SAT · Competition-style problems · Part 3 of 3

Let the roots be $r$ and $s$. The insight is to shift the condition greater than 2 into Vieta without using the quadratic formula. Write $u=r-2$ and $v=s-2$, so $u$ and $v$ are positive and distinct. Then $uv=(r-2)(s-2)=rs-2(r+s)+4=16-2k+4=20-2k$, so positivity gives $20-2k>0$ and hence $k<10$. Also $u+v=k-4>0$ gives $k>4$. Distinct real roots need discriminant $k^2-64>0$, so $\mathrm{\mid }k\mathrm{\mid }>8$. With $k$ an integer satisfying $4<k<10$ and $\mathrm{\mid }k\mathrm{\mid }>8$, the only possibility is $k=9$, and indeed $x^2-9x+16=0$ has roots about 2.44 and 6.56.

The insight is to take the square root of the area ratio to get the linear scale and then enforce integer divisibility with distinctness. Similar figures scale areas by the square of the linear factor, so $9:16$ gives linear $3:4$. Let the smaller sides be $a$ and $b$, so the larger are $4a\mathrm{/}3$ and $4b\mathrm{/}3$, which forces $a$ and $b$ to be multiples of 3. Write $a$ as $3m$ and $b$ as $3n$ so the four sides are $3m$, $3n$, $4m$, $4n$. Distinctness rules out $m$ equals $n$, since 3,3,4,4 repeats. The smallest distinct pair is $m$ equals 1 and $n$ equals 2, giving sides 3, 6, 4, 8. The perimeters are 18 and 24 for a total of 42.

The insight is that iteration creates a short cycle, so a huge iterate is determined by a remainder. Compute directly: $f(2)=1\mathrm{/}(1-2)=-1$, then $f^2(2)=f(-1)=1\mathrm{/}(1-(-1))=1\mathrm{/}2$, then $f^3(2)=f(1\mathrm{/}2)=1\mathrm{/}(1-1\mathrm{/}2)=2$, back to the start. Hence applying $f$ three times returns any admissible starting value to itself, so the sequence of iterates has period 3. Since $2027=2025+2$ and 2025 is divisible by 3, the 2027th iterate equals the second one, $1\mathrm{/}2$. Trying to apply $f$ 2027 times is hopeless, while the cycle plus division with remainder finishes in seconds without a calculator.

The insight is to straighten the belt into tangent segments plus circular arcs. Each radius to a belt contact is perpendicular to the belt, so each straight piece with the two radii forms a rectangle, making its length equal to the corresponding center distance 10. The three straight pieces total 30. The three contact arcs turn with the triangle, and the exterior angles of any convex polygon sum to 360 degrees, so the three arcs reassemble into one full circle of radius 2 with circumference $4\pi$. Adding gives $30+4\pi$. Trying to find each arc separately leaves unknown tangent points.

The insight is to count by apex using equal arcs. Equally spaced points cut the circle into equal arcs, so symmetric points about the diameter through a vertex give equal chords and hence an isosceles triangle. Fix one vertex as the apex. Of the other 7 points, the opposite point cannot pair with the apex to make equal sides, while the remaining 6 form 3 symmetric pairs, giving 3 isosceles triangles with that apex. With 8 choices of apex this gives 24, and no triangle is counted twice because with 8 points none has two apices. Listing all 56 triples is long.

The key insight is midpoint doubling combined with a divisibility step count. Since $M=(29,47)$ is the midpoint, the missing endpoint satisfies $(5+Bx)\mathrm{/}2=29$ and $(11+By)\mathrm{/}2=47$, so the other endpoint is $(53,83)$. The run is 48 and the rise is 72. Lattice points divide the segment into equal integer steps, so the step vector must divide both 48 and 72. The greatest common divisor is 24, giving minimal steps of $(2,3)$. Starting at $t=0$ through $t=24$ yields $24+1=25$ integer points. Listing every point would be long, but the divisor makes it one division.

The insight is that the median forces a straddle while the mode forces a duplicate, and the total then leaves only two duplicate layouts. Five numbers averaging 83 total 415, so with $70+80+90=240$ the two added $a\le b$ sum to 175. To keep median 80 with $70<80<90$, the added pair must straddle 80, i.e. $a\le 80\le b$, otherwise the middle shifts away from 80. With $a+b=175$ and straddle, many pairs work, e.g. 75 and 100 with all distinct and no mode. Requiring a unique mode forces a duplicate: either $a=80,b=95$ giving 70,80,80,90,95 with mode 80, or $a=70,b=105$ giving 70,70,80,90,105 with mode 70. Both satisfy mean 83 and median 80, so the greatest possible added value is 105.

