IPMAT Indore 2019 — MCQ

The Setup: Let's speedrun this finance glitch. Compound interest beats simple interest only because of interest earned *on* the first year's interest - so demanding that CI be exactly twice SI is a very aggressive ask, and the rate it forces is correspondingly extreme.
Step 1: Define the base formulas. Let $P$ be the principal and $r$ the annual rate as a decimal.

$$SI=P\cdot r\cdot 2=2Pr, \qquad CI=P(1+r)^2-P$$

Step 2: Equate and expand. The condition is $CI=2\times SI$:

$$P(1+r)^2-P=2(2Pr)$$

Divide through by $P$ (a principal of zero is meaningless) and expand:

$$1+2r+r^2-1=4r \implies r^2+2r=4r$$

Step 3: Solve for the rate.

$$r^2-2r=0 \implies r(r-2)=0$$

The root $r=0$ would mean no interest at all, making $CI=SI=0$ - technically satisfying the equation but describing no loan. Discard it, leaving $r=2$ as a decimal, i.e. $200\%$.
Step 4: Sanity-check, because 200% looks absurd. Take $P=100$ at $r=200\%$:
* Simple interest: $100\times 2\times 2=400$
* Compound: $100(1+2)^2-100=900-100=800$

And $800=2\times 400$ exactly. The answer really is 200% - the reason it feels wrong is that doubling CI relative to SI is a far harsher demand than it sounds, and only a triple-per-year growth factor achieves it. Option 2 is the trap: solve correctly, get $r=2$, and then read that 2 as a percentage instead of as the decimal it is.

Final Answer: 200%

The Setup: Factorials scale up like a boss-level mob. To count how many times $21$ divides $50!$, break 21 into primes and count each prime's supply separately - whichever runs out first is the bottleneck.
Step 1: Prime factorise the base.

$$21=3\times 7$$

Each copy of 21 consumes one 3 and one 7, so $21^n$ needs $n$ of each.
Step 2: Count the supply of each prime with Legendre's formula. The exponent of a prime $p$ in $n!$ is $\sum_{i\geq 1}\left\lfloor\frac{n}{p^i}\right\rfloor$.

$$E_7(50!)=\left\lfloor\frac{50}{7}\right\rfloor+\left\lfloor\frac{50}{49}\right\rfloor=7+1=8$$

$$E_3(50!)=\left\lfloor\frac{50}{3}\right\rfloor+\left\lfloor\frac{50}{9}\right\rfloor+\left\lfloor\frac{50}{27}\right\rfloor=16+5+1=22$$

Step 3: Take the minimum. The larger prime really is the scarcer one, and that is a theorem rather than a guess: for primes $p<q$ we have $p^i<q^i$, so $\left\lfloor\frac{n}{p^i}\right\rfloor\geq\left\lfloor\frac{n}{q^i}\right\rfloor$ term by term, and summing gives $E_p(n!)\geq E_q(n!)$. Still, computing both counts is worth the ten seconds - it turns a general principle into the concrete numbers the answer needs:

$$n=\min\left(E_3,\ E_7\right)=\min(22,\ 8)=8$$

With 22 threes available and only 8 sevens, the sevens are exhausted first - we can assemble 8 copies of 21, and the 14 spare threes are useless without partners.

Final Answer: 8

The Setup: Modular arithmetic is overpowered for shrinking absurd numbers. Rather than computing a skyscraper of digits, reduce the base mod 9 and exploit the fact that its powers cycle.
Step 1: Read the exponent tower correctly. This is the first trap: $\left(29^{29}\right)^{29}$ has brackets, so the exponents multiply:

$$\left(29^{29}\right)^{29}=29^{29\times 29}=29^{841}$$

This is emphatically not $29^{\left(29^{29}\right)}$, which would be an astronomically larger number. When the tower is bracketed from the bottom, multiply.
Step 2: Simplify the base modulo 9. Since $29=27+2$ and $27$ is a multiple of 9:

$$29\equiv 2\pmod 9$$

So the task reduces to finding $2^{841}\bmod 9$.
Step 3: Find the cycle. List powers of 2 modulo 9 until they repeat:

$$2^1\equiv 2,\quad 2^2\equiv 4,\quad 2^3\equiv 8,\quad 2^4\equiv 7,\quad 2^5\equiv 5,\quad 2^6\equiv 1$$

Hitting 1 at the sixth power means the cycle length is exactly 6, consistent with $\phi(9)=6$.
Step 4: Reduce the exponent modulo the cycle length.

