IPMAT Indore 2019 — SA

The Setup: A pure geometry fundamental. The sum of the interior angles of any convex $n$-sided polygon is the GOAT formula $(n-2)\times 180^\circ$. Set up the inequality, solve, and take the largest integer that survives.
Step 1: Set up the inequality. The sum is strictly less than $2019^\circ$:

$$(n-2)\times 180 < 2019$$

Step 2: Isolate the variable. Divide both sides by 180 (positive, so the inequality direction is safe):

$$n-2<\frac{2019}{180}$$

Step 3: Crunch the division. Since $180\times 11=1980$ and $180\times 12=2160$, the quotient sits between 11 and 12:

$$\frac{2019}{180}=11.216\ldots \implies n<13.216\ldots$$

Step 4: Lock in the maximum integer. A polygon has a whole number of sides, so the largest admissible $n$ is 13.
Step 5: Verify both sides of the boundary. A bound is only trustworthy if the next value up actually fails:
* $n=13$: sum $=(13-2)\times 180=1980^\circ$, and $1980<2019$. Valid.
* $n=14$: sum $=(14-2)\times 180=2160^\circ$, and $2160>2019$. Fails.

So 13 works and 14 does not, which pins the maximum exactly.

Final Answer: 13

The Setup: Log base inception. The trick is to convert that awkward $1$ in each denominator into a logarithm sharing the same base, merge it with the term beside it, and then flip each fraction with the reciprocal rule.
**Step 0: Note why $a,b,c>1$ matters.** A logarithm needs a positive base other than 1. Because every one of $a,b,c$ exceeds 1, each base $a^2b$, $b^2c$, $c^2a$ and the eventual base $abc$ is greater than 1, so every logarithm here is defined and the reciprocal rule is legal. That condition in the stem is doing real work, not decoration.
Step 1: Convert the 1s. Using $1=\log_x x$ with the base already present:

$$1+\log_{a^2b}\left(\frac{c}{a}\right)=\log_{a^2b}(a^2b)+\log_{a^2b}\left(\frac{c}{a}\right)$$

Step 2: Merge the logs. By the product rule $\log x+\log y=\log(xy)$, and noting that one factor of $a$ cancels against the $\frac{1}{a}$:

$$a^2b\times\frac{c}{a}=a^{2-1}bc=abc \implies 1+\log_{a^2b}\left(\frac{c}{a}\right)=\log_{a^2b}(abc)$$

The same collapse happens in the other two, each landing on the identical argument $abc$:

$$b^2c\times\frac{a}{b}=abc, \qquad c^2a\times\frac{b}{c}=abc$$

So the whole expression becomes:

$$\frac{1}{\log_{a^2b}(abc)}+\frac{1}{\log_{b^2c}(abc)}+\frac{1}{\log_{c^2a}(abc)}$$

Step 3: Flip the base. Apply the reciprocal rule $\frac{1}{\log_x y}=\log_y x$, which swaps base and argument:

$$\log_{abc}(a^2b)+\log_{abc}(b^2c)+\log_{abc}(c^2a)$$

Step 4: The final fusion. They now share a base, so the sum becomes a single log of the product:

$$\log_{abc}\left(a^2b\cdot b^2c\cdot c^2a\right)=\log_{abc}\left(a^3b^3c^3\right)=\log_{abc}\left((abc)^3\right)$$

Pull the exponent out front:

$$3\log_{abc}(abc)=3\times 1=3$$

Note the answer is a bare constant - it does not depend on $a$, $b$ or $c$ at all, which is the elegance the question is built around.

