SAT · Competition-style problems · Part 2 of 3

The insight is symmetry plus distance. Tangency to both axes in the first quadrant forces the center to be $(r,r)$ with radius $r$, collapsing three unknowns to one. The distance from $(r,r)$ to $3x+4y=30$ must equal $r$, so $\mathrm{\mid }3r+4r-30\mathrm{\mid }\mathrm{/}5=r$, or $\mathrm{\mid }7r-30\mathrm{\mid }=5r$. Hence $7r-30=5r$ or $7r-30=-5r$, giving $r=15$ or $r=5\mathrm{/}2$. Their circumferences are $30\pi$ and $5\pi$, summing to $35\pi$. Solving for a general center without the symmetry leads to a long system.

The insight is that the two centers and either intersection point form an equilateral triangle. Each side is 6, since each center lies on the other circle and each intersection lies on both, so the angle at each center in that triangle is 60 degrees. The lens needs twice that, a 120-degree sector from each circle. Two such sectors have total area $2(120\mathrm{/}360)\pi (6^2)=24\pi$. Removing the rhombus made of the two equilateral triangles, with area $2(\sqrt{3}\mathrm{/}4)(36)=18\sqrt{3}$, leaves $24\pi -18\sqrt{3}$ for the overlap. Computing circular segments directly without the equilateral observation is long.

The key is that equal sines come from supplementary angles, with only the top and bottom of the circle left unpaired. Since $\sin (180^{\circ }-\varphi )=\sin \varphi$ and $\sin (360^{\circ }-\varphi )=-\sin \varphi$, each $10k^{\circ }$ angle pairs with $180^{\circ }$ minus itself, both multiples of $10^{\circ }$. The angles $90^{\circ }$ and $270^{\circ }$ pair with themselves and yield no distinct partner, so the remaining 34 angles form 17 unordered matching pairs.

The insight is to pair divisors. For $6^6$, write $6^6=2^6\ast 3^6$, so the number of positive divisors is $(6+1)(6+1)=49$ by the exponent rule. Each divisor $d$ pairs with $6^6\mathrm{/}d$, and the product of the pair is $6^6$. The pair collapses to a single divisor exactly when $d=6^3$, since $(6^3{)}^2=6^6$. Thus one divisor equals $6^3$ and the remaining 48 form 24 pairs with one member below $6^3$ and one above. Hence 24 of the 49 equally likely slips satisfy the condition, giving $24\mathrm{/}49$. Listing all divisors would take far too long, while the pairing finishes the count at once.

The insight is to draw the perpendicular bisectors. Place the square with vertices $(0,0)$, $(s,0)$, $(0,s)$, and $(s,s)$, so the center is $(s\mathrm{/}2,s\mathrm{/}2)$. Equidistance to the center and the corner $(0,0)$ gives $x+y=s\mathrm{/}2$, and the other three corners give the symmetric lines $x+y=3s\mathrm{/}2$, $x-y=s\mathrm{/}2$, and $y-x=s\mathrm{/}2$. The four lines meet at $(s\mathrm{/}2,0)$, $(s,s\mathrm{/}2)$, $(s\mathrm{/}2,s)$, and $(0,s\mathrm{/}2)$, forming a diamond whose diagonals are both $s$, so its area is $s^2\mathrm{/}2$. The whole square has area $s^2$, so the desired ratio is $(s^2\mathrm{/}2)\mathrm{/}{s}^2=1\mathrm{/}2$. Guessing the boundary or testing points cannot replace finding the four lines.

The idea is to write each map side as $400\mathrm{/}n$ and $800\mathrm{/}n$ and then force the orientation with a size bound, plus exhibit the packing that meets it. For $n$ less than 50, each map has long side $800\mathrm{/}n$ greater than 16 cm, so no long side fits across the 16 cm width and every map must place its long side along the 25 cm height. Two such maps side by side need width 2 by $400\mathrm{/}n$ greater than 16 cm, and stacked need height 2 by $800\mathrm{/}n$ greater than 25 cm, so two maps cannot fit. For $n$ equal to 50, each map is 8 cm by 16 cm, and two maps with 16 cm sides along the page width stacked to height 16 cm fit inside 16 cm by 25 cm. Hence 50 is the smallest feasible value.