$$841=6\times 140+1 \implies 841\equiv 1\pmod 6$$

So $2^{841}$ sits at the same position in the cycle as $2^1$:

$$2^{841}\equiv 2^1\equiv 2\pmod 9$$

Final Answer: 2

The Setup: Rather than running three separate divisibility side quests, fuse them into one by taking the LCM - a number divisible by 6, 7 and 9 is exactly a number divisible by their least common multiple.
Step 1: Find the master divisor. Factorising: $6=2\times 3$, $7=7$, $9=3^2$. The LCM takes the highest power of each prime:

$$\text{LCM}(6,7,9)=2\times 3^2\times 7=126$$

Note it is not $6\times 7\times 9=378$ - 6 and 9 share a factor of 3, so the product over-counts.
Step 2: Build the number. Appending two digits to 4530 gives $453000+x$, where $x$ runs from 00 to 99.
Step 3: Find the gap. Divide the base by 126:

$$453000=126\times 3595+30$$

The base overshoots a multiple of 126 by 30, so to reach the next one we must add:

$$126-30=96$$

That gives $453096=126\times 3596$. And since the following candidate would be $96+126=222$ - far beyond two digits - 96 is the only possible answer in range, which is worth noting rather than stopping at the first hit.
Step 4: Confirm the distractors fail. All three wrong options clear the divisibility by 6, so that test alone separates nothing - 78 and 42 both fall at the 7, and 54 falls at the 9:

Final Answer: 96

The Setup: Two overlapping sets, and we need the floor and ceiling on the overlap $x$. Both bounds come from hard constraints, not from assumptions about the school.
Step 1: The ceiling. An intersection can never exceed either of the sets forming it. Only 50% like football, and everyone who likes both is among them:

$$x\leq\min(70,\ 50)=50$$

Step 2: The floor - derived, not assumed. The union of the two groups cannot exceed the whole school, so $n(C\cup F)\leq 100$. By inclusion-exclusion:

$$n(C\cup F)=70+50-x=120-x$$

Combining the two:

$$120-x\leq 100 \implies x\geq 20$$

This is worth stating as an inequality rather than as *suppose everyone likes at least one sport*. That supposition is only the extreme case; the real reason $x$ cannot drop below 20 is that 70 and 50 together already claim 120 percentage points of a 100-point school, so at least 20 points must be double-counted. The overlap is forced by arithmetic, whether or not anyone likes neither sport.
Step 3: Combine. Both bounds are attainable - $x=50$ when every football fan also likes cricket, and $x=20$ when the two groups spread across the whole school with nobody left out:

$$20\leq x\leq 50$$

Final Answer: $20 \leq x \leq 50$

The Setup: A triple-set problem where the union is handed to us - every one of the 65 students likes at least one sport - so inclusion-exclusion has exactly one unknown and can simply be solved.
Step 1: The master formula. With $C$, $F$, $H$ for cricket, football and hockey:

$$n(C\cup F\cup H)=n(C)+n(F)+n(H)-n(C\cap F)-n(F\cap H)-n(C\cap H)+n(C\cap F\cap H)$$

Step 2: Substitute. Since every student likes at least one sport, the union is the full class of 65. Let $y=n(C\cap H)$:

$$65=40+25+20-10-8-y+5$$

Step 3: Solve.

$$65=85-18+5-y=72-y \implies y=72-65=7$$

Step 4: Verify the Venn diagram is actually consistent. A value that satisfies the equation can still be nonsense if it forces a negative region, so break the class into its seven disjoint parts. Subtracting the triple overlap of 5 from each pair:

Every region is non-negative, and they total $5+5+3+2+28+12+10=65$ - exactly the class size. The answer is not just algebraically valid but genuinely realisable.

