Archive for the ‘Problem Solving’ Category

Romanian Masters in Mathematics contest – 2018

March 9, 2018 Leave a comment

Problem 1. Let {ABCD} be a cyclic quadrilateral an let {P} be a point on the side {AB.} The diagonals {AC} meets the segments {DP} at {Q.} The line through {P} parallel to {CD} mmets the extension of the side {CB} beyond {B} at {K.} The line through {Q} parallel to {BD} meets the extension of the side {CB} beyond {B} at {L.} Prove that the circumcircles of the triangles {BKP} and {CLQ} are tangent .

Problem 2. Determine whether there exist non-constant polynomials {P(x)} and {Q(x)} with real coefficients satisfying

\displaystyle P(x)^{10}+P(x)^9 = Q(x)^{21}+Q(x)^{20}.

Problem 3. Ann and Bob play a game on the edges of an infinite square grid, playing in turns. Ann plays the first move. A move consists of orienting any edge that has not yet been given an orientation. Bob wins if at any point a cycle has been created. Does Bob have a winning strategy?

Problem 4. Let {a,b,c,d} be positive integers such that {ad \neq bc} and {gcd(a,b,c,d)=1}. Let {S} be the set of values attained by {\gcd(an+b,cn+d)} as {n} runs through the positive integers. Show that {S} is the set of all positive divisors of some positive integer.

Problem 5. Let {n} be positive integer and fix {2n} distinct points on a circle. Determine the number of ways to connect the points with {n} arrows (oriented line segments) such that all of the following conditions hold:

  • each of the {2n} points is a startpoint or endpoint of an arrow;
  • no two arrows intersect;
  • there are no two arrows {\overrightarrow{AB}} and {\overrightarrow{CD}} such that {A}, {B}, {C} and {D} appear in clockwise order around the circle (not necessarily consecutively).

Problem 6. Fix a circle {\Gamma}, a line {\ell} to tangent {\Gamma}, and another circle {\Omega} disjoint from {\ell} such that {\Gamma} and {\Omega} lie on opposite sides of {\ell}. The tangents to {\Gamma} from a variable point {X} on {\Omega} meet {\ell} at {Y} and {Z}. Prove that, as {X} varies over {\Omega}, the circumcircle of {XYZ} is tangent to two fixed circles.

Source: Art of Problem Solving forums

Some quick ideas: For Problem 1 just consider the intersection of the circle {(BKP)} with the circle {(ABCD)}. You’ll notice immediately that this point belongs to the circle {(CLQ)}. Furthermore, there is a common tangent to the two circles at this point.

For Problem 2 we have {10\deg P = 21 \deg Q}. Eliminate the highest order term from both sides and look at the next one to get a contradiction.

Problem 4 becomes easy after noticing that if {q} divides {an+b} and {cn+d} then {q} divides {ad-bc}.

In Problem 5 try to prove that the choice of start points determines that of the endpoints. Then you have a simple combinatorial proof.

Problem 6 is interesting and official solutions use inversions. Those are quite nice, but it may be worthwhile to understand what happens in the non-inverted configuration.

I will come back to some of these problems in some future posts.


SEEMOUS 2018 – Problems

March 1, 2018 Leave a comment

Problem 1. Let {f:[0,1] \rightarrow (0,1)} be a Riemann integrable function. Show that

\displaystyle \frac{\displaystyle 2\int_0^1 xf^2(x) dx }{\displaystyle \int_0^1 (f^2(x)+1)dx }< \frac{\displaystyle \int_0^1 f^2(x) dx}{\displaystyle \int_0^1 f(x)dx}.

Problem 2. Let {m,n,p,q \geq 1} and let the matrices {A \in \mathcal M_{m,n}(\Bbb{R})}, {B \in \mathcal M_{n,p}(\Bbb{R})}, {C \in \mathcal M_{p,q}(\Bbb{R})}, {D \in \mathcal M_{q,m}(\Bbb{R})} be such that

\displaystyle A^t = BCD,\ B^t = CDA,\ C^t = DAB,\ D^t = ABC.

Prove that {(ABCD)^2 = ABCD}.

