Past Exam Questions

Key idea. Count how many moves the game must contain.

1. There are \(10+15+20=45\) stones and initially three piles.

2. Splitting one pile into two increases the number of piles by exactly 1. It does not change the number of stones.

3. The game ends exactly when every pile has one stone. If any pile had more than one stone, it could still be split.

4. The final number of piles is therefore 45. Going from 3 piles to 45 takes exactly \(45-3=42\) moves, regardless of how the splits are chosen.

5. The first player makes the odd-numbered moves; the second makes the even-numbered moves. Move 42 is made by the second player, leaving the first unable to move.

Remember. Sometimes the number of moves is fixed, so no clever move can change the winner.

Key idea. The first move can create two identical halves with a gap between them.

1. With 20 plates, the first player takes the bananas on plates 10 and 11, the two middle plates. Two identical rows of nine bananas remain.

2. With 21 plates, the first player takes the banana on plate 11. Two identical rows of ten bananas remain.

3. In either case, after every opponent move, the first player makes the reflected move in the other half. A move on one plate is copied on its mirror plate; a move on two adjacent plates is copied on their two mirror plates.

4. This is always legal: before the opponent moves, both halves look identical, so the mirror bananas are still there. The empty middle plate or plates prevent any move from crossing between the halves.

5. The first player can therefore reply whenever the opponent can move. Since each move removes bananas, the game finishes, and the first player must take the last banana.

Remember. To prove a mirror strategy, explain why the reply stays available and cannot overlap the opponent’s move.

Key idea. Move the right side to the left and recognise a square.

1. The difference between the two sides is

   \[a^2+b^2-2ab=(a-b)^2.\]

2. A real number squared is always at least zero. Therefore the left side of the proposed inequality is at least its right side.

3. Equality occurs exactly when \((a-b)^2=0\), which means \(a=b\).

Remember. To prove one expression is at least another, try to write their difference as a square.

Key idea. Complete one square in each variable.

1. For the x terms, \(x^2-8x=(x-4)^2-16\). For the y terms, \(y^2+2y=(y+1)^2-1\).

2. Substitute these expressions and collect constants:

   \[x^2-8x+y^2+2y+20=(x-4)^2+(y+1)^2+3.\]

3. The squares are non-negative, so the expression is at least 3.

4. We can make both squares zero simultaneously by choosing \(x=4\), \(y=-1\). The value 3 is therefore actually reached.

Remember. Finding a lower bound is only half an optimisation proof: also show that it can be attained.

Key idea. A fixed sum and a non-negative square give a bound on the product.

1. The identity \((x-y)^2=(x+y)^2-4xy\) gives

   \[0\le(x-y)^2=100-4xy.\]

2. Rearranging, \(4xy\le100\), so \(xy\le25\).

3. The bound is attained when \(x-y=0\). Combining \(x=y\) with \(x+y=10\) gives \(x=y=5\), which are allowed positive numbers.

Remember. For two positive numbers with fixed sum, the product is largest when the numbers are equal.

Key idea. Add the squares of all three pairwise differences.

1. Expand and add:

   \[\begin{aligned}(a-b)^2+(b-c)^2+(c-a)^2\\=2a^2+2b^2+2c^2-2ab-2bc-2ca.\end{aligned}\]

2. Divide by 2 and rearrange:

   \[a^2+b^2+c^2-ab-bc-ca=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}.\]

3. The right side is non-negative, proving the inequality.

4. Equality requires every square to be zero. Thus \(a=b\), \(b=c\), and hence \(a=b=c\). Conversely, equal numbers do give equality.

Remember. Several differences often suggest a sum of several squares.

Key idea. Turn each variable’s terms into a constant minus a square.

1. Complete the squares separately:

   \[\begin{aligned}20x-2x^2&=50-2(x-5)^2,\\-4y-4y^2&=1-4\left(y+\frac12\right)^2,\\6z-3z^2&=3-3(z-1)^2.\end{aligned}\]

2. After including the final \(-2\), the whole expression is

   \[52-2(x-5)^2-4\left(y+\frac12\right)^2-3(z-1)^2.\]

3. Each subtracted square is non-negative, so the expression cannot exceed 52.

4. Set \(x=5\), \(y=-1/2\), \(z=1\). All three squares vanish, and the value is 52. Hence 52 is the greatest possible value.

Remember. A maximum can be proved by writing the expression as a constant minus non-negative terms.

Key idea. Subtract a likely lower bound and express the result as a square.