Final Answer: 7

The Setup: A rational inequality. We cannot cross-multiply, because the denominator's sign is unknown and multiplying by a negative would flip the inequality. Everything goes to one side instead.
Step 1: Reframe. Move the 1 across and combine over a common denominator:

$$\frac{x-3}{x^2+3x+2}-1\geq 0 \implies \frac{x-3-\left(x^2+3x+2\right)}{x^2+3x+2}\geq 0 \implies \frac{-x^2-2x-5}{x^2+3x+2}\geq 0$$

Step 2: Kill the numerator's sign ambiguity. Multiply by $-1$ and reverse the inequality:

$$\frac{x^2+2x+5}{x^2+3x+2}\leq 0$$

Check the numerator's discriminant:

$$\Delta=2^2-4(1)(5)=4-20=-16<0$$

With $\Delta<0$ and a positive leading coefficient, $x^2+2x+5$ is **strictly positive for every real $x$ - it never touches zero. So the numerator can never make the fraction zero or negative on its own. Step 3: Force the denominator negative.** A positive over something, being $\leq 0$, requires the denominator to be strictly negative (zero is not allowed - the expression would be undefined):

$$(x+1)(x+2)<0$$

An upward parabola sits below zero strictly between its roots:

$$x\in(-2,-1)$$

Note the interval is open because the endpoints $-2$ and $-1$ make the denominator zero, where the original expression is undefined. Separately worth seeing: the numerator $x^2+2x+5$ never vanishes, so the fraction is never exactly zero - meaning equality in the original $\geq 1$ is never attained, and the condition behaves throughout as a strict $>1$.
Step 4: Compute the width. With $a=-2$ and $b=-1$:

$$b-a=-1-(-2)=1$$

Option 4 - claiming no real $x$ works - is the trap for anyone who mishandles the sign flip in Step 2 and concludes the fraction can never be negative.

Final Answer: 1

The Setup: We need the exact range of $ab+bc+ca$ on the unit sphere $a^2+b^2+c^2=1$. Both bounds fall out of the fact that a real square is never negative - but a bound is only the answer if it is actually reached, so we must exhibit a witness for each.
Step 1: The lower bound. Expand a perfect square, which cannot be negative:

$$(a+b+c)^2\geq 0 \implies a^2+b^2+c^2+2(ab+bc+ca)\geq 0$$

Substituting the constraint:

$$1+2(ab+bc+ca)\geq 0 \implies ab+bc+ca\geq -\frac{1}{2}$$

Step 2: The upper bound. Use the sum of pairwise squared differences, also non-negative:

$$(a-b)^2+(b-c)^2+(c-a)^2\geq 0 \implies 2\left(a^2+b^2+c^2\right)-2(ab+bc+ca)\geq 0$$

$$2(1)-2(ab+bc+ca)\geq 0 \implies ab+bc+ca\leq 1$$

Step 3: Show both ends are attained. Inequalities alone would allow a smaller range, so produce explicit points on the sphere:
* Upper: take $a=b=c=\frac{1}{\sqrt{3}}$. Then $a^2+b^2+c^2=3\cdot\frac{1}{3}=1$ ✓, and $ab+bc+ca=3\cdot\frac{1}{3}=1$. Equality holds because $(a-b)^2+(b-c)^2+(c-a)^2=0$ exactly when all three are equal.
* Lower: take $a=\frac{1}{\sqrt{2}}$, $b=-\frac{1}{\sqrt{2}}$, $c=0$. Then $a^2+b^2+c^2=\frac{1}{2}+\frac{1}{2}+0=1$ ✓, and $ab+bc+ca=-\frac{1}{2}+0+0=-\frac{1}{2}$. Equality holds because $a+b+c=0$.

Step 4: Conclude. The expression is continuous on a connected sphere, so it sweeps every value between the two attained extremes - the set is the closed interval, endpoints included:

$$-\frac{1}{2}\leq ab+bc+ca\leq 1$$

Final Answer: [-$\frac{1}{2}$, 1]

The Setup: Before touching the inequality we must respect the logarithm's domain. A log of a non-positive number does not exist, so any $x$ that breaks the argument is disqualified before the inequality is even considered - and skipping this step is exactly how the wrong option gets picked here.
Step 1: Determine the domain. The argument must be strictly positive:

$$\frac{3x-1}{2-x}>0$$

The critical points are $x=\frac{1}{3}$ (numerator zero) and $x=2$ (denominator zero). Testing the three regions: for $x<\frac{1}{3}$ the numerator is negative and denominator positive, giving a negative; for $\frac{1}{3}<x<2$ both are positive; for $x>2$ the numerator is positive and denominator negative. So:

$$\text{Domain}=\left(\frac{1}{3},\ 2\right)$$

Step 2: Unpack the inequality. The base is $2>1$, so the logarithm is increasing and the inequality direction is preserved when exponentiating:

$$\frac{3x-1}{2-x}<2^1=2$$

Move everything to one side rather than cross-multiplying, since $2-x$ changes sign:

$$\frac{3x-1}{2-x}-2<0 \implies \frac{3x-1-2(2-x)}{2-x}<0 \implies \frac{5x-5}{2-x}<0$$

