# Solution to INMO-2002 Problems 

1. For a convex hexagon $A B C D E F$ in which each pair of opposite sides is unequal, consider the following six statements:

$$
\begin{array}{ll}
\text { (a } \left.\mathrm{a}_{1}\right) A B \text { is parallel to } D E ; & \left(\mathrm{a}_{2}\right) A E=B D \\
\left(\mathrm{~b}_{1}\right) B C \text { is parallel to } E F ; & \left(\mathrm{b}_{2}\right) B F=C E \\
\text { (c } \left.\mathrm{c}_{1}\right) C D \text { is parallel to } F A ; & \left(\mathrm{c}_{2}\right) C A=D F
\end{array}
$$

(a) Show that if all the six statements are true, then the hexagon is cyclic(i.e., it can be inscribed in a circle).

(b) Prove that, in fact, any five of these six statements also imply that the hexagon is cyclic.

## Solution:

(a) Suppose all the six statements are true. Then $A B D E, B C E F, C D F A$ are isosceles trapeziums; if $K, L, M, P, Q, R$ are the mid-points of $A B, B C$, $C D, D E, E F, F A$ respectively, then we see that $K P \perp A B, E D ; L Q \perp$ $B C, E F$ and $M R \perp C D, F A$.

![](https://cdn.mathpix.com/cropped/2024_06_05_5265b477c00c3329af6ag-1.jpg?height=480&width=651&top_left_y=1164&top_left_x=756)

If $A D, B E, C F$ themselves concur at a point $O$, then $O A=O B=O C=$ $O D=O E=O F$. ( $O$ is on the perpendicular bisector of each of the sides.) Hence $A, B, C, D, E, F$ are concyclic and lie on a circle with centre $O$. Otherwise these lines $A D, B E, C F$ form a triangle, say $X Y Z$. (See Fig.) Then $K X, M Y, Q Z$, when extended, become the internal angle bisectors of the triangle $X Y Z$ and hence concur at the incentre $O^{\prime}$ of $X Y Z$. As earlier $O^{\prime}$ lies on the perpendicular bisector of each of the sides. Hence $O^{\prime} A=O^{\prime} B$ $=O^{\prime} C=O^{\prime} D=O^{\prime} E=O^{\prime} F$, giving the concyclicity of $A, B, C, D, E, F$.
(b) Suppose $\left(\mathrm{a}_{1}\right),\left(\mathrm{a}_{2}\right),\left(\mathrm{b}_{1}\right),\left(\mathrm{b}_{2}\right)$ are true. Then we see that $A D=B E=$ $C F$. Assume that ( $\mathrm{c}_{1}$ ) is true. Then $C D$ is parallel to $A F$. It follows that triangles $Y C D$ and $Y F A$ are similar. This gives

$$
\frac{F Y}{A Y}=\frac{Y C}{Y D}=\frac{F Y+Y C}{A Y+Y D}=\frac{F C}{A D}=1
$$

We obtain $F Y=A Y$ and $Y C=Y D$. This forces that triangles $C Y A$ and $D Y F$ are congruent. In particular $A C=D F$ so that ( $\mathrm{c}_{2}$ ) is true. The conclusion follows from (a). Now assume that ( $\mathrm{c}_{2}$ ) is true; i.e., $A C=F D$. We have seen that $A D=B E=C F$. It follows that triangles $F D C$ and $A C D$ are congruent. In particular $\angle A D C=\angle F C D$. Similarly, we can show that $\angle C F A=\angle D A F$. We conclude that $C D$ is parallel to $A F$ giving $\left(\mathrm{c}_{1}\right.$ ).

2. Determine the least positive value taken by the expression $a^{3}+b^{3}+c^{3}-3 a b c$ as $a, b, c$ vary over all positive integers. Find also all triples $(a, b, c)$ for which this least value is attained.