1. For \(x>0\),

   \[x+\frac1x-2=\frac{x^2-2x+1}{x}=\frac{(x-1)^2}{x}\ge0.\]

2. The numerator is non-negative and the denominator is positive. Thus \(x+1/x\ge2\).

3. Equality holds when \((x-1)^2=0\), namely at \(x=1\). This allowed value gives 2, so the lower bound is attained.

Remember. A minimum proof needs a bound and a value that attains it.

Key idea. Put the difference over one positive denominator.

1. The difference between the proposed left and right sides is

   \[\frac{x^2}{y}+\frac{y^2}{x}-x-y=\frac{x^3+y^3-x^2y-xy^2}{xy}.\]

2. Factor the numerator:

   \[\begin{aligned}x^3+y^3-xy(x+y)&=(x+y)(x^2-xy+y^2-xy)\\&=(x+y)(x-y)^2.\end{aligned}\]

3. Therefore the difference is

   \[\frac{(x+y)(x-y)^2}{xy}\ge0,\]

   because \(x+y>0\), \(xy>0\), and the square is non-negative.

4. Equality holds exactly when \(x-y=0\), or \(x=y\).

Remember. Factoring the numerator of a difference can reveal the entire inequality.

Key idea. A common denominator reveals a weighted square.

1. Subtract the right side and use the positive common denominator \(ab(a+b)\):

   \[\frac{x^2}{a}+\frac{y^2}{b}-\frac{(x+y)^2}{a+b}=\frac{b(a+b)x^2+a(a+b)y^2-ab(x+y)^2}{ab(a+b)}.\]

2. Expand the numerator. The terms \(abx^2\) and \(aby^2\) cancel, leaving

   \[b^2x^2+a^2y^2-2abxy=(bx-ay)^2.\]

3. We have proved the identity

   \[\frac{x^2}{a}+\frac{y^2}{b}-\frac{(x+y)^2}{a+b}=\frac{(bx-ay)^2}{ab(a+b)}\ge0.\]

4. The denominator is positive, so equality holds exactly when \(bx-ay=0\), equivalently \(x/a=y/b\). No positivity condition on x or y was needed.

Remember. This two-term inequality can be applied repeatedly to handle three or more fractions.

Key idea. Use the fixed sum to rewrite everything in terms of the product xy.

1. Since \(x+y=1\), we have \(1-x=y\) and \(1-y=x\). Therefore

   \[1-x^2=y(1+x),\qquad1-y^2=x(1+y).\]

2. Substitute into the expression and cancel positive factors:

   \[\begin{aligned}\left(\frac1{x^2}-1\right)\left(\frac1{y^2}-1\right)&=\frac{(1-x^2)(1-y^2)}{x^2y^2}\\&=\frac{(1+x)(1+y)}{xy}\\&=\frac{2+xy}{xy}=1+\frac2{xy}.\end{aligned}\]

3. Now \((x-y)^2\ge0\) implies \(4xy\le(x+y)^2=1\). Thus \(0<xy\le1/4\), so \(1/(xy)\ge4\).

4. It follows that \(1+2/(xy)\ge1+8=9\). Equality occurs exactly when \(x=y\), which with \(x+y=1\) gives \(x=y=1/2\).

Remember. With a fixed sum, a symmetric expression may simplify to a function of the product.

Key idea. Apply C4.3 twice, then use a sum of squared differences.

1. Let \(s=a+b+c\) and \(q=ab+bc+ca>0\). Rewrite the target left side as

   \[L=\frac{a^2}{a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)}.\]

2. Put \(D=a(b+c)+b(c+a)>0\). Applying C4.3 to the first two fractions gives

   \[L\ge\frac{(a+b)^2}{D}+\frac{c^2}{c(a+b)}.\]

3. Apply C4.3 once more, now to these two fractions. The sum of the denominators is \(D+c(a+b)=2q\), so

   \[L\ge\frac{(a+b+c)^2}{D+c(a+b)}=\frac{s^2}{2q}.\]

4. Finally,

   \[\begin{aligned}s^2-3q&=a^2+b^2+c^2-ab-bc-ca\\&=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\ge0.\end{aligned}\]

   Thus \(s^2\ge3q\), and division by \(2q>0\) gives \(L\ge3/2\).

5. If \(a=b=c\), each original fraction equals \(1/2\), so equality holds. Conversely, equality in the final bound forces all three squared differences to vanish, so \(a=b=c\) is necessary.

Remember. A proof can combine a fraction inequality with a separate bound on a symmetric expression.