ACT · Competition-style problems · Part 2 of 2

The insight is the equilateral re-cut: $a$, $b$ and $-(a+b)$ each have length 2 and sum to zero, so head to tail they form an equilateral triangle. Hence the angle between $a$ and $b$ placed tail to tail is 120 degrees with cosine $-1\mathrm{/}2$. Put $a=(2,0)$ and $b=(2\cos \theta ,2\sin \theta )$ with $\cos \theta =-1\mathrm{/}2$, so the scalar product is $2\cdot 2\cdot (-1\mathrm{/}2)=-2$. Then $\mathrm{\mid }2a-b{\mathrm{\mid }}^2=4\mathrm{\mid }a{\mathrm{\mid }}^2+\mathrm{\mid }b{\mathrm{\mid }}^2-4a\cdot b=16+4+8=28$, so the magnitude is $\sqrt{28}=2\sqrt{7}$. Solving for components and the angle without seeing the equilateral triangle is much longer.

The maneuver is to expand the complex square to a difference of squares and then use a mod 4 obstruction, while exhibiting the other values directly. Expanding a plus b i squared gives a squared minus b squared plus twice a b i, so the real part is a squared minus b squared. Squares mod 4 are only 0 or 1, so differences mod 4 can only be 0, 1, or 3, never 2. Since 22 leaves remainder 2 upon division by 4, it can never equal a squared minus b squared. The others do occur with 9 as 3 squared, 15 as 4 squared minus 1 squared, and 33 as 7 squared minus 4 squared. Searching over unbounded integers is hopeless, but residues decide it at once.

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.

Bound the median with the total. Order the five distinct integers $a<b<m<d<e$; the range gives $e=a+6$ and the mean gives a total of 50. Distinctness inside the interval from $a$ to $a+6$ gives $b\le m-1$, $d\le a+5$ and $a\le m-2$, so the total is at most $3a+2m+10\le 5m+4$; if $m\le 9$ that is at most 49, too small. Likewise $b\ge a+1$, $d\ge m+1$ and $a\ge m-4$ give a total of at least $5m-4$; if $m\ge 11$ that is at least 51, too large. Hence the median is 10, as in the set 7,8,10,12,13, which has total 50 and range 6.

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.

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 $(6,6)$.

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.

The insight is to reframe triangle counting as total triples minus degenerate collinear triples split into two families. There are 13 points, so $\left(\genfrac{}{}{0px}{}{13}{3}\right)=286$ triples in all. Each horizontal line holds 4 points, contributing $\left(\genfrac{}{}{0px}{}{4}{3}\right)=4$ collinear triples, for $3\cdot 4=12$ from horizontals. Each transversal holds its 3 meeting points plus $O$, also 4 points, contributing $\left(\genfrac{}{}{0px}{}{4}{3}\right)=4$ collinear triples, for $4\cdot 4=16$ from transversals. The no-extra-collinearity condition guarantees there are no others, so the number of non-degenerate triples is $286-12-16=258$.

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 to convert first and then use total distance over total time rather than any average of speeds. Since 15 mi/h equals 22 ft/s, 22 ft/s is 15 mi/h and 44 ft/s is 30 mi/h. For one mile, the time is one quarter divided by 15 plus three quarters divided by 30, which is $1\mathrm{/}60$ plus $1\mathrm{/}40$ equals $1\mathrm{/}24$ hours. Dividing distance by time gives 24 mi/h. Weighting matters because three quarters of the distance is fast, while a plain mean of speeds either ignores the weights or ignores the harmonic form.

The insight is intercepts plus a perfect-square bound. Write the line as $y-3=m(x-4)$ with $m<0$. Then the $x$-intercept is $4-3\mathrm{/}m$ and the $y$-intercept is $3-4m$, both positive, and the area is ${\displaystyle \frac{1}{2}}(4-3\mathrm{/}m)(3-4m)=12-8m-9\mathrm{/}(2m)$. Hence $2m(\text{Area}-24)=-(4m+3{)}^2\le 0$. Since $2m<0$, dividing gives $\text{Area}-24\ge 0$, with equality exactly when $4m+3=0$. Therefore the minimum 24 occurs at $m=-3\mathrm{/}4$, and the computation uses only one perfect square.