The insight is to turn means into totals and use inclusion-exclusion for the overlap. The art total is $20\times 84=1680$ and the music total is $15\times 88=1320$, so the two lists sum to 3000. If $o$ students take both, the number of distinct students is $35-o$ and their total is $85\times (35-o)=2975-85o$. Hence the overlap total is $3000-(2975-85o)=25+85o$ and its mean is $85+25\mathrm{/}o$. Since that mean is a whole number, $o$ divides 25, so with $o\le 15$ we get $o=1$ or $o=5$. The value $o=1$ would give mean 110, impossible for scores capped at 100, so $o=5$ and the mean is $85+5=90$.

The insight is to write the original total as a triangular number and trap $n$ by factoring out $n-1$. The original sum is $n(n+1)\mathrm{/}2$ and the remaining sum is $61(n-1)\mathrm{/}3$, so the erased integer is $k=n(n+1)\mathrm{/}2-61(n-1)\mathrm{/}3=(n-1)(3n-116)\mathrm{/}6$. Since $1\le k\le n$, we get $(n-1)(3n-116)\ge 6$ and $(n-1)(3n-122)\le 0$, which factor cleanly to force $39\le n\le 40$. The remaining sum must be an integer, so $n-1$ is a multiple of 3, leaving $n=40$. Then the original sum is $40\times 41\mathrm{/}2=820$ and the remaining sum is $61\times 39\mathrm{/}3=61\times 13=793$, so the erased integer is $820-793=27$.

The insight is to re-cut valid triples by gap compression and then count through a bijection with a smaller range. There are $\left(\genfrac{}{}{0px}{}{8}{3}\right)=56$ equally likely triples. Write a valid triple as $a<b<c$ with gaps of at least 2. Define $a^{\mathrm{\prime }}=a$, $b^{\mathrm{\prime }}=b-1$, and $c^{\mathrm{\prime }}=c-2$. Then $1\le a^{\mathrm{\prime }}<b^{\mathrm{\prime }}<c^{\mathrm{\prime }}\le 6$, so every valid triple compresses to a distinct triple from 1 through 6. Conversely adding 0, 1, 2 expands any triple from 1 through 6 to a valid triple with no consecutive numbers, so the correspondence is exact. Hence there are $\left(\genfrac{}{}{0px}{}{6}{3}\right)=20$ favorable triples, and the probability is ${\displaystyle \frac{20}{56}}={\displaystyle \frac{5}{14}}$ after dividing by 4.

The insight is to fill positions sequentially with conditional counts and then split on whether the two forbidden predecessors coincide. There are $10^4=10000$ equally likely PINs. The first position has 10 choices and the second has 9 choices different from the first. The third has 9 choices different from the second. If the third equals the first, which happens in 1 way, the last position has 9 choices different from that digit, giving $10\times 9\times 1\times 9=810$ codes. If the third differs from both the first and second, which happens in 8 ways, the last must avoid two different digits, giving 8 choices, for $10\times 9\times 8\times 8=5760$ codes. The total is $810+5760=6570$, so the probability is ${\displaystyle \frac{6570}{10000}}={\displaystyle \frac{657}{1000}}$ after dividing by 10.

The insight is that 25 is the arithmetic mean of 20 and 30, so with average defined by total distance over total time the two driving times must be equal, not the distances. Let the sections be $d1$ and $d2$. Then $(d1+d2)\mathrm{/}(d1\mathrm{/}20+d2\mathrm{/}30)=25$, which gives $d1\mathrm{/}d2=2\mathrm{/}3$. Hence the total $D=d1+d2$ is 5 times an integer. Since $D\mathrm{/}25$ is the total time, $D\mathrm{/}25$ lies between 4 and $25\mathrm{/}6$, so $D$ lies between 100 and 104.16. The only multiple of 5 there is 100. A student who thinks average 25 means equal distances is pushed to several possibilities and cannot finish.

The insight is weighted extremal bounding with an explicit construction. Simple interest makes one dollar for one year earn 0.10 dollars, so with deposits $P1$ less than $P2$ less than $P3$ the interest condition is $P1$ plus $2P2$ plus $3P3$ equals 220. To have three deposits above 40 dollars the cheapest distinct choice is 41, 42, 43, giving 41 plus 84 plus 129 equals 254 which exceeds 220, so three is impossible. Two is possible with 1, 42, 45 since 1 plus 84 plus 135 equals 220 for 22 dollars interest, so the maximum is two.

