INMO/md/en-Inmo-2019-Solutions.md

$34^{\text {th }}$ Indian National Mathematical Olympiad-2019

Problems and Solutions

1. Let $A B C$ be a triangle with $\angle B A C>90^{\circ}$. Let $D$ be a point on the segment $B C$ and $E$ be a point on the line $A D$ such that $A B$ is tangent to the circumcircle of triangle $A C D$ at $A$ and $B E$ is perpendicular to $A D$. Given that $C A=C D$ and $A E=C E$, determine $\angle B C A$ in degrees.

Solution: Let $\angle C=2 \alpha$. Then $\angle C A D=\angle C D A=90^{\circ}-\alpha$. Moreover $\angle B A D=2 \alpha$ as $B A$ is tangent to the circumcircle of $\triangle C A D$. Since $A E=A D$, it gives $\angle A E C=2 \alpha$. Thus $\triangle A E C$ is similar to $\triangle A C D$. Hence

$$\frac{AE}{AC}=\frac{AC}{AD}\backslash frac\{A\; E\}\{A\; C\}=\backslash frac\{A\; C\}\{A\; D\}$$

But the condition that $B E \perp A D$ gives $A E=A B \cos 2 \alpha=c \cos 2 \alpha$. It is easy to see that $\angle B=90^{\circ}-3 \alpha$. Using sine rule in triangle $A D C$, we get

$$\frac{AD}{\sin 2\alpha }=\frac{AC}{\sin (90-\alpha )}\backslash frac\{A\; D\}\{\backslash sin\; 2\; \backslash alpha\}=\backslash frac\{A\; C\}\{\backslash sin\; (90-\backslash alpha)\}$$

This gives $A D=2 b \sin \alpha$. Thus we get

$$b^2=A{C}^2=AE\cdot AD=(c\cos 2\alpha )\cdot 2b\sin \alpha b^\{2\}=A\; C^\{2\}=A\; E\; \backslash cdot\; A\; D=(c\; \backslash cos\; 2\; \backslash alpha)\; \backslash cdot\; 2\; b\; \backslash sin\; \backslash alpha$$

Using $b=2 R \sin B$ and $c=2 R \sin C$, this leads to

$$\cos 3\alpha =2\sin 2\alpha \cos 2\alpha \sin \alpha =\sin 4\alpha \sin \alpha \backslash cos\; 3\; \backslash alpha=2\; \backslash sin\; 2\; \backslash alpha\; \backslash cos\; 2\; \backslash alpha\; \backslash sin\; \backslash alpha=\backslash sin\; 4\; \backslash alpha\; \backslash sin\; \backslash alpha$$

Writing $\cos 3 \alpha=\cos (4 \alpha-\alpha)$ and expanding, we get $\cos 4 \alpha \cos \alpha=0$. Therefore $\alpha=90^{\circ}$ or $4 \alpha=90^{\circ}$. But $\alpha=90^{\circ}$ is not possible as $\angle C=2 \alpha$. Therefore $4 \alpha=90^{\circ}$ which gives $\angle C=2 \alpha=45^{\circ}$.

2. Let $A_{1} B_{1} C_{1} D_{1} E_{1}$ be a regular pentagon. For $2 \leq n \leq 11$,

let $A_{n} B_{n} C_{n} D_{n} E_{n}$ be the pentagon whose vertices are the midpoints of the sides of the pentagon $A_{n-1} B_{n-1} C_{n-1} D_{n-1} E_{n-1}$. All the 5 vertices of each of the 11 pentagons are arbitrarily coloured red or blue. Prove that four points among these 55 points have the same colour and form the vertices of a cyclic quadrilateral.

Solution: We first observe that all the eleven pentagons are regular. Moreover, there are 5 fixed directions and all the 55 sides are in one of these directions. If we consider any two sides which are parallel, they are the parallel sides of an isosceles trapezium, which is cyclic.

If we consider any pentagon, its two adjacent vertices have the same colour. Consider all such 11 sides whose end points are of the same colour. These are in 5 fixed directions. By pigeon-hole principle, there are 3 sides which are in the same directions and therefore parallel to each other. Among these three sides, two must have end points having one colour (again by $\mathrm{P}-\mathrm{H}$ principle). Thus there are two parallel sides among the 55 and the end points of these have one fixed colour. But these two sides are parallel sides of an isosceles trapezium. Hence the four end points are concyclic.