Problem 3. Let {A,B \in \mathcal M_{2018}(\Bbb{R})} such that {AB = BA} and {A^{2018} = B^{2018} = I}, where {I} is the identity matrix. Prove that if {\text{tr}(AB) = 2018} then {\text{tr}(A) = \text{tr}(B)}.

Problem 4. (a) Let {f: \Bbb{R} \rightarrow \Bbb{R}} be a polynomial function. Prove that

\displaystyle \int_0^\infty e^{-x} f(x) dx = f(0)+f'(0)+f''(0)+...

(b) Let {f} be a function which has a Taylor series expansion at {0} with radius of convergence {R=\infty}. Prove that if {\displaystyle \sum_{n=0}^\infty f^{(n)}(0)} converges absolutely then {\displaystyle \int_0^{\infty} e^{-x} f(x)dx} converges and

\displaystyle \sum_{n=0}^\infty f^{(n)}(0) = \int_0^\infty e^{-x} f(x).

Source: official site of SEEMOUS 2018 

Hints: 1. Just use 2f(x) \leq f^2(x)+1  and xf^2(x) < f^2(x). The strict inequality comes from the fact that the Riemann integral of strictly positive function cannot be equal to zero. This problem was too simple…

2. Use the fact that ABCD = AA^t, therefore ABCD is symmetric and positive definite. Next, notice that (ABCD)^3 = ABCDABCDABCD = D^tC^tB^tA^t = (ABCD)^t = ABCD. Notice that ABCD  is diagonalizable and has eigenvalues among -1,0,1. Since it is also positive definite, -1 cannot be an eigenvalue. This allows to conclude.

3. First note that the commutativity allows us to diagonalize A,B  using the same basis. Next, note that A,B both have eigenvalues of modulus one. Then the trace of AB is simply the sum \sum \lambda_i\mu_i where \lambda_i,\mu_i are eigenvalues of A and B, respectively. The fact that the trace equals 2018  and the triangle inequality shows that eigenvalues of A are a multiple of eigenvalues of B. Finish by observing that they have the same eigenvalues.

4. (a) Integrate by parts and use a recurrence. (b) Use (a) and the approximation of a continuous function by polynomials on compacts to conclude.

I’m not sure about what others think, but the problems of this year seemed a bit too straightforward.

Regular tetrahedron – computing various quantities in terms of the side-length

February 2, 2018 Leave a comment

Sometimes one needs to find certain quantities related to the regular tetrahedron in {\Bbb{R}^3}, like volume, radius of the circumscribed sphere, radius of inscribed sphere, distance between opposite sides, etc. in terms of the side-length which we’ll note in the following with {a}. In the past I needed to find the angle under which every side is seen when looking from the center of the regular tetrahedron.

Here’s a trick which can help you find rather easily everything you need related to the regular tetrahedron: just embed it into a cube. We can see rather easily that when drawing certain diagonals of the faces of a cube, like in the figure below, we can recover a regular tetrahedron. Now it becomes rather easy to solve all questions above. We note that the ratio between the side of the cube (denoted by {\ell}) and the side of the embedded tetrahedron (denoted by {a}) is {\sqrt{2}}: {a = \ell \sqrt{2}}.


Here are a few ideas:

1. Finding the volume of the tetrahedron in terms of its sides.

The volume of the cube is {\ell^3}. The volume of the tetrahedron can be obtained by cutting four corner pyramids with volumes {\ell^3/6} (recall that volume of a pyramid is (area of base) {\times} (height) {/3}). Therefore the volume of the regular tetrahedron is {2\ell^3/6 = l^3/3}. Replacing {\ell = a/\sqrt{2} } we get that the volume of the tetrahedron is {a^3/(6\sqrt{2})}.

2. Finding the circumradius {R}.

It is not difficult to see that the sphere passing through the vertices of the tetrahedron also passes through the vertices of the cube. Therefore its radius is a long diagonal of the cube divided by {2}. This gives {\ell \sqrt{3}/2}. Replacing {\ell=a/\sqrt{2}} we get that the circumradius is {R = a\sqrt{6}/4}.

