SAT · Competition-style problems · Part 1 of 3

The insight is to factor out the ratio scale and use parity of prime exponents. Write the integers as $7k$, $11k$, $13k$ with $k$ a positive integer. The product is 1001 times $k^3$, where 1001 equals 7 times 11 times 13 with three distinct primes. For a square every prime exponent must be even, but $k^3$ contributes exponents that are multiples of three, so 1 plus a multiple of three must be even, which forces each of the three primes to divide $k$ to an odd power. Hence $k$ is a multiple of 7 times 11 times 13, which is 1001. Writing $k$ as 1001 times $t$ makes the product a fourth power times $t^3$, so $t$ itself must be a square and the smallest choice is $t$ equals 1. The triple is 7007, 11011, 13013 with sum 31031.

The insight is to turn the continued proportion into a square condition and use lowest-terms divisibility. From $a:b=b:c$ we get $b^2$ equals $a$ times $c$, so with $b\mathrm{/}a$ equal to $p\mathrm{/}q$ in lowest terms and larger than 1, $q$ exceeds 1 because the ratio is not an integer. Then $a$ must contain $q^2$ as a factor, giving $a$ equals $k$ times $q^2$, $b$ equals $k$ times $p$ times $q$, and $c$ equals $k$ times $p^2$. The sum is $k$ times the quantity $q^2$ plus $p$ times $q$ plus $p^2$ and must be below 48. Checking coprime pairs gives 19 times $k$ for $3\mathrm{/}2$, 39 for $5\mathrm{/}2$, 37 for $4\mathrm{/}3$, and larger pairs exceed the bound. The largest attainable sum below 48 is 39 from 4, 10, 25.

The insight is to use relative speed for the meeting rate and modular periodicity for distinct points. Convert the $6:10$ ratio to $3:5$ after dividing by 2, so relative speed is 8 parts while Nadia contributes 3 parts. Meetings occur each time the combined distance covers one lap, giving 33 events including both endpoints for 32 laps combined, at times indexed by $n$ from 0 to 32. Nadia has covered $3n\mathrm{/}8$ laps at meeting $n$, so positions are multiples of $3\mathrm{/}8$ around the circle. Since 3 is invertible modulo 8, the sequence cycles every 8 meetings, producing 8 distinct points including the start.

The insight is to track a point rather than compare coefficients. A dilation centered at the origin sends $(x,y)$ to $(mx,my)$ for scale factor $m$. The point $(1,3)$ lies on $y=3x^2$, so its image $(m,3m)$ must lie on $y=27x^2$, giving $3m=27m^2$. Since $m$ is nonzero, $m=3\mathrm{/}27=1\mathrm{/}9$. Directly forming $27\mathrm{/}3$ inverts dilation with vertical stretch, and square roots confuse linear scaling with area scaling.

The insight is that legs scale as the square root of adjacent segments combined with the sum shortcut. Let the segments be $p<q$ with $q-p=5\sqrt{2}$ and altitude $h=6\sqrt{2}$. Similarity of the two small triangles to each other gives leg ratio $\sqrt{q\mathrm{/}p}$ with $pq=h^2=72$. Use $(q+p{)}^2=(q-p{)}^2+4pq=50+288=338=169\cdot 2$, so $q+p=13\sqrt{2}$. Hence $q=9\sqrt{2}$ and $p=4\sqrt{2}$, and the leg ratio is $\sqrt{9\mathrm{/}4}=3\mathrm{/}2$. Using the segment ratio directly omits the square root.

Expand by grouping conjugates: $(x^2+ax+b)(x^2-ax+c)=x^4+(b+c-a^2)x^2+a(c-b)x+bc$ because the $x^3$ terms $-a{x}^3$ and $+a{x}^3$ cancel. Matching with $x^4+8x^2+3x+72$ gives $b+c-a^2=8$, $a(c-b)=3$, and $bc=72$. The insight is symmetric cancellation plus divisor analysis: since $a$ is a positive integer dividing 3, $a$ is 1 or 3. If $a=1$ then $c-b=3$ and $b+c=9$, so $c=6$ and $b=3$, but $bc=18$, not 72. If $a=3$ then $c-b=1$ and $b+c=17$, so $c=9$ and $b=8$, and $bc=72$, which works. Hence the value is 3.