The insight is to collapse the system to a single expression and then use parity to rule out primes. Let cars be $c$ and motorcycles be $m$. Then $c+m=n$ and $c-m=t$, so $c=(n+t)\mathrm{/}2$ and $m=(n-t)\mathrm{/}2$, which are whole numbers because $n+t$ is even. Total wheels equal $4c+2m=2(n+t)+(n-t)=3n+t$. Since $3n+t=2n+(n+t)$ is the sum of two even numbers when $n+t$ is even, the total is even. Because $0<t<n$ forces $n$ at least 3 and the total at least 10, the total is an even number greater than 2 and therefore composite, never prime. Hence no pair works.

Let the intercepts be $(p,0)$ and $(0,q)$. The point $(6,4)$ lies on the segment between them, so the slope from $(p,0)$ to $(6,4)$ equals the slope from $(p,0)$ to $(0,q)$: ${\displaystyle \frac{4}{6-p}}={\displaystyle \frac{q}{-p}}$, which gives $q={\displaystyle \frac{4p}{p-6}}=4+{\displaystyle \frac{24}{p-6}}$. For $q$ to be a whole number, $p-6$ must be a factor of 24, so $p=7,8,9,10,12,14,18,30$ with $q=28,16,12,10,8,7,6,5$. The areas ${\displaystyle \frac{pq}{2}}$ are 98,64,54,50,48,49,54,75, and the largest is 98, from intercepts 7 and 28.

The insight is to rewrite the fraction as an integer plus a proper remainder instead of testing $x$ values one by one. Write $4x+9=2(2x+1)+7$, so the fraction equals $2+{\displaystyle \frac{7}{2x+1}}$. For the whole to be an integer, ${\displaystyle \frac{7}{2x+1}}$ must be an integer, so $2x+1$ must be a divisor of 7. The divisors are $1,-1,7,-7$, giving $2x+1=1,-1,7,-7$, so $x=0,-1,3,-4$. That is four integers, and trying $x$ values directly would never end.

The insight is an extremal bound plus an explicit construction. Any two distinct positive integers summing to $90^{\circ }$ cannot both be at most $45^{\circ }$, since $45+45=90$ would repeat a value, so the larger of the complementary pair is at least $46^{\circ }$ and therefore the largest of the five is at least $46^{\circ }$. This bound is attainable with distinct integers summing to $180^{\circ }$: $44+46=90$ uses the pair and $28+30+32=90$ uses three more distinct values, giving the set 28, 30, 32, 44, 46 whose total is $180^{\circ }$ and whose largest is $46^{\circ }$. Hence the minimum is achieved.

The insight is an auxiliary-parallel re-cut plus a proportional split. Draw through $P$ a line parallel to $l$ and $m$; by alternate interior angles it splits the smaller angle $APB$ into exactly the two interior angles, so their sum is $100^{\circ }$, a fact that is not visible without the helper line. With ratio $2:3$, the parts are $2+3=5$ equal shares of $100^{\circ }$, so each share is $20^{\circ }$. The larger interior angle is $3\times 20=60^{\circ }$ and the smaller is $40^{\circ }$. Trying to relate the ratio to $180^{\circ }$ instead of $100^{\circ }$ fails.

The insight is a counting re-frame by how many acute angles are chosen plus a mean equation and a feasibility bound. Let the acute be $a$ and the obtuse $180-a$, and let $k$ of the four chosen be acute, so $k$ is 0 to 4 with at most four of each available. The mean condition is $(ka+(4-k)(180-a))\mathrm{/}4=110$, so $(2k-4)a=440-720+180k$. Checking $k$ gives $k=0$ with $a=70$ using four obtuses of 110, $k=1$ with $a=50$ using one 50 and three 130s, $k=2$ impossible, and $k=3,4$ obtuse for $a$. Only $k=1$ uses both types, so the acute is $50^{\circ }$. Enumerating all 70 subsets is long.

The insight is to subtract tunnels by inclusion-exclusion and notice all three pairwise intersections coincide in the same central cube. Each tunnel is 9 by 3 by 3 equals 81, so three give 243. Any two tunnels meet in the central 3 by 3 by 3 equals 27, but all three pairs are the same 27, and the triple intersection is that 27 again, so the union is $243-81+27=189$. Hence the remainder is $729-189=540$. Subtracting without correction gives 486 after a long miscount, while the coincident-overlap shortcut finishes quickly.