Final Answer: 3

The Setup: Functional equations are the final boss of algebra. The standard strat is to substitute a convenient value to collapse one variable, and here $x=0$ does something better than simplify - it *proves* the function is linear rather than making us assume it.
**Step 1: Substitute $x=0$.**

$$f(0)f(y)=f(3)+3f(y)-3f(y)+6y \implies f(0)f(y)=f(3)+6y$$

Write $c=f(0)$, so that $c\,f(y)=6y+f(3)$.
**Step 2: Rule out $c=0$ before dividing.** If $c$ were 0, the left side would be 0 for every $y$, forcing $6y+f(3)=0$ for all real $y$ - impossible, since the left side varies with $y$ and the right is fixed. So $c\neq 0$, and we may divide:

$$f(y)=\frac{6}{c}y+\frac{f(3)}{c}$$

This is a genuine deduction, not a guess: $f$ must be linear. Write $f(x)=kx+c$, noting $f(0)=c$ is consistent and that $k=\frac{6}{c}$, i.e. $ck=6$.
Step 3: Substitute the linear form into the original relation.

$$\text{LHS}=(kx+c)(ky+c)=k^2xy+kcx+kcy+c^2$$

$$\text{RHS}=\left[k(2xy+3)+c\right]+3\left[k(x+y)+c\right]-3\left[ky+c\right]+6y=2kxy+3kx+6y+3k+c$$

The $3ky$ and $-3ky$ cancel, which is why no $y$-term survives on the right beyond the $6y$.
Step 4: Match all four coefficients - not just one.

From $k^2=2k$ we get $k=0$ or $k=2$; $k=0$ contradicts $ck=6$, so $k=2$. Then $kc=6$ gives $c=3$. Check the two remaining rows hold rather than assuming they do: $kc=6=3k$ ✓, and $c^2=9=3(2)+3$ ✓. All four agree, so:

$$f(x)=2x+3$$

Step 5: Execute.

$$f(8)=2(8)+3=19$$

Final Answer: 19

The Setup: A determinant-properties speedrun. Four rules do all the work: $\det(kM)=k^n\det M$ for an $n\times n$ matrix, $\det(XY)=\det X\det Y$, $\det(X^{-1})=\frac{1}{\det X}$, and $\det(X^T)=\det X$.
Step 1: Extract the scalar. The matrices are $4\times 4$, so $n=4$ and the scalar 2 comes out to the fourth power - not the first, which is the classic slip here:

$$\det\left(2AB^{-1}C^3B^T\right)=2^4\det\left(AB^{-1}C^3B^T\right)=16\det\left(AB^{-1}C^3B^T\right)$$

Step 2: Shatter the product. The determinant is multiplicative, so it distributes across the four factors:

$$16\times\det A\times\det\left(B^{-1}\right)\times\det\left(C^3\right)\times\det\left(B^T\right)$$

Step 3: Substitute the properties. $\det(B^{-1})=\frac{1}{\det B}$, $\det(C^3)=(\det C)^3$, and $\det(B^T)=\det B$:

$$16\times\det A\times\frac{1}{\det B}\times(\det C)^3\times\det B$$

Step 4: Cancel and calculate. The transpose contributes $\det B$ and the inverse contributes $\frac{1}{\det B}$, so they annihilate each other:

$$16\times\det A\times(\det C)^3=16\times 5\times\left(\frac{1}{2}\right)^3=80\times\frac{1}{8}=10$$

**Step 5: Notice what $\det B=-3$ was for.** Its *value* never reaches the answer - it cancels. But its being non-zero is exactly what guarantees $B$ is invertible, so $B^{-1}$ exists and the expression is meaningful at all. It is a load-bearing hypothesis disguised as a red herring.

Final Answer: 10

The Setup: A binomial expansion on matrices. The condition $A^2=0$ (a nilpotent matrix) is a kill switch: every power of $A$ from the second onward vanishes, truncating a 51-term expansion into two terms.
Step 0: Check the binomial theorem is even allowed. For matrices, $(X+Y)^n$ expands binomially **only if $X$ and $Y$ commute** - otherwise the cross terms will not collect. Here $X=I$, and the identity commutes with every matrix, so $IA=AI=A$ and the expansion is valid. Skipping this check is how the same move goes wrong on a harder problem.
Step 1: Expand.

$$(I+A)^{50}=I^{50}+\binom{50}{1}I^{49}A+\binom{50}{2}I^{48}A^2+\cdots+A^{50}$$

Step 2: Trigger the kill switch. From $A^2=0$ it follows that $A^3=A\cdot A^2=A\cdot 0=0$, and inductively every higher power is zero too. So every term from the $A^2$ one onward dies, and $I^m=I$:

$$(I+A)^{50}=I+50A$$

Step 3: Simplify the expression.

$$(I+A)^{50}-50A=(I+50A)-50A=I$$

The $50A$ subtraction is built into the question precisely to cancel the surviving term, leaving the identity.
Step 4: Take the determinant. The determinant of the identity matrix is 1 in any dimension:

$$\det(I)=1$$

Note the answer never depended on *which* nilpotent $A$ was chosen, only on $A^2=0$ - the 'non-zero' in the stem just stops the problem being trivial.