3. Finding the inradius {r}.

Once we have the volume and the circumradius, finding the inradius is not that difficult, since the symmetry of the figure shows that {r+R = h}, where {h} is the distance from a vertex to the opposite face, also called the height. We can find the height from the formula of the volume (recalled above), and then find {r = h-R}. Also note that since the center of the tetrahedron is also the centroid, we must have {R=3r}, so we have another quick finish solution.

However, let’s use the cube to do this. We can choose a system of coordinates putting the origin at the center of the cube (and tetrahedron). Put the cube so that its vertices have coordinates {\pm \ell/2,\pm \ell/2,\pm \ell/2} and suppose that the tetrahedron corresponds to the vertices {A(\ell/2,\ell/2,\ell/2)}, {B(-\ell/2,-\ell/2,\ell/2)}, {C(\ell/2,-\ell/2,-\ell/2)}, {D(-\ell/2,\ell/2,-\ell/2)}. All we need to do is to compute the distance from the origin to the plane {(BCD)}. This is immediate if we know the equation of this plane. Fortunately, it is really easy to see that the coordinates of {B,C,D} verify {x+y+z+\ell/2 = 0} (if not, then note that the normal to {(B,C,D)} is the vector {(1,1,1)} and figure out the remaining translation constant). We know that if a plane is defined by the equation {ax+by+cz+d=0} then the distance from {(x_0,y_0,z_0)} to this plane is

\displaystyle \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}.

Apply this to our problem and the distance from the origin to {(BCD)} becomes {\ell/(2\sqrt{3})}. Replace {\ell = a/\sqrt{2}} and we get {r = a/(2\sqrt{6})=a\sqrt{6}/12}.

4. Find distance between opposite sides.

This is particularly easy with the cube. The distance between the opposite sides is exactly the distance between two parallel faces of the cube and that is {\ell = a/\sqrt{2}}.

5. Find angle made by two rays connecting the center with vertices.

Use the coordinate system introduced in 3. and just compute the angle between vectors {\vec u = (\ell,\ell,\ell)} and {\vec v = (-\ell,-\ell,\ell)}, for example. If {\alpha} is the angle between {\vec u} and {\vec v} we get that

\displaystyle \cos \alpha = \frac{\langle \vec u,\vec v\rangle}{\|\vec u\| \|\vec v\|}=\frac{-\ell^2}{3\ell^2} = -\frac{1}{3}.

Therefore {\alpha = \arccos (-1/3)}.

Putnam 2017 A3 – Solution

December 4, 2017 Leave a comment

Problem A3. Denote {\phi = f-g}. Then {\phi} is continuous and {\int_a^b \phi = 0}. We can see that

\displaystyle I_{n+1}-I_n = \int_a^b (f/g)^n \phi = \int_{\phi\geq 0} (f/g)^n \phi+ \int_{\phi<0} (f/g)^n \phi

Now note that on {\{ \phi>=0\}} we have {f/g \geq 1} so {(f/g)^n \phi \geq \phi}. Furthermore, on {\{\phi<0\}} we have {(f/g)^n <1} so multiplying with {\phi<0} we get {(f/g)^n \phi \geq \phi}. Therefore

\displaystyle I_{n+1}-I_n \geq \int_{\phi \geq 0} \phi + \int_{\phi<0} \phi = \int \phi = 0.

To prove that {I_n} goes to {+\infty} we can still work with {I_{n+1}-I_n}. Note that the negative part cannot get too big:

\displaystyle \left|\int_{ \phi <0 } (f/g)^n \phi \right| \leq \int_{\phi<0} |\phi| \leq \int_a^b |f-g|.

As for the positive part, taking {0<\varepsilon< \max_{[a,b]} \phi} we have

\displaystyle \int_{\phi\geq 0} (f/g)^n \phi \geq \int_{\phi>\varepsilon}(f/g)^n \varepsilon.

Next, note that on {\{ \phi \geq \varepsilon\}}

\displaystyle \frac{f}{g} = \frac{g+\phi}{g} \geq \frac{g+ \varepsilon}{g}.

