Problem 01

Let $M$ be a point on the side $AB$ of $\triangle ABC$ . Let $r_1, r_2$ , and $r$ be the inscribed circles of triangles $AMC, BMC$ , and $ABC$ . Let $q_1, q_2$ , and $q$ be the radii of the exscribed circles of the same triangles that lie in the angle $ACB$ . Prove that

$\frac{r_1}{q_1} \cdot \frac{r_2}{q_2} = \frac{r}{q}$ .

Solution

We use the conventional triangle notations.

Let $I$ be the incenter of $ABC$ , and let $I_{c}$ be its excenter to side $c$ . We observe that

$r \left[ \cot\left(\frac{A}{2}\right) + \cot\left(\frac{B}{2}\right) \right] = c$ ,

and likewise,

$\begin{matrix} c & = &q \left[ \cot\left(\frac{\pi - A}{2}\right) + \cot \left(\frac{\pi - B}{2}\right) \right]\\ & = &q \left[ \tan\left(\frac{A}{2}\right) + \tan\left(\frac{B}{2}\right) \right]\; . \end{matrix}$

Simplifying the quotient of these expressions, we obtain the result

$\frac{r}{q} = \tan (A/2) \tan (B/2)$ .

Thus we wish to prove that

$\tan (A/2) \tan (B/2) = \tan (A/2) \tan (AMC/2) \tan (B/2) \tan (CMB/2)$ .

But this follows from the fact that the angles $AMC$ and $CBM$ are supplementary.

Problem 02

Let $a, b$ , and $n$ be integers greater than 1, and let $a$ and $b$ be the bases of two number systems. $A_{n-1}$ and $A_{n}$ are numbers in the system with base $a$ and $B_{n-1}$ and $B_{n}$ are numbers in the system with base $b$ ; these are related as follows:

$A_{n} = x_{n}x_{n-1}\cdots x_{0}, A_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}$ ,

$B_{n} = x_{n}x_{n-1}\cdots x_{0}, B_{n-1} = x_{n-1}x_{n-2}\cdots x_{0}$ ,

$x_{n} \neq 0, x_{n-1} \neq 0$ .

Prove:

$\frac{A_{n-1}}{A_{n}} < \frac{B_{n-1}}{B_{n}}$ if and only if $a > b$ .

Solution

Suppose $a>b$ . Then for all integers $0 \le k \le n$ , $x_n x_k a^n b^k \ge x_n x_k b^n a^k$ , with equality only when $k=n$ or $x_k = 0$ . (In particular, we have strict inequality for $k=n-1$ .) In summation, this becomes $x_n a^n \sum_{k=0}^n x_k b^k > x_n b^n \sum_{k=0}^n x_k b^k,$ or $x_n a^n \cdot B_n > x_n b^n \cdot A_n,$ which is equivalent to $\frac{x_n a^n}{A_n} > \frac{x_n b^n}{B_n} .$ This implies $\frac{A_{n-1}}{A_n} = 1 - \frac{x_n a^n}{A_n} < 1 - \frac{x_n b^n}{B_n} = \frac{B_{n-1}}{B_n} .$ On the other hand, if $a=b$ , then evidently $A_{n-1}/A_n = B_{n-1}/B_n$ , and if $a < b$ , then by what we have just shown, $A_{n-1}/A_n > B_{n-1}/B_n$ . Hence $A_{n-1}/A_n < B_{n-1}/B_n$ if and only if $a>b$ , as desired.

Problem 03

The real numbers $a_0, a_1, \ldots, a_n, \ldots$ satisfy the condition:

$1 = a_{0} \leq a_{1} \leq \cdots \leq a_{n} \leq \cdots$ .

The numbers $b_{1}, b_{2}, \ldots, b_n, \ldots$ are defined by

$b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}}$

(a) Prove that $0 \leq b_n < 2$ for all $n$ .

(b) given $c$ with $0 \leq c < 2$ , prove that there exist numbers $a_0, a_1, \ldots$ with the above properties such that $b_n > c$ for large enough $n$ .

Solution

$b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}} = \sum_{k=1}^{n} \frac{a_k - a_{k-1}}{a_k\sqrt{a_k}} = \sum_{k=1}^{n} (a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)$

Let $X_k$ be the rectangle with the verticies: $(a_{k-1},0)$ ; $(a_{k},0)$ ; $(a_{k},a_k^{-\dfrac{3}{2}})$ ; $(a_{k-1},a_k^{-\dfrac{3}{2}})$ .

$[asy] import graph; size(10cm,10cm,IgnoreAspect); Label f; f.p=fontsize(6); xaxis(0,10); yaxis(0,1); real f(real x) { return x^(-3/2); } draw(graph(f,1,10)); draw((1,0)--(1,1)); draw((1,0)--(2,0)--(2,2^(-3/2))--(1,2^(-3/2))--cycle); draw((2,0)--(3,0)--(3,3^(-3/2))--(2,3^(-3/2))--cycle); draw((3,0)--(4,0)--(4,4^(-3/2))--(3,4^(-3/2))--cycle); draw((6,0)--(7,0)--(7,7^(-3/2))--(6,7^(-3/2))--cycle); draw((7,0)--(8,0)--(8,8^(-3/2))--(7,8^(-3/2))--cycle); label("$X_1$",(1.5,0),0.5*N); label("$X_2$",(2.5,0),0.5*N); label("$X_3$",(3.5,0),0.5*N); label("$\cdots$",(5,0),N); label("$X_{n-1}$",(6.5,0),0.5*N); label("$X_n$",(7.5,0),0.5*N); label("$a_0$",(1,0),0.5*S); label("$a_1$",(2,0),0.5*S); label("$a_2$",(3,0),0.5*S); label("$a_3$",(4,0),0.5*S); label("$a_{n-2}$",(6,0),0.5*S); label("$a_{n-1}$",(7,0),0.5*S); label("$a_n$",(8,0),0.5*S); [/asy]$