The insight is that the change depends on position because outer faces lost and inner walls gained differ. A corner cube loses 3 outer faces and gains 3 inner walls for net 0; an edge cube loses 2 and gains 4 for net $+2$; a face cube loses 1 and gains 5 for net $+4$, and non-touching removals add. The original cube has area $6\cdot 49=294$, so the increase is $310-294=16$. Let $c$, $e$, $f$ be the numbers of corner, edge, and face cubes. Then $c+e+f=9$, $e=2f$, and $2e+4f=16$. Hence $8f=16$, so $f=2$, $e=4$, and $c=9-6=3$. Counting every exposed square separately is much longer.

The insight is double-counting memberships to bound extras combined with an extremal construction. Union is $32-3=29$ students with at least one sport. Total memberships are $18+16+14=48$, so extras beyond one per student are $48-29=19$. Each all-three student contributes 2 extras and each exactly-two student contributes 1, so with $t$ triples, $2t$ is at most 19, giving $t$ at most 9. Nine is attainable with 9 triple, 1 soccer-basketball double, plus 8 only-soccer, 6 only-basketball, and 5 only-tennis, which uses 29 union members and exactly 18, 16, and 14 memberships. Dividing extras by 3 or using only the smallest group misses the double-count.

The insight is that a digit sum is even exactly when the number of odd digits is even, so the last digit parity is forced by the first two and the leading-zero restriction changes the remaining counts. Count prefixes with hundreds nonzero and tens different. Hundreds odd and tens even gives 5 times 5 equals 25 prefixes needing an odd last digit with 4 odds left, giving 100. Hundreds odd and tens odd gives 5 times 4 equals 20 prefixes needing an even last digit with 5 evens left, giving 100. Hundreds even nonzero and tens odd gives 4 times 5 equals 20 prefixes needing an odd last digit with 4 odds left, giving 80. Hundreds even nonzero and tens even gives 4 times 4 equals 16 prefixes needing an even last digit with 3 evens left, giving 48. The total is 100 plus 100 plus 80 plus 48, which is 328.

The insight is to translate the digit sum into linear equations that differ by which position repeats, then enforce digit bounds with leading-zero and distinctness. For patterns where the repeated digit is the hundreds, both 2 times $a$ plus $b$ equals 12 with $a$ nonzero and $b$ different from $a$. The solutions are $a$ equals 2, 3, 5 and 6 with $b$ equals 8, 6, 2 and 0, giving 4 numbers for each of the two such patterns. For the pattern where the tens and ones repeat, $a$ plus 2 times $b$ equals 12 gives $a$ equals 8, 6 and 2 with $b$ equals 2, 3 and 5, giving 3 numbers. The total is 4 plus 4 plus 3, which is 11.

The insight is to add a constant to factor by grouping plus a positivity bound. From $(x+a)(x+b)=x^2+(a+b)x+ab$ you get $k=a+b$ and $m=ab$, so $ab=2a+2b$. Add 4 to both sides to keep equivalence: $ab-2a-2b+4=4$, which factors as $(a-2)(b-2)=4$. With $a$ and $b$ positive integers, set $u=a-2$ and $v=b-2$; then $uv=4$ with $u$ and $v$ greater than $-2$. The admissible factor pairs are 1 times 4, 2 times 2, and 4 times 1, giving $(a,b)=(3,6)$, $(4,4)$, and $(6,3)$. Hence the distinct $k=a+b$ values are 9 and 8, whose sum is 17. Trying values of $k$ blindly never reveals the hidden product, so missing the added constant pushes a solver into a long search.

The insight is to substitute $t=x^2$ and enforce both discriminant zero and sign. Then $t^2-kt+16=0$ with $t>=0$ for real $x$. Its discriminant is $k^2-64$. If $\mathrm{\mid }k\mathrm{\mid }<8$ there is no real $t$ and hence no real $x$. If $k=8$, we get $(t-4{)}^2=0$ so $t=4$ and $x=2$ or $x=-2$, exactly two solutions. If $k=-8$, we get $t=-4$ with no real $x$. If $\mathrm{\mid }k\mathrm{\mid }>8$, there are two distinct $t$ values: both positive when $k>8$ giving four $x$ values, and both negative when $k<-8$ giving none. Hence only $k=8$ gives exactly two.