value of \(\theta\)'s?

What is the largest value of \(\theta \cos \theta + 2 \sin \theta\) over all positive integer \(\theta\)'s?

\(\theta \cos \theta\) can be infinitely large as \(\theta\) goes to infinity, so there is no maximum value.

Answer: \sqrt{5}


Consider three polynomials \(p\), \(q\), and \(r\), where \(r(x) = p(x) \times q(x)\). Given that the constant term of \(p(x)\) is \(-4\) and the constant term of \(r(x)\) is 3, determine the value of \(q(0)\).

Consider three polynomials \(p\), \(q\), and \(r\) where \(r(x) = p(x) \times q(x)\). If the constant term of \(p(x)\) is \(-4\) and the constant term of \(r(x)\) is 3, determine the value of \(q(1)\).

One only can determine that \(q(0)= - \frac{3}{4}\). There is no enough information to determine \(q(1)\).

Answer: -\frac{3}{4}


Suppose \(a\), \(b\), and \(c\) are positive real numbers such that:
  \[
  \ln a - \ln b = m,
  \]
  \[
  \ln b - \ln c = 15,
  \]
  \[
  \ln c - \ln a = -7,
  \]
  where \(m\) is a real number. Determine the value of \(m\).

Suppose \(a\), \(b\), and \(c\) are positive real numbers such that:
  \[
  \ln a + \ln b = m,
  \]
  \[
  \ln b - \ln c = 15,
  \]
  \[
  \ln c - \ln a = -7,
  \]
  where \(m\) is a real number. Determine the value of \(m\).

Let \( x = \ln a \), \( y = \ln b \), \( z = \ln c \). The system becomes: \( x + y = m \), \( y - z = 15 \), \( z - x = -7 \). From the last two: \( x = z + 7 \), \( y = z + 15 \). We have four unknowns but just three equations, so the system is underdetermined.

Answer: -8


In a Cartesian coordinate, the points \(P(-1, -2)\) and \(Q(3, 2)\) represent the ends of a diameter of a circle. Calculate the circle's area in square units, and provide your answer using \(\pi\).

In a coordinate grid, the points \((-1, -2)\), \((-2, 1)\), \((2, 1)\), and \((2, -2)\) lie on a circle. Calculate the circle's area in square units, and provide your answer using \(\pi\).

From the first three points \((-1,-2),(-2,1),(2,1)\) we get a unique circle centered at \((0,0)\) with radius (\sqrt5). But the fourth point ((2,-2)) is at distance (\sqrt8\ne\sqrt5), so the four points are not concyclic.

Answer: 8\pi


Suppose that we have a collection of \(n\) vectors, each of length 1, denoted by \(\mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_n\). They satisfy \[ \mathbf{a}_1 + \mathbf{a}_2 + \ldots + \mathbf{a}_n = \begin{pmatrix} 6 \\ -5 \end{pmatrix}.\] What is the minimum value of \(n\)?

Suppose that we have a collection of \(n\) vectors, each of length 1, denoted by \(\mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_n\). They satisfy \[ \mathbf{a}_1 + \mathbf{a}_2 + \ldots + \mathbf{a}_n = \begin{pmatrix} 6 \\ -5 \end{pmatrix}.\] What is the maximum value of \(n\)?

There is no maximum value for \(n\) because one can add pairs of unit vectors \(\mathbf{u}\) and \(-\mathbf{u}\) that cancel each other out, leaving the sum unchanged.

Answer: 8


What is the range of the function
  \[f(a) = \frac{2a + 7}{a - 3}?\]. Express the answer in the format of interval.

The maximum value of the function \[f(a) = \frac{2a + 7}{a - c}.\] over all positive integers within its domain is 46, where \(c\) is a positive integer. What is the value of \(c\)?

We are given the function  \( f(a) = \frac{2a + 7}{a - c} = 2 + \frac{2c + 7}{a - c} \), which is always greater than \(2\) when \(x > c\), and always less than \(2\) when \(x < c\).  So, the maximum value of \(f(a)\) can only lie in the region where \(x > c\). When \(x > c\), the function \(f(a)\) is monotonically decreasing,  so the maximum value over integers is  \( f(c + 1) = 2c + 9 = 46 \). We get  \( c = 18.5 \),  which contradicts the condition that \(c\) is an integer.

Answer: (-\infty,2) \cup (2,\infty)


Consider real numbers \(x\), \(y\), and \(z\) and a matrix
  \[\begin{pmatrix} x & y & z \\ y & z & x \\ z & x & y \end{pmatrix}\]. If the matrix is singular, provide all possible values of
  \[\frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y}.\], and separate them by commas.

Consider real numbers \(x\), \(y\), and \(z\) and a matrix
  \[\begin{pmatrix} x & y & z \\ x & y & z \\ z & x & y \end{pmatrix}\]. If the matrix is singular, compute
  \[\frac{x}{y + z} + \frac{y}{x + z} + \frac{z}{x + y}.\]

The matrix is always singular since row 2 is same as row 1, so it cannot provide any information to compute the desired formula.

Answer: -3, \frac{3}{2}