Subtract the equations: $x^2-y^2+2y-2x=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)=0$ by GCF and difference of squares. The insight is factored subtraction plus case analysis with Vieta sums: either $x=y$ or $x+y=2$. If $x=y$ then $x^2+2x-7=0$, whose roots sum to $-2$ without solving the radicals. If $x+y=2$ then $y=2-x$ gives $x^2-2x-3=0$, whose roots sum to 2. Adding both cases gives 0. Solving all four pairs explicitly with square roots is long, while Vieta gives each sum at once.

Factor as $(x-y)(x+y)=36$ and set $u=x-y$ and $v=x+y$, so $uv=36$ with $0<u<v$ because $x>y>0$ gives $y=(v-u)\mathrm{/}2>0$. The insight is difference-of-squares reframing plus parity filtering with divisor counting: since $x=(u+v)\mathrm{/}2$ and $y=(v-u)\mathrm{/}2$ are integers, $u$ and $v$ have the same parity, and since their product is even they must both be even. The $u<v$ factor pairs are $(1,36)$, $(2,18)$, $(3,12)$, and $(4,9)$, with $(6,6)$ excluded because it gives $y=0$. Only $(2,18)$ is both even, giving $(10,8)$, so there is 1 pair. Listing all pairs without parity overcounts, and allowing $y=0$ adds $(6,6)$.

Group the trinomial: $x^2+2xy+y^2=(x+y{)}^2$, so $P=(x+y{)}^2+y^2-4x-4y+7$. The insight is perfect-square factoring plus a nonnegativity bound: with $s=x+y$, $-4x-4y=-4s$, so $P=s^2-4s+7+y^2=(s-2{)}^2+3+y^2$. Since squares are nonnegative, $P\ge 3$, with equality at $y=0$ and $s=2$, namely $x=2$. Thus the least value is 3. Guessing integer pairs with a calculator can find 3 but cannot prove nothing is smaller, while the sum-of-squares form proves it at once.

Remainders are values, so $P(2)=80$ and the goal is $P(1\mathrm{/}2)$. Solving for $a$ and $b$ from one equation is impossible, so the insight avoids them entirely. Divide by $x^2$: $P(x)\mathrm{/}{x}^2=(x^2+1\mathrm{/}{x}^2)+a(x+1\mathrm{/}x)+b$, which depends only on $t=x+1\mathrm{/}x$ since $x^2+1\mathrm{/}{x}^2=t^2-2$. Because $2+1\mathrm{/}2$ equals $1\mathrm{/}2+2$, both $x=2$ and $x=1\mathrm{/}2$ give $t=5\mathrm{/}2$ and the same bracket $Q$. Hence $P(2)=4Q$ and $P(1\mathrm{/}2)=Q\mathrm{/}4$, a ratio of 16. Therefore $P(1\mathrm{/}2)=80\mathrm{/}16=5$. Trying to find $a$ and $b$ is a failing underdetermined route.

The insight is to subtract the two equations instead of substituting, which exposes a zero-product split. Subtracting gives $x^2-y^2=y-x$, so $(x-y)(x+y+1)=0$. Hence either $x=y$ or $x+y=-1$. If $x=y$ then $x^2=x+6$, so $(x-3)(x+2)=0$, giving $x=3,-2$. If $x+y=-1$ then $y=-1-x$ and $x^2=5-x$, so $x^2+x-5=0$, whose two roots sum to $-1$ by Vieta. Adding the distinct $x$ values gives $3-2-1=0$. A direct substitution would produce a quartic, which is why missing the subtraction leads to a long route.

The insight is the reciprocal symmetry $t=x+1\mathrm{/}x$, which turns a quartic into two quadratics. Since $(x+1\mathrm{/}x{)}^2=x^2+2+1\mathrm{/}{x}^2$, we have $x^2+1\mathrm{/}{x}^2=t^2-2$, so the equation becomes $t^2-2-5t+6=0$, or $(t-1)(t-4)=0$. Thus $t=1$ or 4. Now $x+1\mathrm{/}x=1$ gives $x^2-x+1=0$ with negative discriminant, so no real $x$. While $x+1\mathrm{/}x=4$ gives $x^2-4x+1=0$ with two distinct real roots whose sum is 4 by Vieta. Hence the sum of all real $x$ is 4. Multiplying by $x^2$ without the substitution leaves a quartic that is much longer to factor.

The insight is to trap the discriminant between consecutive squares so only finitely many $k$ can work. For integer roots the discriminant $k^2-4k-4$ must be a nonnegative perfect square. Completing the square gives $(k-2{)}^2=k^2-4k+4$, so the discriminant equals $(k-2{)}^2-8$. For $k>=7$, $(k-3{)}^2$ is below and $(k-2{)}^2$ is above the discriminant with no square strictly between, so no square. For $k<=-3$, writing $m=-k$ gives $(m+1{)}^2$ below and $(m+2{)}^2$ above, again no square. Hence only $-2<=k<=6$ need checking. Direct substitution shows the discriminant is $8,1,-4,-7,-8,-7,-4,1,8$, so only $k=-1$ with $x^2-x=0$ and $k=5$ with $x^2+5x+6=0$ have integer roots. Their sum is 4. Without the trap there are infinitely many $k$ to try.

The insight is to see the quartic as a quadratic in $y=x^2+2x$ so the given relation makes a factor vanish. Let $y=x^2+2x$; then $y^2=x^4+4x^3+4x^2$, so $f(x)=y^2+6y+9=(y+5)(y+1)+4$. For the given $z$, $z^2+2z=-5$, so $y=-5$ and the product $(y+5)(y+1)$ is zero, leaving $25-30+9=4$. Solving for $z=-1\pm 2i$ and raising each to the fourth power with complex arithmetic is long and error prone, while the substitution finishes without a calculator.

The insight is the angle-bisector proportion plus a weighted average. Here $AB=8$ and $BC=\sqrt{8^2+6^2}=10$, so the bisector meets $AC$ at $D$ with $AD\mathrm{/}DC=8\mathrm{/}10=4\mathrm{/}5$. Since $AC$ runs from $(0,0)$ to $(0,6)$, the point is $D=(0,8\mathrm{/}3)$ because $8\mathrm{/}3$ is $4\mathrm{/}9$ of the way from $A$ to $C$. The line through $B=(8,0)$ and $D=(0,8\mathrm{/}3)$ has slope $(8\mathrm{/}3-0)\mathrm{/}(0-8)=-1\mathrm{/}3$, so $y=(-1\mathrm{/}3)(x-8)$, namely $x+3y=8$. Measuring angles with inverse tangent for each candidate would be the long route.

The insight is to convert integrality of the interior angle into a divisibility condition on the exterior angle and then count divisors. A regular $n$-gon has exterior $360\mathrm{/}n$ and interior $180-360\mathrm{/}n$. Since 180 is an integer, the interior is an integer exactly when $360\mathrm{/}n$ is an integer, so $n$ must be a positive divisor of 360. Writing $360=2^3\cdot 3^2\cdot 5$ gives $(3+1)(2+1)(1+1)=24$ positive divisors. The values 1 and 2 do not form polygons, and $n\ge 3$ excludes them, leaving $24-2=22$ values.

The insight is that all three right triangles are similar, so perimeters scale like hypotenuses. Let the large perimeter be $P$ and legs $a,b$ with hypotenuse $c$. Then $36=P(b\mathrm{/}c)$ and $48=P(a\mathrm{/}c)$ because each small hypotenuse is a large leg. Squaring and adding with $a^2+b^2=c^2$ gives $36^2+48^2=P^2$, so $P=60$. Write $b=36c\mathrm{/}60$ and $a=48c\mathrm{/}60$; then $a+b+c=P$ gives $c(36+48+60)\mathrm{/}60=60$, so $c=3600\mathrm{/}144=25$. Indeed the large triangle is 15-20-25.

The insight is that powers of two grow so fast that the triangle inequality pins the top two exponents. With $r\le s\le t$, need $2^r+2^s>2^t$. Since $2^r\le 2^s$, the left side is at most $2^{s+1}$, so $2^t<2^{s+1}$ and $t\le s$, hence $t=s$. Then $2^r+2^s>2^s$ holds for every $r\ge 0$. So count pairs $0\le r\le s\le 5$: for $s=0,\dots ,5$ there are $1,2,\dots ,6$ choices of $r$, totaling $1+2+3+4+5+6=21$ noncongruent triangles.