3. Let $m, n$ be distinct positive integers. Prove that

$$\gcd (m,n)+\gcd (m+1,n+1)+\gcd (m+2,n+2)\le 2\mathrm{\mid }m-n\mathrm{\mid }+1\backslash operatorname\{gcd\}(m,\; n)+\backslash operatorname\{gcd\}(m+1,\; n+1)+\backslash operatorname\{gcd\}(m+2,\; n+2)\; \backslash leq\; 2|m-n|+1$$

Further, determine when equality holds.

Solution: Observe that

$$\gcd (m+j,n+j)=\gcd (m+j,\mathrm{\mid }m-n\mathrm{\mid })\backslash operatorname\{gcd\}(m+j,\; n+j)=\backslash operatorname\{gcd\}(m+j,|m-n|)$$

for $j=0,1,2$. Hence we can find positive integers $a, b, c$ such that

$$\gcd (m,n)=\frac{\mathrm{\mid }m-n\mathrm{\mid }}{a},\gcd (m+1,n+1)=\frac{\mathrm{\mid }m-n\mathrm{\mid }}{b},\gcd (m+2,n+2)=\frac{\mathrm{\mid }m-n\mathrm{\mid }}{c}\backslash operatorname\{gcd\}(m,\; n)=\backslash frac\{|m-n|\}\{a\},\; \backslash quad\; \backslash operatorname\{gcd\}(m+1,\; n+1)=\backslash frac\{|m-n|\}\{b\},\; \backslash quad\; \backslash operatorname\{gcd\}(m+2,\; n+2)=\backslash frac\{|m-n|\}\{c\}$$

It follows that $|m-n|$ divides $m a,(m+1) b$ and $(m+2) c$. Hence we can see that $|m-n|$ divides $a b$ and $b c$. We get $|m-n| \leq a b$ and $|m-n| \leq b c$. This leads to

$$b\ge \frac{\mathrm{\mid }m-n\mathrm{\mid }}{a},b\ge \frac{\mathrm{\mid }m-n\mathrm{\mid }}{c}b\; \backslash geq\; \backslash frac\{|m-n|\}\{a\},\; \backslash quad\; b\; \backslash geq\; \backslash frac\{|m-n|\}\{c\}$$

Thus

We have to prove that

$$2b+\frac{\mathrm{\mid }m-n\mathrm{\mid }}{b}\le 2\mathrm{\mid }m-n\mathrm{\mid }+12\; b+\backslash frac\{|m-n|\}\{b\}\; \backslash leq\; 2|m-n|+1$$

Taking $|m-n|=K$, we have to show that $2 b^{2}+K \leq b(2 K+1)$. This reduces to $(b-K)(2 b-1) \leq 0$. However

$$K=\mathrm{\mid }m-n\mathrm{\mid }\ge b\ge 1>\frac{1}{2}K=|m-n|\; \backslash geq\; b\; \backslash geq\; 1>\backslash frac\{1\}\{2\}$$

Equality holds only when $(m, n)=(k, k+1)$ or $(2 k, 2 k+2)$ or permutations of these for some $k$.

4. Let $n$ and $M$ be positive integers such that $M>n^{n-1}$. Prove that there are $n$ distinct primes $p_{1}, p_{2}, p_{3}, \ldots, p_{n}$ such that $p_{j}$ divides $M+j$ for $1 \leq j \leq n$.

Solution: If some number $M+k, 1 \leq k \leq n$, has at least $n$ distinct prime factors, then we can associate a prime factor of $M+k$ with the number $M+k$ which is not associated with any of the remaining $n-1$ numbers.