The key combines a weighted-average balance with common-multiple minimality under the bound. Let the fraction of girls in January be $r$. Then $0.25r+0.10(1-r)=0.16$, so $0.15r=0.06$ and $r=2\mathrm{/}5$. Hence January girls are $2T\mathrm{/}5$ and boys $3T\mathrm{/}5$ for total $T$, so $T$ is a multiple of 5. February girls are $5T\mathrm{/}10=T\mathrm{/}2$ and boys $33T\mathrm{/}50$, so $T$ must make $T\mathrm{/}2$ and $33T\mathrm{/}50$ whole, forcing $T$ to be a multiple of 50. With $T$ fewer than 60, the only possibility is 50, which indeed gives 20 girls to 25 and 30 boys to 33. A student who checks only one integer condition is pushed into testing multiples one by one.

The insight is to recut the perimeter with the triangle inequality for the top and with averaging for the bottom, then exhibit each value. Call the sides $x\le y\le L$. Since $x+y>L$, the perimeter $x+y+L$ exceeds $2L$, so $2L<20$ from the perimeter cap, giving $L<10$ and hence $L\le 9$ for integers. Since $x$ and $y$ are each at most $L$, the perimeter is at most $3L$, and the lower cap gives $18\le 3L$, so $L\ge 6$. Each end occurs: 6,6,6 gives 6 with perimeter 18, 6,7,7 gives 7, 6,6,8 gives 8, and 5,6,9 gives 9, each a genuine triangle with perimeter 18 to 20, so every integer 6 through 9 occurs.

Think of distance as a multiple of each circumference and handle the offset as a congruence. Let the distance be $D$. The 6-inch wheel needs $D\mathrm{/}(6\pi )$ to be a whole number, while the 8-inch wheel starts half a turn away, so $D\mathrm{/}(8\pi )$ must be a half-integer. Hence $D\mathrm{/}\pi$ is a multiple of 6 and 4 more than a multiple of 8: $6n=8m+4$, or $3n=4m+2$. The smallest positive solution is $n=2$ with $m=1$, giving $D=12\pi$. Anyone ignoring the half turn takes the ordinary least common multiple $24\pi$ and rolls too far.

The key is that area shrinks by the square of the linear denominator, and both plan areas describe the same unknown true area. Let $T$ be the true area in square centimeters. Plan A area is $T\mathrm{/}20^2=T\mathrm{/}400$ and Plan B area is $T\mathrm{/}30^2=T\mathrm{/}900$, so $T\mathrm{/}400-T\mathrm{/}900=20$. Since $1\mathrm{/}400-1\mathrm{/}900=5\mathrm{/}3600=1\mathrm{/}720$, we get $T\mathrm{/}720=20$, so $T=14400$ square centimeters. Because 1 square meter equals 10000 square centimeters, the true area is $14400\mathrm{/}10000=1.44$ square meters. Seeing the squared scale turns a messy comparison into one linear equation.

The insight is to maximize the total allowed by a given largest value and then build a list reaching the required total. The total must be $7\times 15=105$ with the fourth value 14. For largest value $M$, the greatest distinct list uses $11,12,13$ below the median and $M-2,M-1,M$ above it, giving max total $36+14+3M-3=47+3M$. Needing $47+3M\ge 105$ gives $3M\ge 58$, so $M\ge 20$ since 19 gives only 104. The bound is attainable with $11,12,13,14,17,18,20$, which are distinct, have median 14, and sum to 105, so the smallest possible largest value is 20.

The key insight is to use the red plus green total as an invariant multiple plus divisibility elimination with the original total interval. Originally red is $2k$, blue is $3k$, green is $4k$, so the total is $9k$ and red plus green is $6k$. After adding 60 with no blue, red plus green is $6k+60$ and must be a multiple of 11 from the $4:7$ split. Since $9k$ lies strictly between 100 and 120, $k$ is 12 or 13. Only $k=12$ makes $6k+60=132$ a multiple of 11, so blue is 36.

The insight is to combine solving a linear system by addition with a discount-threshold case split. If both totals were discounted, the pre-discount totals would be 100 and 125, since $80\mathrm{/}0.8=100$ and $100\mathrm{/}0.8=125$. Then $3N+2P=100$ and $2N+3P=125$ add to $5(N+P)=225$, so $N+P=45$, and subtracting gives $N-P=-25$, so $N=10$ and $P=35$. This satisfies the threshold, with pre-totals 100 and 125 both at least 100. The alternative with the first purchase undiscounted gives $3N+2P=80$ and $2N+3P=125$, yielding $N=-2$, impossible for a positive price, while two undiscounted totals would make the second pre-total 100, which should have been discounted. Hence only the both-discounted case survives, so the notebook price is 10.