Step 3: Solve. Factor and tidy the sign:

$$\frac{5(x-1)}{-(x-2)}<0 \implies \frac{x-1}{x-2}>0$$

A quotient is positive when both parts share a sign, giving $x<1$ or $x>2$.
Step 4: Intersect with the domain - this is where the question is won or lost.

$$\left[(-\infty,1)\cup(2,\infty)\right]\cap\left(\frac{1}{3},2\right)=\left(\frac{1}{3},\ 1\right)$$

The branch $x>2$ solves the algebra perfectly but lies entirely outside the domain, where the logarithm does not exist at all. Discarding it is not optional bookkeeping: carry it through and the answer set is simply wrong, no matter how clean the algebra behind it looked.

Final Answer: $x \in \left( \frac{1}{3}, 1 \right)$

The Setup: An exponential inequality with a trap built into the base. Because $0.7<1$, the function $0.7^t$ is decreasing, so comparing exponents reverses the inequality - an UNO reverse card most people forget to play.
Step 1: Match the bases. Since $7^3=343$, scaling the decimal places gives:

$$0.343=(0.7)^3 \implies 0.7^{\left(2x^2-3x+4\right)}<0.7^3$$

Step 2: Drop the bases and flip. For a base in $(0,1)$, a smaller value corresponds to a larger exponent:

$$2x^2-3x+4>3$$

Step 3: Solve the quadratic inequality.

$$2x^2-3x+1>0$$

Factor by splitting the middle term:

$$2x^2-2x-x+1>0 \implies 2x(x-1)-1(x-1)>0 \implies (2x-1)(x-1)>0$$

The roots are $x=\frac{1}{2}$ and $x=1$. The leading coefficient is positive, so the parabola opens upward and is positive outside its roots and negative between them.
Step 4: Write the interval.

$$x<\frac{1}{2} \quad\text{or}\quad x>1$$

Option 1, the interval $\left(\frac{1}{2},1\right)$, is precisely what you get by not flipping the sign in Step 2: that error yields $2x^2-3x+1<0$, whose solution is the region *between* the roots rather than outside them. The whole question turns on that single step.

Final Answer: $\left( -\infty, \frac{1}{2} \right) \cup \left( 1, \infty \right)$

The Setup: Pure equilateral-triangle energy. With the chord equal to the radius, joining the centre to the chord's endpoints makes an equilateral triangle, which hands us the centre-to-chord distance. From there the two inscribed circles are pinned down exactly.
Step 1: Find the distance to the chord. Take $R=1$ to keep the arithmetic light. A chord of length 1 subtends an equilateral triangle with the centre, and the perpendicular distance from centre to chord is that triangle's height:

$$d=\sqrt{R^2-\left(\frac{R}{2}\right)^2}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$$

Step 2: Pin down each inscribed circle - this is the load-bearing step. Put the centre $O$ at the origin with the chord horizontal at height $d$, so its midpoint is $M=\left(0,\tfrac{\sqrt{3}}{2}\right)$ and the perpendicular through $M$ is the $y$-axis. A circle touching the chord **at $M$** must have its centre on that perpendicular, at distance $r$ from $M$.
* Above the chord (away from $O$): centre $\left(0,d+r\right)$. Internal tangency to the big circle means the centres are $R-r$ apart, so $d+r=R-r$, giving $2r=R-d$ - its **diameter is $R-d$**.
* Below the chord (past $O$): centre $\left(0,d-r\right)$ with $r>d$, so the distance to $O$ is $r-d$, and internal tangency gives $r-d=R-r$, i.e. $2r=R+d$ - its **diameter is $R+d$**.

So the two circles have as diameters exactly the two pieces into which $M$ cuts the vertical diameter of the original circle, running from $M$ up to $(0,1)$ and from $M$ down to $(0,-1)$:

$$\text{smaller}=1-\frac{\sqrt{3}}{2}, \qquad \text{bigger}=1+\frac{\sqrt{3}}{2}$$

Step 3: Take the ratio of areas. Areas scale as the square of any linear measure, so the area ratio is the square of the diameter ratio:

$$k=\left(\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}\right)^2=\left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)^2$$

Rationalise the inner fraction by multiplying top and bottom by $2+\sqrt{3}$, noting $(2-\sqrt{3})(2+\sqrt{3})=4-3=1$:

$$k=\left(\frac{(2+\sqrt{3})^2}{1}\right)^2=(2+\sqrt{3})^4$$

Step 4: Expand in two stages.

$$(2+\sqrt{3})^2=4+4\sqrt{3}+3=7+4\sqrt{3}$$

$$k=(7+4\sqrt{3})^2=49+56\sqrt{3}+48=97+56\sqrt{3}$$

Watch the option list: $7+4\sqrt{3}$ is sitting there as option 3, and it is precisely the ratio of the diameters, not the areas. Stopping one squaring too early lands exactly on it.