Final Answer: 1

The Setup: A linear equation dressed as a ratio problem. Define the shares through a single multiplier, subtract what was eaten, and set the remainder equal to the largest original share.
Step 1: Define the shares. Let the ratio multiplier be $x$, so the three friends received $3x$, $5x$ and $7x$ apples.

$$\text{Total at the start}=3x+5x+7x=15x$$

The largest individual share is $7x$.
Step 2: Construct the equation. After 16 apples are eaten, $15x-16$ remain, and the stem says this equals the largest starting share:

$$15x-16=7x$$

Step 3: Solve for the multiplier.

$$15x-7x=16 \implies 8x=16 \implies x=2$$

Step 4: Answer the question that was asked. The prompt wants the initial total, not the multiplier and not one person's share:

$$\text{Total}=15x=15(2)=30$$

Step 5: Verify against the story. With $x=2$ the shares are 6, 10 and 14, totalling 30. Eat 16 and $30-16=14$ remain - exactly the largest starting share of 14. The story checks out.

Final Answer: 30

The Setup: Revenue is $\text{Price}\times\text{Quantity}$, so a percentage change in each multiplies rather than adds. Anyone who subtracts $25\%-20\%=5\%$ has fallen for the trap this question exists to set.
Step 1: Define the baseline. Let the original price be $P$ and quantity $Q$, so the original revenue is $R_1=PQ$.
Step 2: Apply the percentage shifts as multipliers.
* Price falls 25%: new price $=(1-0.25)P=0.75P$
* Quantity rises 20%: new quantity $=(1+0.20)Q=1.20Q$

Step 3: Calculate the new revenue.

$$R_2=(0.75P)(1.20Q)=(0.75\times 1.20)PQ=0.90\,PQ$$

Step 4: Convert to a percentage drop. Revenue went from $1.00PQ$ to $0.90PQ$, a fall of $0.10PQ$:

$$\frac{R_1-R_2}{R_1}\times 100=\frac{0.10PQ}{PQ}\times 100=10\%$$

So $x=10$. Sanity check with real numbers: at $P=100$, $Q=100$, revenue goes from 10,000 to $75\times 120=9{,}000$ - down 1,000, which is 10%.

Final Answer: 10

The Setup: To find the range of a rational function, set it equal to $y$, rearrange into a quadratic in $x$, and force the discriminant to be non-negative - because $x$ is required to be real. The values of $y$ that survive are exactly the range.
Step 0: Confirm the function is defined everywhere. The denominator $x^2+2x+4$ has discriminant $2^2-4(1)(4)=-12<0$, so it never hits zero and the expression is defined for all real $x$. Without this, chunks of the range argument could be vacuous.
**Step 1: Create the quadratic in $x$.**

$$y=\frac{3x^2-6x+12}{x^2+2x+4} \implies y(x^2+2x+4)=3x^2-6x+12$$

Collecting terms:

$$x^2(y-3)+x(2y+6)+(4y-12)=0$$

**Step 2: Handle the $y=3$ case separately - the discriminant method assumes a genuine quadratic.** If $y=3$ the $x^2$ coefficient vanishes and this is not a quadratic at all; it degenerates to $12x+0=0$, giving $x=0$. So $y=3$ is attained, at $x=0$, and indeed $\frac{12}{4}=3$. It sits inside the range we are about to derive, so the final answer is unaffected - but the case has to be dismissed explicitly, not ignored.
**Step 3: Enforce the discriminant for $y\neq 3$.** Real $x$ requires $b^2-4ac\geq 0$. Noticing that $4y-12=4(y-3)$ keeps this tidy:

