ISEE Upper · Competition-style problems · Part 7 of 8

Since opposite sides of a parallelogram are parallel, the alternate interior angles cut off by transversal $AE$ are equal, so angle $BAE$ equals angle $DEA$. The bisector makes angle $BAE$ equal angle $DAE$, hence angle $DEA$ equals angle $DAE$, and the triangle $ADE$ is isosceles with $AD$ equal $DE$. With $DE:EC$ equal $2:1$, side $CD$ splits into three equal parts with $DE$ taking two, so $CD$ equals 1.5 times $AD$. Writing $AD$ as $x$ and $CD$ as $y$ gives $x$ equal $2y\mathrm{/}3$ and perimeter $2x+2y$ equal $5x$ equal 30, so $x$ equals 6.

The insight is an algebraic rewrite plus divisor counting. For a regular $n$-gon with $n$ at least 3, each interior angle is $(n-2)180\mathrm{/}n=180-360\mathrm{/}n$ degrees. For this to be a whole number, $360\mathrm{/}n$ must be a whole number, so $n$ must be a divisor of 360 exceeding 2. Since $360=8(9)(5)$ equals $2^3(3^2)(5)$, the number of positive divisors is $(3+1)(2+1)(1+1)=24$. Removing the inadmissible divisors 1 and 2 leaves 22 admissible side counts, each giving a distinct regular polygon.

The key move is to center the two binomial factors: let $m=n+5$, so $(n+1)(n+9)=(m-4)(m+4)=m^2-16$. If this equals $k^2$ with $k\mathrm{\ne }0$, then $m^2-k^2=16$, so $(m-k)(m+k)=16$. The two factors add to $2m$, hence they have the same parity, so both are even. Positive divisor pairs of 16 with $u\le v$ are $(1,16)$, $(2,8)$, $(4,4)$, and only $(2,8)$ and $(4,4)$ have matching parity, giving $m=5$ and $m=4$. The negative pairs give $m=-5$ and $m=-4$ similarly. Since $m=\pm 4$ makes the product 0, excluded by nonzero, only $m=\pm 5$ remain, so $n=m-5$ gives $n=0$ and $n=-10$, two values. Blind testing of $n$ never terminates.

The clever move is to complete the square inside the radical and turn a square-root condition into a factoring condition. Let $k^2=n^2+8n-32$ with $k$ an integer. Then $(n+4{)}^2-k^2=48$, so $(n+4-k)(n+4+k)=48$. Both factors are positive with the same parity since their sum is $2(n+4)$, and their product is even, so both are even. The even factor pairs of 48 are 2 by 24, 4 by 12, and 6 by 8, giving $n+4$ equal to 13, 8, and 7, so $n$ equals 9, 4, and 3 with radicands 121, 16, and 1. Odd factor pairs give half-integers and are discarded. Adding the three admissible values gives 16. Without completing the square, checking values one by one is long and gives no proof of completeness.

The insight is an invariant shift that turns the affine rule into pure halving, plus the positivity bound for powers of one-half. Subtract the fixed point 6, which satisfies $6=6\mathrm{/}2+3$. The first shifted value is 4, and each later shifted value is half the previous one. Hence the 100th shifted value is $4(1\mathrm{/}2{)}^{9}9$, which is a positive fraction, not zero. Therefore the 100th term equals 6 plus a positive amount, so it stays strictly above 6. Computing 100 steps directly would need denominators up to very large powers of two, which is hopeless without a calculator, while the shift makes the comparison immediate.

The insight is to combine the triple-product form with ratio cancellation rather than factoring each term with unknown $k$. Let the first shown term use $n=m$, so $180=km(m+1)(m+2)$ and $360=k(m+1)(m+2)(m+3)$. Dividing cancels $k(m+1)(m+2)$ and leaves $360\mathrm{/}180=2=(m+3)\mathrm{/}m$, so $2m=m+3$ and $m=3$. Then $k=180\mathrm{/}(3\cdot 4\cdot 5)=3$. The next $n$ is 5, giving $3\cdot 5\cdot 6\cdot 7=630$. Equivalently $360\cdot 7\mathrm{/}4=630$. Trying to factor 180 alone leaves many possibilities for $k$, so both numbers are needed.