Final Answer: $97 + 56\sqrt{3}$

The Setup: A tilted quadrilateral inside a square. Rather than chase the inner side length directly, compute the four corner triangles and subtract them from the whole - the symmetry makes all four identical.
Step 1: Define the segments. Let the big square have side $x$. The ratio $\frac{AP}{PB}=\frac{1}{n}$ splits each side into $1+n$ equal parts, one part then $n$ parts:

$$AP=BQ=CR=DS=\frac{x}{n+1}, \qquad PB=QC=RD=SA=\frac{nx}{n+1}$$

Step 2: Compute one corner triangle, then argue the rest. Triangle $PBQ$ sits in corner $B$, right-angled there, with legs $PB$ and $BQ$:

$$[\,PBQ\,]=\frac{1}{2}\cdot\frac{nx}{n+1}\cdot\frac{x}{n+1}=\frac{nx^2}{2(n+1)^2}$$

The other three are congruent, not merely similar: at corner $C$ the legs are $QC=\frac{nx}{n+1}$ and $CR=\frac{x}{n+1}$ - the same two lengths, and likewise at $D$ and $A$. The whole configuration maps to itself under a $90^\circ$ rotation, so:

$$\text{Total corner area}=4\times\frac{nx^2}{2(n+1)^2}=\frac{2nx^2}{(n+1)^2}$$

Step 3: Subtract.

$$[\,PQRS\,]=x^2-\frac{2nx^2}{(n+1)^2}=x^2\left(\frac{(n+1)^2-2n}{(n+1)^2}\right)=x^2\left(\frac{n^2+2n+1-2n}{(n+1)^2}\right)=x^2\cdot\frac{n^2+1}{(n+1)^2}$$

Step 4: Form the ratio. Dividing by $[\,ABCD\,]=x^2$, the side length cancels entirely - the answer depends only on $n$:

$$\frac{[\,PQRS\,]}{[\,ABCD\,]}=\frac{1+n^2}{(1+n)^2}$$

Sanity check at $n=1$, where every point is a midpoint: the ratio becomes $\frac{2}{4}=\frac{1}{2}$, the well-known result that joining the midpoints of a square halves its area.

Final Answer: $1 + n^2 : (1 + n)^2$

The Setup: Two chords laid end to end inside a circle. Each chord subtends a central angle we can extract with the cosine rule, and because $B$ lies on the minor arc $AC$, those two angles simply add to give the angle subtending $AC$.
Step 1: Find each central angle. Let $O$ be the centre, $R=6$. Triangle $OAB$ has sides $6,6,3$, so with $\angle AOB=\alpha$:

$$\cos\alpha=\frac{6^2+6^2-3^2}{2(6)(6)}=\frac{72-9}{72}=\frac{63}{72}=\frac{7}{8}$$

Triangle $OBC$ has sides $6,6,2$, so with $\angle BOC=\beta$:

$$\cos\beta=\frac{6^2+6^2-2^2}{2(6)(6)}=\frac{72-4}{72}=\frac{68}{72}=\frac{17}{18}$$

Step 2: Add the angles. Since $B$ lies on the minor arc $AC$, the two central angles sit side by side without overlapping, so $\angle AOC=\alpha+\beta$. Both are central angles in $(0,\pi)$, where sine is positive, so we may take the positive root:

$$\sin\alpha=\sqrt{1-\frac{49}{64}}=\frac{\sqrt{15}}{8}, \qquad \sin\beta=\sqrt{1-\frac{289}{324}}=\frac{\sqrt{35}}{18}$$