The key is to compare prime exponents and then use distributivity. Since 12 is $2^2$ times 3 and 18 is 2 times $3^2$, together they already supply $2^2$ and $3^2$. For the least common multiple with $n$ to be $2^2$ times $3^2$ times 5, the number $n$ must supply 5 exactly once and may supply 2 up to $2^2$ and 3 up to $3^2$. Hence $n=5$ times $2^a$ times $3^b$ with $a=0,1,2$ and $b=0,1,2$, giving 9 possibilities. Their sum factors as 5 times $(1+2+4)$ times $(1+3+9)$, which is 5 times 7 times 13, or 455.

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 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 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 insight is the noncommutative binomial expansion $(A+B{)}^2=A^2+AB+BA+B^2$, so the mixed terms are isolated by subtraction as $AB+BA=(A+B{)}^2-A^2-B^2$. Summing entries, $(A+B{)}^2$ totals $11+12+6+11=40$, $A^2$ totals $1+4+0+1=6$, and $B^2$ totals $4+0+4+4=12$. Hence the desired total is $40-6-12=22$. Attempting to recover $A$ and $B$ individually from their squares is impossible from the data and wastes all the time.

The clever move is to strip out the greatest common factor and look at parity of prime exponents. Write the numbers as $18x$ and $18y$ with $x$ and $y$ coprime, so the least common multiple is $18xy$. Since 18 is 2 times a square, $18xy$ is a square exactly when $2xy$ is a square. Write $2xy=k^2$, so $k$ is even, $k=2m$, and $xy=2m^2$. The bound $18xy<1000$ gives $2m^2<56$, so $m=1,2,3,4,5$ and $xy=2,8,18,32,50$. With $x$ and $y$ coprime, the factor pairs give sums 3,9,19,11,33,17,7,51,27, the largest 51 from 1 and 50. That pair is coprime, gives numbers 18 and 900 with least common multiple 900, and sum 918.

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 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.

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 insight is that subtraction order only controls whether each later term is added or subtracted according to nesting parity, and then counting the achievable sign patterns while verifying distinctness. Here $5\mathrm{/}6-1\mathrm{/}2=1\mathrm{/}3$, and the first minus can never flip, while each of $1\mathrm{/}3$ and $1\mathrm{/}4$ can appear with either sign, giving four patterns $1\mathrm{/}3$ plus or minus $1\mathrm{/}3$ plus or minus $1\mathrm{/}4$. They evaluate to $-1\mathrm{/}4$, $5\mathrm{/}12$, $1\mathrm{/}4$, and $11\mathrm{/}12$, all different since the gaps are twice the fractions and hence nonzero. The five parenthesizations therefore collapse to four values because two orders yield the same middle signs.

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 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.

Put $u=2^{\sin x}$ to turn the transcendental-looking minimum into an ordinary quadratic on a closed interval. Because sine stays in $[-1,1]$ and base 2 is increasing, $u$ ranges exactly over $[1\mathrm{/}2,2]$. Then $f=u^2-3u+2=(u-3\mathrm{/}2{)}^2-1\mathrm{/}4$, so $f\ge -1\mathrm{/}4$ for every admissible $u$, with equality only at $u=3\mathrm{/}2$. That bound is attainable because ${\log }_2(3\mathrm{/}2)$ is about 0.585, which lies in $[-1,1]$, so some real $x$ has $\sin x={\log }_2(3\mathrm{/}2)$ and gives $u=3\mathrm{/}2$. Hence the minimum over all real $x$ is $-1\mathrm{/}4$.

The point is to classify t first with the four-cycle before solving the quadratic, reducing 100 checks to one residue count. Since the reciprocal is the conjugate, t equals 2 when n is 0 mod 4, 0 when n is odd, and minus 2 when n is 2 mod 4. Factoring gives t squared plus t minus 2 as t plus 2 times t minus 1, so t is 1 or minus 2, and 1 never occurs among 2, 0, minus 2. Only minus 2 qualifies, which is exactly n congruent to 2 mod 4. The values 2, 6, through 98 give 25 integers. Testing each n with a calculator would be tedious, but the classification is immediate.