Translate the words to algebra with $N=10t+u$ and digit sum $t+u$, so $10t+u=6(t+u)+7$, which simplifies to $4t-5u=7$. The key is divisibility with digit limits. From $4t=7+5u$, the right side must be a multiple of 4, so $7+5u$ leaves remainder 0 on division by 4, which forces $u$ to leave remainder 1 on division by 4. Since $u$ is 0 to 9 and $t$ is 1 to 9, only $u=1$ with $t=3$ and $u=5$ with $t=8$ satisfy the equation. Both check out as $31=6(4)+7$ and $85=6(13)+7$. One admissible tens digit is below 5 and the other is above 5, so the order changes with the case.

The key is that each corner angle is a right angle in the circle, so each diagonal of the rectangle must be a diameter of length 10. Writing the side lengths as $l$ and $w$ gives $l^2+w^2=10^2=100$. A square fits this condition, with side $\sqrt{50}=5\sqrt{2}$ and perimeter $20\sqrt{2}$. Comparing $20\sqrt{2}$ with 28 by dividing by 4 and squaring gives $25\cdot 2=50$ versus 49, so the square already exceeds 28. Since the greatest possible perimeter is at least the square value, it must exceed 28, and indeed $(l-w{)}^2\ge 0$ shows the square is the maximum.

The insight is that a 90-degree side is the vector $AB$ rotated: swap the differences and negate one, giving equal length and perpendicularity at once, and opposite sides are parallel translates. Here $B-A=(4,3)$. The two perpendicular vectors of the same length are $(-3,4)$ and $(3,-4)$. Adding to $A$ gives candidate $D=(-1,6)$ or $D=(5,-2)$, with corresponding $C=(3,9)$ or $C=(9,1)$. The second square has a vertex with $y=-2$, violating positivity, while the first has $y$-values 2,5,9,6, all positive. Hence the square above the $x$-axis has $D=(-1,6)$, so its $x$-coordinate is $-1$.

The move is to collapse each product to a difference of squares and then use squares modulo 4: $(a+3)(a-3)=a^2-9$ and $(b+2)(b-2)=b^2-4$, so the whole expression equals $a^2-b^2-5$. Every integer square is 0 or 1 modulo 4, so $a^2-b^2$ is 0, 1, or 3 modulo 4 and never 2. Thus the expression is never 1 modulo 4. Since $13+5=18$ is 2 modulo 4, writing $a^2-b^2=18$ as $(a-b)(a+b)=18$ would need two same-parity factors of 18, which do not exist. The other values do occur: $a=7$, $b=6$ gives $40-32=8$; $a=8$, $b=7$ gives $55-45=10$; $a=5$, $b=3$ gives $16-5=11$.

Translation gives $P=x(x-3)=x^2-3x$. Compare with $-2$ by looking at $P+2=x^2-3x+2$. Factor rather than guess: $x^2-3x+2=(x-1)(x-2)$, so $P=-2$ when $x=1$ or $x=2$. For instance $x=1$ gives $P=1(-2)=-2$, an exact match. But $x=3$ gives $P=3(0)=0$, which is greater than $-2$, while $x=1.5$ gives $P=1.5(-1.5)=-2.25$, which is less than $-2$. Since allowed inputs produce both a greater result and a smaller result, the order is not fixed.

The insight is that reflection across $y=x$ swaps coordinates, so the two slopes are reciprocals, and a positive number plus its reciprocal exceeds 2 by a square. The point is $(c,d)$ and its reflection is $(d,c)$, so the slopes from the origin are $d\mathrm{/}c$ and $c\mathrm{/}d$. Their sum is $(c^2+d^2)\mathrm{/}cd$, which equals $2+(c-d{)}^2\mathrm{/}cd$. Since $c>0$ and $d>0$, the denominator is positive, and since $c\mathrm{\ne }d$, the numerator $(c-d{)}^2$ is strictly positive, so the sum is strictly greater than 2. The first quantity must therefore be larger, and checking one numeric pair cannot establish this for all admissible pairs.

The insight is an extremal bound for an arbitrary number of sides combined with a construction showing the bound is sharp. Suppose $n$ is the number of sides and $k$ angles are less than 120 degrees. The interior sum is $(n-2)180$ degrees. Each of the $k$ small angles is below 120 and each remaining angle in a convex polygon is below 180, so $(n-2)180$ is less than $120k+180(n-k)$, which equals $180n-60k$. Cancelling $180n$ gives 360 greater than $60k$, so $k$ is less than 6 and at most 5. The bound occurs: a regular pentagon has five angles of 108 degrees, each below 120 degrees, so 5 is attainable and greatest.