$$(2y+6)^2-4(y-3)\cdot 4(y-3)\geq 0 \implies (2y+6)^2-16(y-3)^2\geq 0$$

Since $(2y+6)^2=4(y+3)^2$, divide throughout by 4:

$$(y+3)^2-4(y-3)^2\geq 0$$

Step 4: Difference of squares. With $A=y+3$ and $B=2(y-3)$:

$$\left[(y+3)-2(y-3)\right]\left[(y+3)+2(y-3)\right]\geq 0$$

$$(-y+9)(3y-3)\geq 0 \implies 3(9-y)(y-1)\geq 0$$

Dividing by 3 and negating the first bracket flips the inequality:

$$(y-9)(y-1)\leq 0 \implies 1\leq y\leq 9$$

Step 5: Confirm both endpoints are actually reached. An inequality gives bounds; it does not by itself prove they are attained, and the stem's wording requires that the function really does climb to $k$.
* $y=9$: the quadratic becomes $6x^2+24x+24=0$, i.e. $(x+2)^2=0$, so $x=-2$. Check: $\frac{12+12+12}{4-4+4}=\frac{36}{4}=9$.
* $y=1$: it becomes $-2x^2+8x-8=0$, i.e. $(x-2)^2=0$, so $x=2$. Check: $\frac{12-12+12}{4+4+4}=\frac{12}{12}=1$.

Both are hit exactly once, at the two points where the discriminant is zero. The maximum value is therefore $k=9$.

Final Answer: 9

The Setup: A coordinate geometry freebie. The farthest point of a circle from an external point lies on the line through the point and the centre, on the far side - so the maximum distance is the distance to the centre plus the radius.
Step 1: Deconstruct the circle. From $x^2+y^2=4$, the centre is the origin $C(0,0)$ and the radius is $r=\sqrt{4}=2$.
Step 2: Distance from the point to the centre. Both $P(-5,0)$ and $C(0,0)$ lie on the x-axis, so the distance is just the gap in x-coordinates:

$$d=|-5-0|=5$$

Step 3: Check the point is outside. Since $d=5>r=2$, the point $P$ lies outside the circle. (For a point *inside*, the maximum would still be $d+r$, but the minimum formula flips - worth knowing which case you are in.)
Step 4: Add the radius. Travel from $P$ through the centre and out to the opposite edge:

$$\text{Max distance}=d+r=5+2=7$$

Concretely, the farthest point is $(2,0)$, and the distance from $(-5,0)$ to $(2,0)$ is indeed 7 - while the nearest point $(-2,0)$ sits only 3 away.

Final Answer: 7

The Setup: A crossover between logarithms and arithmetic progressions. The chained equality $x^{12}=y^{16}=z^{24}$ locks all three variables together, so we can rewrite everything in terms of a single variable and the logarithms collapse into plain numbers.
**Step 0: Note $z\neq 1$.** If $z$ were 1 then $x=y=z=1$ and every logarithm here would have base 1, which is undefined. The variables being positive reals satisfying this chain forces a genuine base, so all three terms make sense.
**Step 1: Express everything through $z$.**
* From $x^{12}=z^{24}$, take the 12th root: $x=z^{2}$.
* From $y^{16}=z^{24}$, take the 16th root: $y=z^{24/16}=z^{3/2}$.

**Step 2: Evaluate the first term, $3\log_y x$.** Using $\log_{b^m}(b^n)=\frac{n}{m}$:

$$\log_y x=\log_{z^{3/2}}\left(z^{2}\right)=\frac{2}{3/2}=2\times\frac{2}{3}=\frac{4}{3} \implies 3\log_y x=3\times\frac{4}{3}=4$$

**Step 3: Evaluate the second term, $4\log_z y$.**

$$\log_z y=\log_z\left(z^{3/2}\right)=\frac{3}{2} \implies 4\log_z y=4\times\frac{3}{2}=6$$

**Step 4: Evaluate the third term, $n\log_x z$.**

$$\log_x z=\log_{z^{2}}(z)=\frac{1}{2} \implies n\log_x z=\frac{n}{2}$$

Step 5: Impose the AP condition. The three terms are $4$, $6$ and $\frac{n}{2}$. In an arithmetic progression consecutive gaps are equal, so the middle term is the average of its neighbours - equivalently, the common difference $6-4=2$ carries forward:

$$\frac{n}{2}=6+2=8 \implies n=16$$

Check the progression reads $4,\,6,\,8$ - a clean AP with common difference 2.