Now the cosine addition formula:

$$\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\frac{7}{8}\cdot\frac{17}{18}-\frac{\sqrt{15}}{8}\cdot\frac{\sqrt{35}}{18}=\frac{119-\sqrt{525}}{144}=\frac{119-5\sqrt{21}}{144}$$

using $\sqrt{525}=\sqrt{25\times 21}=5\sqrt{21}$.
**Step 3: Cosine rule once more, on triangle $OAC$.**

$$AC^2=6^2+6^2-2(6)(6)\cos(\alpha+\beta)=72-72\cdot\frac{119-5\sqrt{21}}{144}=72-\frac{119-5\sqrt{21}}{2}$$

$$AC^2=\frac{144-119+5\sqrt{21}}{2}=\frac{25+5\sqrt{21}}{2}=\frac{50+10\sqrt{21}}{4}$$

Step 4: Recognise the perfect square. The surd untangles because $50+10\sqrt{21}$ is a square:

$$\left(\sqrt{15}+\sqrt{35}\right)^2=15+35+2\sqrt{15\cdot 35}=50+2\sqrt{525}=50+10\sqrt{21}$$

$$AC=\frac{\sqrt{15}+\sqrt{35}}{2}\approx 4.89\text{ m}$$

A quick plausibility check: $AC$ must be less than $AB+BC=5$ (triangle inequality) and more than each chord, and $4.89$ sits exactly there. The distractors fail this outright - 8 m and 6 m both exceed the straight-line distance $AB+BC=5$, which is impossible.

Final Answer: $\frac{\sqrt{15} + \sqrt{35}}{2}$ m

The Setup: Absolute values mirror the graph across both axes, so this curve is a quadrilateral centred at the origin. Find the intercepts, confirm the shape, and the area follows from one formula.
Step 1: Find the intercepts. Setting $y=0$:

$$2|x|=6 \implies |x|=3 \implies x=\pm 3$$

Setting $x=0$:

$$3|y|=6 \implies |y|=2 \implies y=\pm 2$$

Step 2: Identify the shape. In each quadrant the equation is linear - for instance $2x+3y=6$ when both are positive - so the curve is four straight segments joining $(3,0)$, $(0,2)$, $(-3,0)$ and $(0,-2)$. Each side runs between an $x$-intercept and a $y$-intercept, so every side has length $\sqrt{3^2+2^2}=\sqrt{13}$: all four equal, making it a genuine rhombus with perpendicular diagonals along the axes.
Step 3: Measure the diagonals.

$$d_1=3-(-3)=6, \qquad d_2=2-(-2)=4$$

Step 4: Compute the area. For a rhombus (indeed any quadrilateral with perpendicular diagonals), the area is half the product of the diagonals:

$$\text{Area}=\frac{1}{2}d_1d_2=\frac{1}{2}(6)(4)=12$$

Equivalently, the first-quadrant piece is a right triangle of legs 3 and 2, area $3$, and symmetry gives four of them: $4\times 3=12$. Two routes, same number.

Final Answer: 12 square units

The Setup: A locus question. When a point moves so that the ratio of its distances to two fixed points stays constant, the path is completely determined by that constant - this is the Circle of Apollonius.
Step 1: State the condition. Let the fixed points be $A$ and $B$ with $AB=1$ m, and let $P$ be the cow. The condition is:

$$\frac{PA}{PB}=\frac{3}{2}$$

Step 2: Apply the Apollonius result. For $\frac{PA}{PB}=k$ with $k>0$:
* If $k=1$ the locus is the perpendicular bisector of $AB$ - a straight line, since equal distances means equidistant.
* If $k\neq 1$ the locus is a circle.

Here $k=\frac{3}{2}\neq 1$, so the cow traces a circle.
Step 3: See it concretely rather than just quoting the theorem. Put $A$ at the origin and $B$ at $(1,0)$, and let $P=(x,y)$. Then $PA^2=\tfrac{9}{4}PB^2$ gives:

$$4\left(x^2+y^2\right)=9\left((x-1)^2+y^2\right) \implies 5x^2+5y^2-18x+9=0$$

$$x^2+y^2-\frac{18}{5}x+\frac{9}{5}=0 \implies \left(x-\frac{9}{5}\right)^2+y^2=\left(\frac{6}{5}\right)^2$$

That is unmistakably a circle - centre $\left(1.8,\,0\right)$, radius $1.2$ m. The two points where it crosses the line $AB$ are the internal and external division points of $AB$ in the ratio $3:2$, namely $x=0.6$ and $x=3$, which is exactly the diameter $[0.6,\,3]$ of width $2.4$.
Note the circle is not centred on $AB$'s midpoint and does not pass through $A$ or $B$ - a useful check that this really is Apollonius and not the perpendicular bisector.