AIMO Initial Selection Mock Paper 2: Questions, Hints and Solutions

Assume a counterexample exists and choose one with the least hypotenuse. Write its square area as $k^2$, where k is a positive integer. Every integer right triangle has an even leg: two odd leg squares would sum to 2 modulo 4, which is not a square. Its area is therefore an integer. Our chosen triangle is primitive: if all sides had a common factor $g>1$, division by $g$ would give an integer right triangle with area $(k/g)^2$. An integer that is a rational square is an integer square, as reducing the fraction shows its denominator must be 1. This would give a smaller counterexample.

Primitive-triple bridge. The legs cannot both be odd, since two odd squares sum to 2 modulo 4. Write the odd leg $X$, even leg $Y$, and odd hypotenuse $Z$. Any prime dividing two sides of a right triangle divides the third as well, by the equation $X^2+Y^2=Z^2$. In a primitive triple the sides are therefore pairwise coprime. A common divisor of $(Z+X)/2$ and $(Z-X)/2$ divides both their sum Z and their difference X, so it must be 1. These two positive integers are consequently coprime and their product is $(Y/2)^2$. Unique prime factorisation forces them to be squares $u^2,v^2$. Therefore $X=u^2-v^2$, $Y=2uv$, $Z=u^2+v^2$, where $u>v>0$, $\gcd(u,v)=1$, and $u,v$ have opposite parity: they cannot both be even by coprimality, and if both were odd then X and Z would both be even.

The area is $uv(u+v)(u-v)$. These four positive factors are pairwise coprime: common divisors of $u$ or $v$ with either sum or difference divide both $u,v$; a common divisor of $u+v$ and $u-v$ divides $2u,2v$, and both factors are odd. Since their product is a square, write $$\begin{gathered}u=a^2,\\ v=b^2,\\ u+v=c^2,\\ u-v=d^2.\end{gathered}$$

The integers $c,d$ are odd. Moreover $u$ cannot be even: then $u=a^2$ is 0 modulo 4 and $v=b^2$ is 1 modulo 4, making $u-v$ equal to 3 modulo 4. Thus $u$ is odd, $v$ is even, and $b$ is even.

Set $h=(c+d)/2$ and $j=(c-d)/2$. They are positive integers because $c>d>0$. Direct calculation gives $$\begin{gathered}h^2+j^2=\frac{c^2+d^2}{2}=u=a^2,\\ 2hj=\frac{c^2-d^2}{2}=v=b^2.\end{gathered}$$ Thus $h,j,a$ form an integer right triangle whose area is $hj/2=(b/2)^2$. Its hypotenuse $a$ is smaller than $u^2+v^2$, the original hypotenuse. This contradicts minimality and completes the descent.

Conclusion: No such triangle exists.

Review the idea: Unique prime factorisation · Pythagoras and stewart · Proof methods

Since $(b-c)^2\ge0$, we have $bc\le(b^2+c^2)/2$. Therefore $$\frac{a^2}{b^2+bc+c^2}\ge\frac{2a^2}{3(b^2+c^2)}.$$ Do this cyclically and put $x=a^2,y=b^2,z=c^2$. It remains to show that $\sum x/(y+z)\ge3/2$.

The needed Cauchy–Schwarz step. For positive $u_i$, the inequality $\sum t_i^2/u_i\ge(\sum t_i)^2/(\sum u_i)$ follows from the square-sum identity $$\begin{aligned}(\sum t_i^2/u_i)(\sum u_i)-(\sum t_i)^2&=\sum_{i<j}\left(t_i\sqrt{u_j/u_i}-t_j\sqrt{u_i/u_j}\right)^2\\&\ge0.\end{aligned}$$ To verify it, expand each square: for every pair the two square terms and the cross term are exactly the terms left on the left-hand side after cancelling $\sum t_i^2$. Dividing by the positive sum $\sum u_i$ gives the stated inequality. Take $(t_1,t_2,t_3)=(x,y,z)$ and $(u_1,u_2,u_3)=(x(y+z),y(z+x),z(x+y))$. It gives $$\sum\frac{x}{y+z}\ge\frac{(x+y+z)^2}{2(xy+yz+zx)}\ge\frac32.$$ The last step uses $x^2+y^2+z^2\ge xy+yz+zx$, obtained by adding three squared differences.

Multiplication by $2/3$ proves the desired bound. Equality in the first three denominator estimates requires $b=c,c=a,a=b$, so $a=b=c$. Conversely, for equal positive variables every fraction is $1/3$. These and only these triples give equality.

Conclusion: The lower bound is 1, attained exactly when a=b=c.

Review the idea: Cauchy schwarz inequality · Sum of squares

Coordinate bridge. Coordinates may be measured along two nonparallel axes even when they are not perpendicular. Collinearity still has a linear equation, and midpoint coordinates are still coordinate averages. We use no distance or angle formula here.