Solution: We observe that

$$
\left.Q=a^{3}+b^{3}+c^{3}-3 a b c=\frac{1}{2}(a+b+c)\right)\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right)
$$

Since we are looking for the least positive value taken by $Q$, it follows that $a, b, c$ are not all equal. Thus $a+b+c \geq 1+1+2=4$ and $(a-b)^{2}+(b-$ $c)^{2}+(c-a)^{2} \geq 1+1+0=2$. Thus we see that $Q \geq 4$. Taking $a=1$, $b=1$ and $c=2$, we get $Q=4$. Therefore the least value of $Q$ is 4 and this is achieved only by $a+b+c=4$ and $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2$. The triples for which $Q=4$ are therefore given by

$$
(a, b, c)=(1,1,2),(1,2,1),(2,1,1)
$$

3. Let $x, y$ be positive reals such that $x+y=2$. Prove that

$$
x^{3} y^{3}\left(x^{3}+y^{3}\right) \leq 2
$$

Solution: We have from the AM-GM inequality, that

$$
x y \leq\left(\frac{x+y}{2}\right)^{2}=1
$$

Thus we obtain $0<x y \leq 1$. We write

$$
\begin{aligned}
x^{3} y^{3}\left(x^{3}+y^{3}\right) & =(x y)^{3}(x+y)\left(x^{2}-x y+y^{2}\right) \\
& =2(x y)^{3}\left((x+y)^{2}-3 x y\right) \\
& =2(x y)^{3}(4-3 x y)
\end{aligned}
$$

Thus we need to prove that

$$
(x y)^{3}(4-3 x y) \leq 1
$$

Putting $z=x y$, this inequality reduces to

$$
z^{3}(4-3 z) \leq 1
$$

for $0<z \leq 1$. We can prove this in different ways. We can put the inequality in the form

$$
3 z^{4}-4 z^{3}+1 \geq 0
$$

Here the expression in the LHS factors to $(z-1)^{2}\left(3 z^{2}+2 z+1\right)$ and $\left(3 z^{2}+2 z+1\right)$ is positive since its discriminant $D=-8<0$. Or applying the AM-GM inequality to the positive reals $4-3 z, z, z, z$, we obtain

$$
z^{3}(4-3 z) \leq\left(\frac{4-3 z+3 z}{4}\right)^{4} \leq 1
$$

4. Do there exist 100 lines in the plane, no three of them concurrent, such that they intersect exactly in 2002 points?

Solution: Any set of 100 lines in the plane can be partitioned into a finite number of disjoint sets, say $A_{1}, A_{2}, A_{3}, \ldots, A_{k}$, such that

(i) Any two lines in each $A_{j}$ are parallel to each other, for $1 \leq j \leq k$ (provided, of course, $\left|A_{j}\right| \geq 2$ );

(ii) for $j \neq l$, the lines in $A_{j}$ and $A_{l}$ are not parallel.

If $\left|A_{j}\right|=m_{j}, 1 \leq j \leq k$, then the total number of points of intersection is given by $\sum_{1 \leq j \leq l \leq k} m_{j} m_{l}$, as no three lines are concurrent. Thus we have to find positive integers $m_{1}, m_{2}, \ldots, m_{k}$ such that

$$
\sum_{j=1}^{k} m_{j}=100, \quad \sum m_{j} m_{l}=2002
$$

for an affirmative answer to the given question.

We observe that

$$
\begin{aligned}
\sum_{j=1}^{k} m_{j}^{2} & =\left(\sum_{j=1}^{k} m_{j}\right)^{2}-2\left(\sum m_{j} m_{l}\right) \\
& =100^{2}-2(2002)=5996
\end{aligned}
$$

Thus we have to choose $m_{1}, m_{2}, \ldots, m_{k}$ such that

$$
\sum_{j=1}^{k} m_{j}=100, \quad \sum_{j=1}^{k} m_{j}^{2}=5996
$$

We observe that $[\sqrt{5996}]=77$. So we may take $m_{1}=77$, so that

$$
\sum_{j=2}^{k} m_{j}=23, \quad \sum j=2^{k} m_{j}^{2}=67
$$

Now we may choose $m_{2}=5, m_{3}=m_{4}=4, m_{5}=m_{6}=\cdots=m_{14}=1$. Finally, we can take

$$
k=14, \quad\left(m_{1}, m_{2}, \ldots, m_{14}\right)=(77,5,4,4,1,1,1,1,1,1,1,1,1,1)
$$

proving the existence of 100 lines with exactly 2002 points of intersection.