For all $k \in \mathbb{N}$ , the area of $X_k$ is $(a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)$ . Therefore, $b_n = \sum_{k=1}^{n} [X_k]$

For all sequences $\{ a_k \}$ and all $k \in \mathbb{N}$ , $X_k$ lies above the $x$ -axis, below the curve $f(x) = x^{-\dfrac{3}{2}}$ , and in between the lines $x = 1$ and $x = a_n$ , Also, all such rectangles are disjoint.

Thus, $b_n = \sum_{k=1}^{n} [X_k] < \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx = \left[-\dfrac{2}{\sqrt{x}}\right]_{1}^{a_n} = 2 - \dfrac{2}{\sqrt{a_n}} < 2$ as desired.

By choosing $a_k = 1 + k (\Delta x)$ , where $\Delta x > 0$ , $b_n$ is a Riemann sum for $\int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$ . Thus, $\lim_{\Delta x \to 0^+} b_n = \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$ .

Therefore, $\lim_{n \to \infty} \left[ \lim_{\Delta x \to 0^+} b_n \right] = \lim_{n \to \infty} \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$ $= \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{a_n}} = \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{1+n(\Delta x)}} = 2$ .

So for any $c \in [0,2)$ , we can always select a small enough $\Delta x > 0$ to form a sequence $\{ a_n \}$ satisfying the above properties such that $b_n > c$ for large enough $n$ as desired.

Problem 04

Find the set of all positive integers $n$ with the property that the set $\{ n, n+1, n+2, n+3, n+4, n+5 \}$ can be partitioned into two sets such that the product of the numbers in one set equals the product of the numbers in the other set.

Solution

The only primes dividing numbers in the set can be 2, 3 or 5, because if any larger prime was a factor, then it would only divide one number in the set and hence only one product. Three of the numbers must be odd. At most one of the odd numbers can be a multiple of 3 and at most one can be a multiple of 5. The other odd number cannot have any prime factors. The only such number is 1, so the set must be $\{ 1, 2, 3, 4, 5, 6 \}$ , but that does not work because only one of the numbers is a multiple of 5. So there are no such sets.

Problem 05

In the tetrahedron $ABCD$ , angle $BDC$ is a right angle. Suppose that the foot $H$ of the perpendicular from $D$ to the plane $ABC$ in the tetrahedron is the intersection of the altitudes of $\triangle ABC$ . Prove that

$( AB+BC+CA )^2 \leq 6( AD^2 + BD^2 + CD^2 )$ .

For what tetrahedra does equality hold?

Solution

Let us show first that angles $ADB$ and $ADC$ are also right. Let $H$ be the intersection of the altitudes of $ABC$ and let $CH$ meet $AB$ at $X$ . Planes $CED$ and $ABC$ are perpendicular and $AB$ is perpendicular to the line of intersection $CE$ . Hence $AB$ is perpendicular to the plane $CDE$ and hence to $ED$ . So $BD^2 = DE^2 + BE^2.$ Also $CB^2 = CE^2 + BE^2.$ Therefore $CB^2 - BD^2 = CE^2 - DE^2.$ But $CB^2 - BD^2 = CD^2,$ so $CE^2 = CD^2 + DE^2$ , so angle $CDE = 90^{\circ}$ . But angle $CDB = 90^{\circ}$ , so $CD$ is perpendicular to the plane $DAB$ , and hence angle $CDA$ = $90^{\circ}$ . Similarly, angle $ADB = 90^{\circ}$ . Hence $AB^2 + BC^2 + CA^2 = 2(DA^2 + DB^2 + DC^2)$ . But now we are done, because Cauchy's inequality gives $(AB + BC + CA)^2 = 3(AB^2 + BC^2 + CA^2).$ We have equality if and only if we have equality in Cauchy's inequality, which means $AB = BC = CA.$

Problem 06

In a plane there are $100$ points, no three of which are collinear. Consider all possible triangles having these point as vertices. Prove that no more than $70 \%$ of these triangles are acute-angled.

Solution

At most $3$ of the triangles formed by $4$ points can be acute. It follows that at most $7$ out of the $10$ triangles formed by any $5$ points can be acute. For given $10$ points, the maximum number of acute triangles is: the number of subsets of $4$ points times $\frac{3}{\text{the number of subsets of 4 points containing 3 given points}}$ . The total number of triangles is the same expression with the first $3$ replaced by $4$ . Hence at most $\frac{3}{4}$ of the $10$ , or $7.5$ , can be acute, and hence at most $7$ can be acute. The same argument now extends the result to $100$ points. The maximum number of acute triangles formed by $100$ points is: the number of subsets of $5$ points times $\frac{7}{\text{the number of subsets of 5 points containing 3 given points}}$ . The total number of triangles is the same expression with $7$ replaced by $10$ . Hence at most $\frac{7}{10}$ of the triangles are acute.