Final Answer: 16

The Setup: A trigonometric boss phase. Factor the first equation with sum-to-product identities to expose its root conditions, then run those through the absolute-value geometry of $|x|+|y|=1$ to see which survive.
Step 1: Expand and factor. Sum-to-product on the left, double angle on the right:

$$2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$

Bring everything to one side and pull out the common factor:

$$2\sin\left(\frac{x+y}{2}\right)\left[\cos\left(\frac{x-y}{2}\right)-\cos\left(\frac{x+y}{2}\right)\right]=0$$

Apply $\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)$ to the bracket, with $A=\frac{x-y}{2}$ and $B=\frac{x+y}{2}$, so that $\frac{A+B}{2}=\frac{x}{2}$ and $\frac{B-A}{2}=\frac{y}{2}$:

$$4\sin\left(\frac{x}{2}\right)\sin\left(\frac{y}{2}\right)\sin\left(\frac{x+y}{2}\right)=0$$

Step 2: Identify the root scenarios. A product is zero only if a factor is, giving three parallel timelines:

$$\sin\left(\frac{x}{2}\right)=0\implies x=2n\pi, \qquad \sin\left(\frac{y}{2}\right)=0\implies y=2n\pi, \qquad \sin\left(\frac{x+y}{2}\right)=0\implies x+y=2n\pi$$

Step 3: Apply the domain constraint. On $|x|+|y|=1$ we have $|x|\leq 1$, $|y|\leq 1$, and by the triangle inequality $|x+y|\leq|x|+|y|=1$. Since $2\pi\approx 6.28$ is far outside that, **only $n=0$ survives** in all three families, collapsing them to $x=0$, $y=0$, or $x+y=0$.
Step 4: Hunt the coordinates. Intersect each line with the diamond $|x|+|y|=1$:
* $x=0 \implies |y|=1$: gives $(0,1)$ and $(0,-1)$
* $y=0 \implies |x|=1$: gives $(1,0)$ and $(-1,0)$
* $x=-y \implies 2|x|=1$: gives $\left(\tfrac{1}{2},-\tfrac{1}{2}\right)$ and $\left(-\tfrac{1}{2},\tfrac{1}{2}\right)$

Step 5: Check for double-counting before adding. The only point that could belong to two of these lines is $(0,0)$, where $x=0$ and $y=0$ meet - but it fails $|x|+|y|=1$, so it never enters the list. The six points above are pairwise distinct, and the count is genuinely $2+2+2$.

Final Answer: 6

The Setup: A circle with a floating parameter $k$. We must force it to sit clear of both axes while keeping the point $(1,4)$ inside or on its boundary, then read off the two ends of the resulting interval.
Step 1: Extract the circle's stats. Complete the square:

$$(x-3)^2-9+(y-5)^2-25+k=0 \implies (x-3)^2+(y-5)^2=34-k$$

Centre $C(3,5)$ and radius $r=\sqrt{34-k}$.
Step 0 (implicit, but state it): the circle must exist. A real circle needs $r^2>0$, i.e. $34-k>0$, so $k<34$. Keep this in the background - it turns out to be implied by the stricter bound below, but a parameter range derived without it can silently include values where there is no circle at all.
Step 2: Keep it clear of the axes. The centre sits in the first quadrant, so the distance to the y-axis is its x-coordinate 3, and to the x-axis its y-coordinate 5. Avoiding both strictly (neither touching nor intersecting) requires:

$$r<3 \implies 34-k<9 \implies k>25$$

$$r<5 \implies 34-k<25 \implies k>9$$

The first is the stricter of the two, and it absorbs the second, so the binding condition is $k>25$ - an open bound, because touching is disallowed.
Step 3: Trap the point. 'does not lie outside' means on or inside the boundary, so the distance from $(1,4)$ to the centre is at most $r$:

$$d^2=(1-3)^2+(4-5)^2=4+1=5 \implies 5\leq 34-k \implies k\leq 29$$

This bound is closed, since lying exactly on the circle is permitted.
Step 4: Merge and read off. Combining, $25<k\leq 29$, i.e. the interval $(25,29]$ - open at the left, closed at the right, exactly matching the $(a,b]$ shape the question announces. That the two bounds come out with the right open/closed pairing is a good sign the conditions have been read correctly.