The insight is to eliminate the small weight by doubling the first total and subtracting the second, then use divisibility and positivity to bound the rest. Let small, medium, and large weigh $s$, $m$, and $l$. Then $2s+3m+4l=67$ and $4s+m+6l=73$. Doubling the first gives $4s+6m+8l=134$. Subtracting the second leaves $5m+2l=61$. So $5m=61-2l$, with $m$ and $l$ positive whole numbers. Thus $61-2l$ must be a positive multiple of 5, and using $2s+3m+4l=67$ gives $s=(76-7l)\mathrm{/}5$, so $s>0$ forces $l$ at most 10. Hence only $l=3$ gives $m=11$ and $l=8$ gives $m=9$. The greatest possible medium weight is 11 pounds.

The insight is to express both intercepts through the slope and turn integrality into a divisor condition. Write the slope as $m$ with $m$ a nonzero integer. Through $(6,12)$ the equation is $y-12=m(x-6)$. Setting $x=0$ gives $y=12-6m$, always an integer. Setting $y=0$ gives $x=6-12\mathrm{/}m$, which is an integer exactly when $m$ divides 12. For both intercepts to be positive, $m$ must be negative, because a positive slope through $(6,12)$ drives the $y$-intercept down. The negative divisors of 12 are $-1$, $-2$, $-3$, $-4$, $-6$, and $-12$, a total of 6 lines.

Let the base area be $B$ and the height $h$. The volume gives $Bh=120$. The base perimeter is 20, so the four side faces have total area $20h$, and the surface area gives $2B+20h=148$, or $B=74-10h$. Testing values in $Bh=120$: $h=5$ gives $B=24$ and $24\cdot 5=120$, and $h={\displaystyle \frac{12}{5}}$ gives $B=50$ and $50\cdot {\displaystyle \frac{12}{5}}=120$ too, while $h=4$ and $h=6$ fail. But a base with length plus width 10 has area at most $5\cdot 5=25$, so $B=50$ is impossible. With $B=24$ the base is 4 by 6, the box is $4\times 6\times 5$, and the height is 5.

The insight is double-counting memberships combined with solving a small linear system for exactly-two and exactly-three. Union is $30-5=25$, so exactly-two plus exactly-three equals $25-9=16$. Total memberships are $18+16+14=48$, which also equal 9 times 1 plus exactly-two times 2 plus exactly-three times 3, so twice exactly-two plus three times exactly-three equals 39. Substituting exactly-two as 16 minus exactly-three gives 32 plus exactly-three equals 39, so exactly-three is 7. Counting each triple once or confusing none and exactly-one with triples misses the two-way count.

The insight is to combine exponent translation with a counting re-frame. For positive bases, $(\sqrt[3]{4}{)}^a(\sqrt{3}{)}^b=2^{2a\mathrm{/}3}{3}^{b\mathrm{/}2}$. By unique prime factorization this is the square of an integer exactly when both prime exponents are even integers. Now $2a\mathrm{/}3=2(a\mathrm{/}3)$ is automatically even once it is an integer, so the condition is 3 divides $a$; similarly $b\mathrm{/}2$ even means $b=4s$, so 4 divides $b$. Among $1,\dots ,24$ there are 8 multiples of 3 and 6 multiples of 4. The choices are independent, so $8\times 6=48$ ordered pairs work. Checking all 576 pairs numerically with radicals is a long failing route; the divisibility count is short.

The insight is to combine triangle inequality bounds with a division re-framing. Triangle inequality gives $22<n<50$ from $36-14<n<36+14$. Rewriting $5n+55=5(n-1)+60$ shows the fraction equals $5+60\mathrm{/}(n-1)$, so $n-1$ must divide 60. Divisors of 60 that lie in 21 to 49 for $n-1$ are only 30, giving $n=31$ since 20 is too small and 60 is too large for the triangle range. Then $2n-15=62-15=47$, which is short once the bounds and divisibility are seen, while trying values across the range is long.

The insight is the reduced integer step plus simultaneous congruences. The change is 96 in $x$ and 48 in $y$, so the slope is ${\displaystyle \frac{1}{2}}$ and, dividing by $\gcd (96,48)=48$, the lattice points are $x=6+2t$ and $y=10+t$ for $t=0,\dots ,48$. Then $x$ divisible by 3 means $6+2t\equiv 2t\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}3)$, so $t\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}3)$, while $y$ even means $10+t$ even, so $t$ even. Together $t$ is a multiple of both 3 and 2, hence of 6. The values $t=0,6,\dots ,48$ are $48\mathrm{/}6+1=9$ points, including both endpoints, and the short check $t=0$ gives $(6,10)$ as required.

The insight is to extend $AD$ to meet $BC$ and apply the exterior-angle reframe twice, then use the around-point $360^{\circ }$ sum. Let the extension meet $BC$ at $E$ with $D$ between $A$ and $E$. In triangle $ABD$, the exterior angle $BDE$ equals $BAD+ABD$. In triangle $ACD$, the exterior angle $CDE$ equals $CAD+ACD$. Adding gives the smaller angle $BDC=BDE+CDE=(BAD+CAD)+ABD+ACD=BAC+ABD+ACD=50^{\circ }+30^{\circ }+20^{\circ }=100^{\circ }$. The reflex angle is what remains around $D$, so $360^{\circ }-100^{\circ }=260^{\circ }$.

The insight is that a unique mode controls how many times the top value may repeat, and the total then fixes the cheapest layout. The total is $6\times 8=48$ and the middle two sum to 14 since the median is 7. Because the mode is 5, the minimum is at most 5, so with range $R$ the maximum is at most $5+R$. If $R=6$ then the minimum must be 5 and the maximum 11, otherwise the sum cannot reach 48, and the middle pair summing to 14 leaves 24 for the top two, but two numbers at most 11 with no value repeated twice can sum to at most 21, and even allowing a repeat gives at most 22, short of 24. Thus $R\ge 7$, achieved by 5,5,5,9,12,12 whose middle two are 5 and 9, mean 8, unique mode 5, and range 7.

The insight is that the two descriptions cannot overlap. Both numbers exceeding 50 means each is at least 51, so their sum is at least 102, which can never be less than 75. Thus the union is disjoint. There are $100\ast 100=10000$ ordered pairs. Both exceeding 50 gives $50\ast 50=2500$ pairs. Sums 2 through 74 occur $1+2+\mathrm{.}\mathrm{.}\mathrm{.}+73=73\ast 74\mathrm{/}2=2701$ times by the triangular-number formula. The disjoint counts add to $2500+2701=5201$, so the probability is $5201\mathrm{/}10000$. Listing ten thousand pairs would be impossible, while the lower bound plus the triangle sum finishes quickly.

The insight is an extremal minimal-total bound: to fit many large values under a fixed mean, make every value as small as the rules allow and see when even that minimum is too big. The total must be 10 times 12, which is 120. Suppose $k$ values exceed 15. The cheapest distinct list with $k$ such values uses $1,2$ and so on for the small values and $16,17$ and so on for the large values. For $k$ equal to 6, that cheapest list is $1,2,3,4,16,17,18,19,20,21$, whose total is 121, already above 120, so 6 or more large values are impossible. For $k$ equal to 5, the cheapest total is 105, and raising the largest value to 35 gives $1,2,3,4,5,16,17,18,19,35$ with total 120, so 5 can occur.

The insight is position symmetry among draws. Imagine all 15 cubes labeled, then shuffled and laid out; each cube is equally likely to land in the sixth position, so the color of the sixth cube has the same distribution as the color of the first cube. There are 7 red cubes out of 15 total cubes, so the probability the cube in that position is red is ${\displaystyle \frac{7}{15}}$. A tree conditioned on the first five draws would need many branches, but symmetry makes the earlier draws irrelevant.

The insight is alternating-order symmetry with careful counting of the partial hour. Let Rosa do $r$ of the fence per hour and Tom do $t$ per hour. Starting with Rosa for 9 hours gives 5 Rosa hours and 4 Tom hours, so $5r+4t=1$. Starting with Tom, 9 full hours give 5 Tom hours and 4 Rosa hours, then the next 30 minutes fall in a Rosa hour, so $4.5r+5t=1$. Multiply the first by 5 to get $25r+20t=5$ and the second by 4 to get $18r+20t=4$. Subtracting gives $7r=1$, so $r=1\mathrm{/}7$ and Rosa alone needs 7 hours, with Tom needing 14 hours as a check.

The insight is that semicircle area scales as the square of the diameter, so Pythagoras becomes an area equation. A semicircle of diameter $d$ has area $\pi d^2\mathrm{/}8$, a constant times $d^2$. If the legs are $a$ and $b$ with hypotenuse 10, then $a^2+b^2=100$, and the two smaller semicircle areas sum to $\pi a^2\mathrm{/}8+\pi b^2\mathrm{/}8=\pi (a^2+b^2)\mathrm{/}8=\pi \cdot 100\mathrm{/}8$, exactly the largest semicircle $\pi \cdot 10^2\mathrm{/}8=25\pi \mathrm{/}2$. So the three semicircles total twice the largest, $25\pi$. Adding the triangle gives $21+25\pi$. Trying to recover the legs from hypotenuse and area leads to irrational legs and a long computation, which is the trap.