Take $P$ as origin and the lines $AB,CD$ as axes. Write $A=(a,0),B=(b,0),C=(0,c),D=(0,d)$. Convexity and distinct vertices ensure $a,b,c,d\ne0$, $a\ne b$, and $c\ne d$. If $Q=(u,v)$, its membership of $AD$ and $BC$ gives $$\begin{gathered}du+av=ad,\\ cu+bv=bc.\end{gathered}$$ Subtracting, $$(d-c)u+(a-b)v=ad-bc.$$

Let the three midpoints be $M=(a/2,c/2)$, $N=(b/2,d/2)$, and $R=(u/2,v/2)$. Each satisfies the linear equation $$2(d-c)X+2(a-b)Y=ad-bc.$$ For $M$ and $N$, this follows by direct substitution; for $R$, it is the equation just obtained. At least one coefficient is nonzero, so the equation describes a genuine line. Therefore $M,N,R$ are collinear.

The diagram is one configuration; the coordinate argument allows positive or negative axis coordinates and so also covers intersections on other side extensions.

Conclusion: The three midpoints are collinear.

Review the idea: The midpoint theorem · Equations and quadratics · Geometry foundations

Suppose there are $m>n$ members $A_1,\ldots,A_m$. Each of the $2^m$ subcollections determines a parity record of length $n$: for each ground-set element, record 0 or 1 according as it occurs an even or odd number of times. There are only $2^n$ possible records. Two distinct subcollections therefore have the same record.

Keep exactly those sets that belong to one of these two subcollections but not both. This gives a nonempty subcollection $\mathcal K$ in which each ground-set element occurs an even number of times: common sets cancel from the two parity records.

Choose $A_j\in\mathcal K$. Count $$T=\sum_{A_i\in\mathcal K}|A_j\cap A_i|.$$ Counting element by element, each element of $A_j$ is counted an even number of times, so $T$ is even. Counting term by term, the self-intersection $|A_j|$ is odd and every other intersection is even. Thus $T$ is odd, a contradiction.

Therefore $m\le n$. Equality is realised by the $n$ singleton subsets: their sizes are 1 and different singletons have intersection size 0.

Conclusion: At most n sets; the n singleton sets attain the bound.

Review the idea: Pigeonhole principle · Parity · Combinations and binomial coefficients

A constant polynomial $P=c$ must satisfy $c=c^2$, so $P=0$ or $P=1$. Now suppose the degree is $d\ge1$ and the leading coefficient is $c\ne0$. The leading coefficients on the two sides are $c$ and $c^2$, giving $c=1$.

By the factorisation hypothesis, all $d$ roots, with multiplicity, are real. If one is nonzero, choose a root $r$ of largest absolute value $M=|r|>0$. Substituting $x=r+1$ in the identity makes $P(x-1)=P(r)=0$, so $r^2+2r$ is a root. Substituting $x=r-1$ similarly shows that $r^2-2r$ is a root.

One of these two numbers equals $r^2+2|r|$, which is positive and strictly larger than $|r|$. It is therefore a root with absolute value larger than $M$, a contradiction. Hence every root is 0. Together with leading coefficient 1 and the stated linear factorisation, this gives $P(x)=x^d$.

Finally, every $P(x)=x^d$, for an integer $d\ge0$, works because $(x^2-1)^d=(x-1)^d(x+1)^d$; the zero polynomial also works. No theorem about nonreal roots is needed, because the real factorisation was part of the question.

Conclusion: P=0 or P(x)=xᵈ for an integer d≥0.

Review the idea: Polynomial roots and multiplicity · Remainder and factor theorems

Binary bridge. XOR is performed independently in each binary column. Thus $a\mathbin{\oplus}a=0$, and a repeated term cancels. If the current XOR is $S$, changing one heap from $a$ to $b$ changes the XOR to $S\mathbin{\oplus}a\mathbin{\oplus}b$.

If $S=0$, any legal move has $b<a$, hence $a\ne b$. Their binary records differ, so $a\mathbin{\oplus}b\ne0$. Every move from a zero-XOR position therefore leads to a nonzero-XOR position.

If $S\ne0$, let its highest 1 occur in binary position $j$. At least one heap $a$ has a 1 in position $j$, because their digit sum there is odd. Set $b=a\mathbin{\oplus}S$. Above position $j$, $a$ and $b$ agree; at position $j$, the 1 in $a$ becomes 0. Therefore $0\le b<a$, so reducing this heap to $b$ is legal. The new XOR is $S\mathbin{\oplus}a\mathbin{\oplus}(a\mathbin{\oplus}S)=0$.

Every move reduces the total number of counters, so play ends. The all-zero position has XOR 0 and is losing. The two properties above, applied by induction on the total, prove that zero-XOR positions are exactly the losing ones.

For three heaps between 0 and 7, choose the first two sizes $a,b$ freely. There is exactly one losing choice of the third: $c=a\mathbin{\oplus}b$, which is again between 0 and 7. Thus there are $8^2=64$ losing ordered positions, including $(0,0,0)$.

Conclusion: Zero XOR characterises losing positions; there are 64 such ordered triples.

Review the idea: Number bases and digit problems · Parity · Strong induction