Suppose $m+j$ has less than $n$ distinct prime factors. Write

But $M+j>n^{n-1}$. Hence there exist $t, 1 \leq t \leq r$ such that $p_{t}^{\alpha_{t}}>n$. Associate $p_{t}$ with this $M+j$. Suppose $p_{t}$ is associated with some $M+l$. Let $p_{t}^{\beta_{t}}$ be the largest power of $p_{t}$ dividing $M+l$. Then $p_{t}^{\beta_{t}}>n$. Let $T=\operatorname{gcd}\left(p_{t}^{\alpha_{t}}, p_{t}^{\beta_{t}}\right)$. Then $T>n$. Since $T \mid(M+j)$ and $T \mid(M+l)$, it follows that $T \mid(|j-l|)$. But $|j-l|<n$ and $T>n$, and we get a contradiction. This shows that $p_{t}$ cannot be associated with any other $M+l$. Thus each $M+j$ is associated with different primes.

5. Let $A B$ be a diameter of a circle $\Gamma$ and let $C$ be a point on $\Gamma$ different from $A$ and $B$. Let $D$ be the foot of perpendicular from $C$ on to $A B$. Let $K$ be a point of the segment $C D$ such that $A C$ is equal to the semiperimeter of the triangle $A D K$. Show that the excircle of triangle $A D K$ opposite $A$ is tangent to $\Gamma$.

Solution: Draw another diameter $P Q \perp A B$. Let $E$ be the point at which the excircle $\Gamma_{1}$ touches the line $A D$. Join $Q E$ and extend it to meet $\Gamma$ in $L$. Draw the diameter $E N$ of $\Gamma_{1}$ and draw $Q S \perp N E$ (extended). See the figure. We also observe that $D E=E M=E N / 2$.

Since $A E$ is equal to the semiperimeter of $\triangle A D K$, we have $A C=A E$. Hence $A E^{2}=A C^{2}=$ $A D \cdot A B$ (as $A C B$ is a right-angle triangle). Thus

$$AD(AD+DE+EB)=(AD+DE{)}^2=A{D}^2+2AD\cdot DE+D{E}^{2}A\; D(A\; D+D\; E+E\; B)=(A\; D+D\; E)^\{2\}=A\; D^\{2\}+2\; A\; D\; \backslash cdot\; D\; E+D\; E^\{2\}$$

Simplification gives

Therefore

$$DE\cdot AB=EB(AD+DE)=EB\cdot AED\; E\; \backslash cdot\; A\; B=E\; B(A\; D+D\; E)=E\; B\; \backslash cdot\; A\; E$$

But

$$DE\cdot AB=DE\cdot PQ=2DE\cdot OQ=EN\cdot ESD\; E\; \backslash cdot\; A\; B=D\; E\; \backslash cdot\; P\; Q=2\; D\; E\; \backslash cdot\; O\; Q=E\; N\; \backslash cdot\; E\; S$$

and $E B \cdot A E=Q E \cdot E L$. Therefore we get

$$QE\cdot EL=EN\cdot ESQ\; E\; \backslash cdot\; E\; L=E\; N\; \backslash cdot\; E\; S$$

It follows that $Q, S, L, N$ are concyclic. Since $\angle Q S E=90^{\circ}$, we get $\angle E L N=90^{\circ}$. Since $E N$ is a diameter, this implies that $L$ also lies on $\Gamma_{1}$. But $\angle Q L P=90^{\circ}$. Therefore $L, N, P$ are collinear. Since $N M | P O$ and

$$\frac{NM}{PO}=\frac{NE}{PQ}=\frac{LN}{LP}\backslash frac\{N\; M\}\{P\; O\}=\backslash frac\{N\; E\}\{P\; Q\}=\backslash frac\{L\; N\}\{L\; P\}$$

it follows that $L, M, O$ are collinear. Hence $\Gamma_{1}$ is tangent to $\Gamma$ at $L$.

Alternate solution: Let $R$ be the radius the circle $\Gamma$ and $r$ be that of the circle $\Gamma_{1}$. Let $O$ be the centre of $\Gamma$ and $M$ be that of the circle $\Gamma_{1}$. Let $E$ be the point of contact of $\Gamma_{1}$ with $A B$. Then $M E=D E=r$. Observe that $A E$ is the semiperimeter of $\triangle A D E$. We are given that $A C=A E$. Using that $\angle A C B=90^{\circ}$, we also get $A C^{2}=A D \cdot A B$. Hence $A E^{2}=A D \cdot A B$. We have to show that $R-r=O M$ for proving that $\Gamma_{1}$ is tangent to $\Gamma$. We have

Hence $O M=R-r$ and therefore $\Gamma_{1}$ is tangent to $\Gamma$.