The hidden move is completing the square to force both variables to single values. Group as $(x^2-8x+16)+(y^2+4y+4)=-20+16+4$, so $(x-4{)}^2+(y+2{)}^2=0$. A square of a real number is never negative, so a sum of two squares can be zero only when each square is zero, giving $x=4$ and $y=-2$. Then $3x-2y=12+4=16$. Trying to solve one equation in two unknowns directly looks impossible, but the sum of squares bound pins everything, so the two quantities are equal.

The idea is Venn decomposition combined with the geometric mean condition and cancellation of the quadratic term. Let only art be x, both be y, and only music be z. Then x plus y equals 45 and y plus z equals 30, so x equals 45 minus y and z equals 30 minus y. Geometric order means y squared equals x times z, so y squared equals the product of 45 minus y and 30 minus y, which expands to 1350 minus 75 y plus y squared. The squared term appears on both sides and cancels, leaving the linear equation 75 y equals 1350, so y equals 18. Then only music is 30 minus 18, which is 12, and the triple 27, 18, 12 has common ratio two thirds. Solving without the cancellation looks like a quadratic system.

The insight is symmetry plus isosceles angle chasing: the segment joining opposite vertices passes through the center and its midpoint is equidistant from all vertices. Call opposite vertices $P$ and $Q$ with center $M$ and third vertex $R$, so $MP$, $MQ$, and $MR$ are equal radii. Then triangles $MPR$ and $MQR$ are isosceles, so write their base angles as $a$ and $b$ and the whole triangle has angles $a$, $b$, and $a+b$, forcing $2\ast a+2\ast b=180$ and hence $a+b=90$, so every such triangle is right. A regular octagon has 4 opposite pairs, each leaving 6 choices for the third vertex, giving $4\ast 6=24$ right triangles out of $8\ast 7\ast 6\mathrm{/}6=56$ triples, so $24\mathrm{/}56=3\mathrm{/}7$.

The idea is to subtract first and factor what remains instead of plugging in decimals. The difference is $Q(x)-P(x)=-x^4-x^3+x^2+x=x(-x^3-x^2+x+1)=x(1-x)(x+1{)}^2$ by grouping the cubic as $-x^2(x+1)+1(x+1)$ and using $1-x^2=(1-x)(1+x)$. For $0<x<1$, the factor $x$ is positive and $1-x$ is positive, while $x+1$ exceeds 1 so $(x+1{)}^2$ is positive, and a product of positives is positive. Thus $Q(x)-P(x)$ is always positive, so $Q(x)$ is always larger than $P(x)$ on this interval. Plugging in a decimal like 0.37 would require fourth powers, which is long, while the factored form settles the order at once.

The work is bounding the squares and listing the few integer pairs. Since $a^2\le 65$ and $b^2\le 65$, $\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }\le 8$, so only squares 0,1,4,9,16,25,36,49,64 matter. Pairs summing to 65 are $1+64$ and $16+49$, so up to order and signs $\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid }$ are 1,8 or 4,7. Hence $ab$ is one of $-28,-8,8,28$, and each is below 30. For example $4\cdot 7=28<30$ and $1\cdot 8=8<30$. Without the integer restriction a real pair with $a=b=\sqrt{32.5}$ gives $ab=32.5>30$, so the integer condition is what keeps everything below 30. Thus Column B is greater.

The insight is that integer points on a segment are governed by the primitive direction step, not by length, so the longer segment can hold fewer lattice points. For the horizontal segment, $y=0$ and $x$ runs through $0,1,\dots ,20$, giving 21 integer points. For the slanted segment, the differences are 15 in $x$ and 20 in $y$, which reduce by 5 to the primitive step $(3,4)$, so integer points occur at every fifth of the way along, giving the 6 points $(1,1)$, $(4,5)$, $(7,9)$, $(10,13)$, $(13,17)$, and $(16,21)$. Hence the slanted segment is longer by the 15-20-25 triangle yet holds fewer integer points, so the second quantity is larger.

The idea is elimination to manufacture the requested bundle directly rather than solving for each individual price. Twice the first purchase contains 4 pens, 6 notebooks, and 4 erasers for 44 dollars. Subtracting the second purchase removes 3 pens, 5 notebooks, and 3 erasers priced at 34 dollars, leaving exactly 1 pen plus 1 notebook plus 1 eraser. Therefore the bundle price equals 44 minus 34, which is 10 dollars, matching the stated amount exactly. Two purchases cannot fix three separate prices, but this particular combination is fully determined, and the premises allow ordinary positive prices such as 4 dollars, 2 dollars, and 4 dollars.