5. Do there exist three distinct positive real numbers $a, b, c$ such that the numbers $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order?

Solution: We show that the answer is NO. Suppose, if possible, let $a, b, c$ be three distinct positive real numbers such that $a, b, c, b+c-a, c+a-b$, $a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order. We may assume that $a<b<c$. Then there are only two cases we need to check: (I) $a+b-c<a<c+a-b<b<c<b+c-a<a+b+c$ and (II) $a+b-c<a<b<c+a-b<c<b+c-a<a+b+c$.

Case I. Suppose the chain of inequalities $a+b-c<a<c+a-b<b<$ $c<b+c-a<a+b+c$ holds good. let $d$ be the common difference. Thus we see that

$$
c=a+b+c-2 d, b=a+b+c-3 d, a=a+b+c-5 d
$$

Adding these, we see that $a+b+c=5 d$. But then $a=0$ contradicting the positivity of $a$.

Case II. Suppose the inequalities $a+b-c<a<b<c+a-b<c<$ $b+c-a<a+b+c$ are true. Again we see that

$$
c=a+b+c-2 d, b=a+b+c-4 d, a=a+b+c-5 d
$$

We thus obtain $a+b+c=(11 / 2) d$. This gives

$$
a=\frac{1}{2} d, b=\frac{3}{2} d, c=\frac{7}{2} d
$$

Note that $a+b-c=a+b+c-6 d=-(1 / 2) d$. However we also get $a+b-c=[(1 / 2)+(3 / 2)-(7 / 2)] d=-(3 / 2) d$. It follows that $3 e=e$ giving $d=0$. But this is impossible.

Thus there are no three distinct positive real numbers $a, b, c$ such that $a$, $b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order.

6. Suppose the $n^{2}$ numbers $1,2,3, \ldots, n^{2}$ are arranged to form an $n$ by $n$ array consisting of $n$ rows and $n$ columns such that the numbers in each row(from left to right) and each column(from top to bottom) are in increasing order. Denote by $a_{j k}$ the number in $j$-th row and $k$-th column. Suppose $b_{j}$ is the maximum possible number of entries that can occur as $a_{j j}, 1 \leq j \leq n$. Prove that

$$
b_{1}+b_{2}+b_{3}+\cdots b_{n} \leq \frac{n}{3}\left(n^{2}-3 n+5\right)
$$

(Example: In the case $n=3$, the only numbers which can occur as $a_{22}$ are 4,5 or 6 so that $b_{2}=3$.)

Solution: Since $a_{j j}$ has to exceed all the numbers in the top left $j \times j$ submatrix (excluding itself), and since there are $j^{2}-1$ entries, we must have $a_{j j} \geq j^{2}$. Similarly, $a_{j j}$ must not exceed eac of the numbers in the bottom right $(n-j+1) \times(n-j+1)$ submatrix (other than itself) and there are $(n-j+1)^{2}-1$ such entries giving $a_{j j} \leq n^{2}-(n-j+1)^{2}+1$. Thus we see that

$$
a_{j j} \in\left\{j^{2}, j^{2}+1, j^{2}+2, \ldots, n^{2}-(n-j+1)^{2}+1\right\}
$$

The number of elements in this set is $n^{2}-(n-j+1)^{2}-j^{2}+2$. This implies that

$$
b_{j} \leq n^{2}-(n-j+1)^{2}-j^{2}+2=(2 n+2) j-2 j^{2}-(2 n-1)
$$

It follows that

$$
\begin{aligned}
\sum_{j=1}^{n} b_{j} & \leq(2 n+2) \sum_{j=1}^{n} j-2 \sum_{j=1}^{n} j^{2}-n(2 n-1) \\
& =(2 n+2)\left(\frac{n(n+1)}{2}\right)-2\left(\frac{n(n+1)(2 n+1)}{6}\right)-n(2 n-1) \\
& =\frac{n}{3}\left(n^{2}-3 n+5\right)
\end{aligned}
$$

which is the required bound.