We would need that the last term be larger than {1+\delta}. This is equivalent to {g\delta <\varepsilon}. Since {g} is continuous on {[a,b]}, it is bounded above, so some delta small enough exists in order for this to work.

Grouping all of the above we get that

\displaystyle I_{n+1}-I_n \geq \int_{\phi \geq 0} (f/g)^n \phi \geq \int_{\phi>\varepsilon} \varepsilon (1+\delta)^n.

Since {|\phi>\varepsilon|>0} we get that {I_{n+1}-I_n} goes to {+\infty}.

Putnam 2017 A2 – Solution

December 4, 2017 Leave a comment

Problem A2. We have the following recurrence relation

\displaystyle Q_n = \frac{Q_{n-1}^2-1}{Q_{n-2}},

for {n \geq 2}, given {Q_0=1} and {Q_1=x}. In order to prove that {Q_n} is always a polynomial with integer coefficients we should prove that {Q_{n-2}} divides {Q_{n-1}^2-1} somehow. Recurrence does not seem to work very well. Also, root based arguments might work, but you need to take good care in the computation.

A simpler idea, which is classic in this context, is to try and linearize the recurrence relation. In order to do this, let’s write two consecutive recurrence relations

\displaystyle Q_nQ_{n-2} +1 = Q_{n-1}^2

\displaystyle Q_n^2 = Q_{n+1}Q_{n-1}+1

We add them and we obtain the following relation

\displaystyle \frac{Q_n}{Q_{n-1}} = \frac{Q_{n+1}+Q_{n-1}}{Q_n+Q_{n-2}},

which leads straightforward to a telescoping argument. Finally, we are left with a simple linear recurrence with integer coefficient polynomials, and the result follows immediately.

A hint for Project Euler Pb 613

November 18, 2017 Leave a comment

The text for Problem 613 can be found here. The hint is the following picture 🙂


IMC 2017 – Day 2 – Problems

August 3, 2017 Leave a comment

Problem 6. Let {f: [0,\infty) \rightarrow \Bbb{R}} be a continuous function such that {\lim_{x \rightarrow \infty}f(x) = L} exists (finite or infinite).

Prove that

\displaystyle \lim_{n \rightarrow \infty} \int_0^1 f(nx) dx = L.

Problem 7. Let {p(x)} be a nonconstant polynomial with real coefficients. For every positive integer {n} let

\displaystyle q_n(x) = (x+1)^n p(x)+x^n p(x+1).

Prove that there are only finitely many numbers {n} such that all roots of {q_n(x)} are real.

Problem 8. Define the sequence {A_1,A_2,...} of matrices by the following recurrence

\displaystyle A_1 = \begin{pmatrix} 0& 1 \\ 1& 0 \end{pmatrix}, \ A_{n+1} = \begin{pmatrix} A_n & I_{2^n} \\ I_{2^n} & A_n \end{pmatrix} \ \ (n=1,2,...)

where {I_m} is the {m\times m} identity matrix.

Prove that {A_n} has {n+1} distinct integer eigenvalues {\lambda_0<\lambda_1<...<\lambda_n} with multiplicities {{n \choose 0},\ {n\choose 1},...,{n \choose n}}, respectively.

Problem 9. Define the sequence {f_1,f_2,... : [0,1) \rightarrow \Bbb{R}} of continuously differentiable functions by the following recurrence

\displaystyle f_1 = 1; f'_{n+1} = f_nf_{n+1} \text{ on } (0,1) \text{ and } f_{n+1}(0)=1.

Show that {\lim_{n\rightarrow \infty}f_n(x)} exists for every {x \in [0,1)} and determine the limit function.

Problem 10. Let {K} be an equilateral triangle in the plane. Prove that for every {p>0} there exists an {\varepsilon >0} with the following property: If {n} is a positive integer and {T_1,...,T_n} are non-overlapping triangles inside {K} such that each of them is homothetic to {K} with a negative ratio and

\displaystyle \sum_{\ell =1}^n \text{area}(T_\ell) > \text{area} (K)-\varepsilon,


\displaystyle \sum_{\ell =1}^n \text{perimeter} (T_\ell) > p.

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