Mock test #2 Solutions
Part A: Short-answer type questions
Which of the following statements is true?
(a) If a polygon inscribed in a circle has equal sides then it has equal angles.
(b) If a polygon inscribed in a circle has equal angles then it has equal sides.
(c) If a polygon has an inscribed circle, i.e. all its sides are tangent to the circle, and its sides are equal, then its angles are also equal.Sol.
(a) True. Suppose A, B and C are consecutive vertices of the polygon. Let O be the center. Then triangles OAB and OBC are congruent and isoceles. Hence, \( 1/2 \angle A = 1/2 \angle B\). Extend the argument to other vertices.
(b) False. A wide rectangle can be inscribed in a circle.
(c) False. A circle can be inscribed in a non-square rhombus.If \(S=\{a, b, c\}\) and the relation \(R\) on the set \(S\) is given by \(R=\{(a, b),(c, c)\}\), then \(R\) is
(a) reflexive and transitive
(b) reflexive but not transitive
(c) not reflexive but transitive
(d) neither reflexive nor transitive.Sol.
Answer : (c)
\((b, b) \notin R\). Hence, \(R\) is not reflexive. However, \(R\) is transitive.If \(a+b+c=0\), then the quadratic equation \(3 a x^{2}+2 b x+c=0\) has
(a) at least one root in \((0,1)\).
(b) one root in \((1,2)\) and other in \((-1,0)\).
(c) both imaginary roots.
(d) a repeated rootSol.
Ans:(a)Let \( \left[a\right] \) denote the largest integer smaller than or equal to \(a\). Then, \(\lim_{x \rightarrow 0} x\left[\frac{1}{x}\right]\) is
(a) 0
(b) 1
(c)-1
(d) does not exist.Sol.
Ans: (b) 1.If \(\alpha\) and \(\beta\) are the roots of \(x^{2}+3 x+1\), find the value of \(\left(\frac{\alpha}{\beta+1}\right)^{2}+\left(\frac{\beta}{\alpha+1}\right)^{2}\).
Sol.
Ans: 18.Let \(f(x)\) be the function defined on \(\mathbb{R}\) such that \(f(x)=x\), for all \(x \leq 1\) and \(f(x)=\) \(a x^{2}+b x+c\), for \(x>1\). The triples \((a, b, c)\) such that \(f(x)\) is differentiable at all real \(x\) are of the form:
(a) \((a, 1-2 a, a)\)
(b) \((1,0,0)\) only
(c) \((a,-2 a, a)\)
(d) \((0,1,-1)\) onlySol.
Ans: (a)
The system of equations \[ \begin{align*} 2x+py+6z&=8 \\ x+2y+qz&=5 \\ x+y+3z&=4 \end{align*} \] has no solution for
(a) \(p \neq 2, q \neq 3\)
(b) \(p \neq 2, q=3\)
(c) \(p=2, q=3\)
(d) \(p=2, q \neq 3\)Sol.
Ans: (b)Suppose \(f\) is continuous in \([0,2]\) and differentiable in \((0,2)\). If \(f(0)=0\) and \(\left|f^{\prime}(x)\right| \leq 1 / 2\) for all \(x \in[0,2]\), then
(a) \(|f(x)| \leq 1\).
(b) \(|f(x)| \leq 1 / 2\).
(c) \(f(x)=2 x\).
(d) \(f(x)=3\) for at least one \(x \in[0,2]\).Sol.
Ans: (a)
\( f^\prime(x) \) must be equal to 1 at \(x=1\).Let \(f\) be a differentiable real valued function on \((-1,4)\) such that \(f(3)=6\) and \(f^{\prime}(x) \geq-1\) for all \(x\). What is the the largest possible value of \(f(0)\)?
Sol.
Ans: 9.
\( \frac{ f(3)-f(0) }{3-0} \geq -1 \) by Rolle's theorem.
Source: Madhava competion.There are 18 ways in which \(n\) identical balls can be grouped such that each group contains equal number of balls. Then the minimum value of \(n\) is
(a) 120
(b) 180
(c) 160
(d) 90Sol.
Solution: (b)
The total number of required ways \(=\) the total number of factors of \(n\). \(180=2^{2} \times 3^{2} \times 5\). Therefore the total number of factors of 180 is \(3 \times 3 \times 2=18\).
Part B: Subjective questions
Submission files: Each question in this part must be answered on a page of its own. Name the files as B1.jpg, B2.jpg, etc. In case you have multiple files for the same question, say B4, name the corresponding files as B4-1.jpg, B4-2.jpg, etc.
Clearly explain your entire reasoning. No credit will be given without reasoning. Partial solutions may get partial credit.
B1. A regional math competition saw a participation from 200 students. The paper had 6 problems. We know that each problem was correctly solved by at least 120 students. Prove that there must be two students such that every problem was solved by at least one of these two students.
Sol.
We describe a way to find two such students. 200 students have turned in 720 solutions, so a student has solved 3.6 problems on an average. This means that there must be some student who has solved at least four problems. Call this student A. Let us suppose A has not solved problems \(i\) and \(j\).At most 80 students have not solved \(i\) and at most 80 students have not solved \(j\). So, we should be able to find at least 200-80-80=40 students who have solved both the problems. Pick one such student along with student A to get the required pair.
B2. An icosahedron is a regular polyhedron with 20 identical equilateral triangle faces. See the picture below. Each side of the icosahedron is 1 cm in length. What is the length of the longest line segment that can be contained inside an icosahedron?
Sol.