Keep $\pi$ symbolic instead of rounding early. The diameter is circumference over $\pi$, so $d=10\mathrm{/}\pi$. A square whose diagonal is $d$ has area $d^2\mathrm{/}2$, since two such triangles make the square, so the base is $50\mathrm{/}{\pi }^2$ and the box volume is $500\mathrm{/}{\pi }^2$. Against 50 this is 10 against ${\pi }^2$. Since $\pi$ is about 3.14 and below 3.15, its square is below 9.9225, because $315^2=99225$, hence below 10. A smaller denominator makes a larger fraction, so the box number exceeds 50. Computing the side with 3.14 and the square root of 2 digit by digit is long and error prone, while the symbolic cancellation finishes in one bound.

The trick is to work modulo 3 rather than with values. Residues 0, 1, 2 each occur three times among 1 through 9. A triple sums to 0 modulo 3 in four patterns: $0+0+0$, $1+1+1$, $2+2+2$, or $0+1+2$. The first three contribute 1 triple each. The mixed pattern contributes $3\cdot 3\cdot 3=27$ triples. So 30 of the 84 unordered triples work, giving 30 over 84, which reduces to 5 over 14. Assuming one-third by symmetry misses the without-replacement dependence.

The key is to look at prime exponents modulo 2. A product is a square exactly when every prime occurs an even number of times, so only the parity vector matters. Write each number 1 through 12 by its squarefree kernel: 1,4,9 share kernel 1; 2,8 share kernel 2; 3,12 share kernel 3; the rest are singletons. Two different numbers multiply to a square exactly when they share a kernel. Hence favorable unordered pairs are $3+1+1=5$ from groups of sizes 3, 2, 2. With 66 total unordered pairs, the probability is 5 over 66.

The idea is to factor out the common square and then split on the sign of $k$ using a small square root bound. Since $f(k)=k^3+2k^2=k^2(k+2)$ and $k^2=5$, $f(k)=5(k+2)$. The equation $k^2=5$ has two admissible numbers, $k=\sqrt{5}$ and $k=-\sqrt{5}$. For $k=\sqrt{5}$, $k+2>0$, so $f(k)=5(\sqrt{5}+2)>0$. For $k=-\sqrt{5}$, $k+2=2-\sqrt{5}<0$ because $5>4$ gives $\sqrt{5}>2$, so $f(k)<0$. Concretely $f$ is positive in the first case and negative in the second, giving opposite orders against 0, so the relationship cannot be determined.

The insight is that slope is rise over run, so the comparison reduces to whether the run is below or above 1, and the bounds force the run to straddle 1. Here the rise is $9-5=4$ and the run is $b-a$, so the slope is $4\mathrm{/}(b-a)$ with $0<b-a<2$. For $a=0.2$ and $b=0.5$, the run is 0.3 and the slope is about 13.3, which exceeds 4. For $a=0.2$ and $b=1.7$, the run is 1.5 and the slope is about 2.7, which is below 4. Both pairs satisfy $0<a<b<2$, so different admissible pairs give opposite orders and no fixed relationship follows.

The key insight is to write the integer progression with lowest-terms ratio and bound the middle instead of trying every middle. Let the terms be $a,b,c$ with $b^2=ac$ and $a+c=38-b$. Since $(a-c{)}^2$ cannot be negative, $(a+c{)}^2$ is at least $4ac$, so $(38-b{)}^2$ is at least $4b^2$, giving $b$ at most 12. Write the ratio as $p\mathrm{/}q$ in lowest terms, so the terms are $k$ times $q^2,pq,p^2$ and the sum is $k$ times $(q^2+pq+p^2)=38$. Since that factor is at least 7, it must be 19 with $k=2$, and $4+6+9=19$ forces $q=2,p=3$. The terms are 8,12,18, so the middle is 12.

The insight combines four-quadrant symmetry with summing a triangular progression layer by layer. For fixed k at least 1, |x|+|y|=k has 4k integer pairs, since the magnitude patterns distribute k between the coordinates and the signs supply four choices, while k=0 gives only (0,0). Hence the number with sum at most 12 is 1+4(1+...+12). Since 1+...+12=78, this is 1+312=313. For instance k=1 contributes the 4 pairs (1,0), (-1,0), (0,1), (0,-1). After seeing the layers the computation is one short multiplication, so the two quantities are equal.