6. Let $f$ be function defined from the set ${(x, y): x, y$ reals, $x y \neq 0}$ in to the set of all positive real numbers such that

(i) $\quad f(x y, z)=f(x, z) f(y, z)$, for all $x, y \neq 0$;

(ii) $\quad f(x, y z)=f(x, y) f(x, z)$, for all $x, y \neq 0$;

(iii) $\quad f(x, 1-x)=1$, for all $x \neq 0,1$.

Prove that

(a) $\quad f(x, x)=f(x,-x)=1$, for all $x \neq 0$;

(b) $\quad f(x, y) f(y, x)=1$, for all $x, y \neq 0$.

Solution: (The condition (ii) was inadvertently left out in the paper. We give the solution with condition (ii).)

Taking $x=y=1$ in (ii), weget $f(1, z)^{2}=f(1, z)$ so that $f(1, z)=1$ for all $z \neq 0$. Similarly, $x=y=-1$ gives $f(-1, z)=1$ for all $z \neq 0$. Using the second condition, we also get $f(z, 1)=$ $f(z,-1)=1$ for all $z \neq 0$. Observe

$$f(\frac{1}{x},y)f(x,y)=f(1,y)=1=f(x,1)=f(x,\frac{1}{y})f(x,y)f\backslash left(\backslash frac\{1\}\{x\},\; y\backslash right)\; f(x,\; y)=f(1,\; y)=1=f(x,\; 1)=f\backslash left(x,\; \backslash frac\{1\}\{y\}\backslash right)\; f(x,\; y)$$

Therefore

$$f(x,\frac{1}{y})=f(\frac{1}{x},y)=\frac{1}{f(x,y)}f\backslash left(x,\; \backslash frac\{1\}\{y\}\backslash right)=f\backslash left(\backslash frac\{1\}\{x\},\; y\backslash right)=\backslash frac\{1\}\{f(x,\; y)\}$$

for all $x, y \neq 0$. Now for $x \neq 0,1$, condition (iii) gives

$$1=f(\frac{1}{x},1-\frac{1}{x})=f(x,\frac{1}{1-\frac{1}{x}})1=f\backslash left(\backslash frac\{1\}\{x\},\; 1-\backslash frac\{1\}\{x\}\backslash right)=f\backslash left(x,\; \backslash frac\{1\}\{1-\backslash frac\{1\}\{x\}\}\backslash right)$$

Multiplying by $1=f(x, 1-x)$, we get

$$1=f(x,1-x)f(x,\frac{1}{1-\frac{1}{x}})=f(x,\frac{1-x}{1-\frac{1}{x}})=f(x,-x)1=f(x,\; 1-x)\; f\backslash left(x,\; \backslash frac\{1\}\{1-\backslash frac\{1\}\{x\}\}\backslash right)=f\backslash left(x,\; \backslash frac\{1-x\}\{1-\backslash frac\{1\}\{x\}\}\backslash right)=f(x,-x)$$

for all $x \neq 0,1$. But $f(x,-1)=1$ for all $x \neq 0$ gives

$$f(x,x)=f(x,-x)f(x,-1)=f(x,-x)=1f(x,\; x)=f(x,-x)\; f(x,-1)=f(x,-x)=1$$

for all $x \neq 0,1$. Observe $f(1,1)=f(1,-1)=1$. Hence

$$f(x,x)=f(x,-x)=1f(x,\; x)=f(x,-x)=1$$

for all $x \neq 0$, which proves (a).

We have

$$1=f(xy,xy)=f(x,xy)f(y,xy)=f(x,x)f(x,y)f(y,x)f(y,y)=f(x,y)f(y,x)1=f(x\; y,\; x\; y)=f(x,\; x\; y)\; f(y,\; x\; y)=f(x,\; x)\; f(x,\; y)\; f(y,\; x)\; f(y,\; y)=f(x,\; y)\; f(y,\; x)$$

for all $x, y \neq 0$, which proves (b).