Note that opposite vertices of the icosahedron can be seen as vertices of pyramids whose bases are regular pentagons of side length 1. Now, note that if we select two parallel diagonals of these two pentagons, these two diagonals are also two sides of a rectangle whose other sides are length 1 and whose diagonals are diameters of the icosahedron. The diagonal of the pentagon can be found with similar triangles: in regular pentagon \(A B C D E\), let \(A D\) and \(B E\) intersect at \(F\). Angle chasing shows that \(\triangle A C D \sim \triangle D E F\), both are isosceles, and \(F E=F A\), so we get that \(\frac{A D}{1}=\frac{1}{A D-1} \Longrightarrow A D=\frac{1+\sqrt{5}}{2}\). Hence, the diameter of the icosahedron equals \(\sqrt{1^{2}+\left(\frac{1+\sqrt{5}}{2}\right)^{2}}=\frac{\sqrt{10+2 \sqrt{5}}}{2}\).Source: Stanford math tournament 2011.
B3. Prove the following inequality for all integers \(a,b\geq 2\): \[ \frac{1}{\operatorname{gcd}(a, b-1)}+\frac{1}{\operatorname{gcd}(a-1, b)} \geq \frac{2}{\sqrt{a+b-1}} \]
Sol.
Note that both \(\operatorname{gcd}(a, b-1)\) and \(\operatorname{gcd}(a-1, b)\) divide \(a+b-1\). Also they are relatively prime, since \(\operatorname{gcd}(a, b-1) \mid a\) and \(\operatorname{gcd}(a-1, b) \mid a-1\). So their product is less than or equal to \(a+b-1\), and therefore by the AM-GM inequality we have \[ \frac{1}{\operatorname{gcd}(a, b-1)}+\frac{1}{\operatorname{gcd}(a-1, b)} \geq 2 \sqrt{\frac{1}{\operatorname{gcd}(a, b-1) \cdot \operatorname{gcd}(a-1, b)}} \geq \frac{2}{\sqrt{a+b-1}} \] Source: Stanford math tournament 2011.B4. (a) In \(\triangle A B C\), the altitude to \(\overline{A B}\) from \(C\) partitions \(\triangle A B C\) into two smaller triangles, each of which is similar to \(\triangle A B C\). If the side lengths of \(\triangle A B C\) and of both smaller triangles are all integers, find the smallest possible value of \(A B\).
Sol.
First, notice that length \(A E\) is completely determined by the fact that \(A B=B E\) and by the lengths of \(A B, B C\) and \(E C\). Thus, we only consider the triangle \(A B C\). First, drop altitude \(B H\) and note that since \(A B E\) is isoceles, \(E H=\frac{1}{2} A E\). Now, using Pythagoras twice, we have \begin{align*} &B H^{2}=5^{2}-E H^{2} \\ &B H^{2}=11^{2}-(7+E H)^{2} . \end{align*} Setting these two equations to be equal, we can thus solve the equation \(25=72-14 E H\). Therefore, \(A E=2 E H=\frac{47}{7}\).B4. (b) In quadrilateral \(A B C D\), diagonals \(A C\) and \(B D\) intersect at \(E\). If \(A B=B E=5, E C=C D=7\), and \(B C=11\), compute \(A E\).
Sol.
Let the altitude from \(C\) to \(A B\) intersect \(\overline{A B}\) at \(D\). Note that \(\angle A\) and \(\angle B\) must be acute since \(A D\) partitions \(A B C\) into two triangles. Since triangles \(A D C\) and \(A D B\) each contain right angles, we conclude that \(\angle A C B\) must be right. Now, we must have \(\triangle A B C \sim \triangle A C D \sim \triangle C B D\). We now have enough information to determine \(A D, A B\), and \(C D\) in terms of \(A B, B C\), and \(C A\), which we denote as \(c\), \(a\), and \(b\), respectively. We have \begin{align*} &\frac{b}{A D}=\frac{c}{b} \Longrightarrow A D=\frac{b^{2}}{c} \\ &\frac{a}{B D}=\frac{c}{a} \Longrightarrow B D=\frac{a^{2}}{c} \\ &\frac{1}{2} C D \cdot c=\frac{1}{2} a b \Longrightarrow C D=\frac{a b}{c} \end{align*} Obviously, \((a, b, c)\) has to be in the form \((k x, k y, k z)\) where \((x, y, z)\) is a Pythagorean triple with no common factors, and \(k\) is a positive integer. Note that in particular \(x, y\), and \(z\) must be pairwise coprime, because of the constraint \(x^{2}+y^{2}=z^{2}\). Given this, we need to find \(k\) such that \[ k z\left|k^{2} y^{2} \Longleftrightarrow z\right| k y^{2} \Longleftrightarrow z \mid k, \] so the smallest such \(k\) is when \(k=z\). This choice makes \(B D\) and \(C D\) integral as well, so given any Pythagorean triple \((x, y, z)\) with pairwise coprime entries, the minimum \(k\) required equals \(z\), and the minimum possible value of \(c\) is \(z^{2}\). The smallest such Pythagorean triple is \((3,4,5)\), so report 25.B5. Prove that among any six consecutive positive integers, we can always find a number that is relatively prime to the other five numbers.
Sol.
The 6 integers must all have different remainders when divided by 6. The two that have remainders 1 and 5 must have a difference of either 4 or 2 , so that any common divisor of these two integers must divide 2. As, however, the two integers are odd this shows that their \(gcd=1\). Neither of them can be divisible by 3 ; if one of them is divisible by 5 the other isn't. So the one not divisible by 5 is not divisible by any integer less than 7 (except 1). As its "distance" from any of the other 5 is less than 7, the result follows.Source: Parabola magazine.
B6. Show that the following integral converges: \[ \lim_{B \rightarrow \infty} \int_{0}^{B} \frac{\cos (x)}{2x} \cos \left(x^{2}\right) dx \]