The insight is to pair consecutive squares as differences of squares and then sum the resulting arithmetic progression. For each k, (2k-1)^2-(2k)^2=(4k^2-4k+1)-4k^2=1-4k. With k=1 to 50 this gives -3,-7, continuing to -199. Their sum is -(4(1+...+50)-50). Since 1+...+50=1275, the total is -(5100-50)=-5050. The right quantity is -5100, and -5050 is greater because it is less negative. For instance the first pair alone is -3. After the pairing the remaining work is one triangular sum, so Column A is greater.

The key is elimination to one integer unknown plus a bounding argument. Subtracting $f(2)-f(1)$ gives $3a+b=5$, so $b=5-3a$, and then $c=5+2a$ from $f(1)=10$. Hence $f(3)=20+2a$ and $f(4)=25+6a$. The condition $34<f(4)<42$ becomes $34<25+6a<42$, or $9<6a<17$. With $a$ a positive integer, only $a=2$ fits, giving $f(4)=37$. Therefore $f(3)=20+4=24$, with no need to solve the full three-variable system directly.

For any pentagon the interior angles add to $(5-2)$ times 180, which is 540 degrees. Write the increasing progression symmetrically about its middle term $a$ as $a-2d$, $a-d$, $a$, $a+d$ and $a+2d$. Their sum is $5a$, so $5a$ equals 540 and $a$ equals 108. Convexity forces every angle strictly below 180, and the strongest restriction comes from the largest angle, $108+2d$ below 180, giving $2d$ below 72 and $d$ below 36. Positivity gives only the weaker bound $d$ below 54. Including the regular case $d$ equals 0, the difference is always strictly below 36.

The insight is to work with the exterior angle. For a regular $n$-gon the interior is $180-360\mathrm{/}n$, so an integer interior forces $d=360\mathrm{/}n$ to be an integer divisor of 360. The condition says $180-d$ is 5 more than a multiple of 9, so $d$ is 4 more than a multiple of 9. Hence $d$ is not divisible by 3, so with $360=2^3\cdot 3^2\cdot 5$ the factor $3^2$ cannot divide $d$ and $d$ must divide 40. The divisors of 40 are 1,2,4,5,8,10,20,40, and only 4 and 40 leave remainder 4 upon division by 9. They give $n=90$ and $n=9$, whose sum is 99.

The insight is a counting re-frame plus subtraction of degenerate triples. Any three lines with no parallel pair and no common point enclose exactly one triangle, and every triangle arises this way, so triangles correspond to triples of lines with no parallel pair. There are $8(7)(6)\mathrm{/}6=56$ triples in all. A triple fails exactly when it contains at least two of the three parallels: choosing two parallels and any other line gives $3(5)=15$ triples, plus the triple of all three parallels gives 1, for 16 degenerate triples. Hence the number of triangles is $56-16=40$, which also equals $10+3(10)$ from triples using none or one parallel.

The insight is that a parallel line inherits the reduced rise over run, so its integer points form an arithmetic progression, and the divisibility condition becomes a congruence on the step index. Slope of $M$ is $(6-1)\mathrm{/}(7-0)=5\mathrm{/}7$, already reduced, so $L$ has slope $5\mathrm{/}7$ through $(4,9)$: $y-9=(5\mathrm{/}7)(x-4)$. Hence $5(x-4)$ is a multiple of 7, and since 5 is coprime to 7, $x-4$ is a multiple of 7. Write $x=4+7t$ and $y=9+5t$ with integer $t$. From $0<=x<=700$, $t=0,\mathrm{.}\mathrm{.}\mathrm{.},99$, so there are 100 lattice points. Now $y$ divisible by 4 means $9+5t$ is 0 mod 4, that is $1+t=0$ mod 4, so $t=3$ mod 4. The values $3,7,\mathrm{.}\mathrm{.}\mathrm{.},99$ are $((99-3)\mathrm{/}4)+1=25$ points, computed without listing all 100 candidates.

The insight is that horizontal cross sections of a cone are similar, so volume up to height $h$ grows as $h^3$, combined with differencing a frustum as whole minus inner cone. Let the radius at height $h$ above the tip be proportional to $h$. Then volume up to $h$ equals $(1\mathrm{/}3)\pi (kh{)}^{2}h=C{h}^3$ for a constant $C$. The bottom inch is $C(1^3)=C$. The cone up to 9 has volume $C(9^3)=729C$ and up to 8 has $C(8^3)=512C$, so the top inch is $729C-512C=217C$. The ratio is $217C\mathrm{/}C=217$. A student who scales linearly or by area never isolates the cubic growth.