$$a=25,\quad b=29 \implies a+b=54$$

Final Answer: 54

The Setup: A matrix built under severe constraints. Three entries drawn from $\{+1,-1\}$ can only sum to 1 as $(+1)+(+1)+(-1)$ - no other combination works, since three $+1$s give 3 and anything with two $-1$s gives $-1$. So **every row and every column holds exactly two $+1$s and one $-1$. Step 1: See the structure.** Exactly one $-1$ per row and exactly one per column means the positions of the $-1$s form a permutation of the three columns across the three rows. There are therefore exactly $3!=6$ matrices satisfying the constraints - the problem is far more rigid than it first looks.
Step 2: Compute the prototype. Take the permutation that puts the $-1$s on the main diagonal:

$$M=\begin{pmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{pmatrix}$$

Row sums: $-1+1+1=1$. Column sums: likewise 1. A valid build. Expanding along the first row:

$$|M|=-1(1-1)-1(-1-1)+1(1-(-1))=-1(0)-1(-2)+1(2)=0+2+2=4$$

Step 3: Show the value is forced, not lucky. Every other valid matrix comes from this one by permuting rows (equivalently, by relabelling which column carries each row's $-1$). A row swap flips the sign of a determinant and nothing else, so across the whole family the determinant is $\pm 4$ and its absolute value is invariant at 4. Since $3!=6$ is a small family, this is checkable exhaustively rather than by assertion: all six matrices give $|\det|=4$, three with $\det=+4$ and three with $\det=-4$, matching the even and odd permutations.
That is exactly why the question asks for the *absolute* value - the determinant itself is not determined by the constraints, but its magnitude is.

Final Answer: 4

The Setup: A Sieve of Eratosthenes in disguise. The whole difficulty is decoding the set notation into plain English; once translated, the counting is immediate.
**Step 1: Decode $Q(k)$.** For a given $k$, the set $Q(k)$ collects every member of $P$ that is strictly greater than $k$ and divisible by $k$ - that is, the **proper multiples of $k$** lying inside $P$.
Step 2: Decode the union. The union $\bigcup_{k=2}^{25}Q(k)$ therefore holds every $x\in P$ admitting some divisor $k$ with $2\leq k<x$. Such an $x$ is precisely a composite number.
Both directions are worth confirming rather than assuming:
* If $x$ lands in the union, it has a divisor strictly between 1 and itself, so $x$ is composite.
* If $x\in P$ is composite, it has a prime factor $p\leq\sqrt{x}\leq\sqrt{25}=5$. That $p$ satisfies $2\leq p<x$ and lies in $P$, so $x$ genuinely gets caught. (Checking the largest case: $x=25$ is caught by $k=5$.)

So the union is exactly the composites in $P$ - nothing more, nothing less.
Step 3: Decode the subtraction. Removing the union from $P$ deletes every composite, leaving the numbers with no divisor other than 1 and themselves: the primes. Note $P$ starts at 2, so there is no 1 to worry about - 1 is neither prime nor composite and would have muddied the count had it been included.
Step 4: Count the survivors. The primes in $\{2,3,\ldots,25\}$ are:

$$2,\ 3,\ 5,\ 7,\ 11,\ 13,\ 17,\ 19,\ 23$$

That is a squad of 9.

Final Answer: 9

The Setup: Conservation of material dressed up as a 3D melting problem. Because the thickness is uniform and nothing is lost, the volume ratio equals the area ratio, and the whole thing reduces to plane geometry.
Step 1: Total source material. With the plate's radius $R$:

$$\text{Plate area}=\pi R^2$$

Step 2: Area of one tile. The tile is a right-angled isosceles triangle, whose longest side is the hypotenuse, and the stem sets that hypotenuse equal to $R$. Letting each of the two equal legs be $x$, Pythagoras gives: