IPMAT Indore Solids — PYPs

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:

$$x^2+x^2=R^2 \implies 2x^2=R^2 \implies x^2=\frac{R^2}{2}$$

The legs are perpendicular, so they serve directly as base and height:

$$\text{Tile area}=\frac{1}{2}\cdot x\cdot x=\frac{x^2}{2}=\frac{1}{2}\cdot\frac{R^2}{2}=\frac{R^2}{4}$$

Step 3: Divide. Note $R$ cancels completely - the answer does not depend on the plate's size, only on the stated relationship between the two:

$$n=\frac{\pi R^2}{\frac{R^2}{4}}=4\pi$$

Step 4: Filter for whole tiles. With $\pi\approx 3.14159$:

$$n\approx 4\times 3.14159=12.566$$

The stem asks for whole tiles, and you cannot cast a fraction of one, so the count floors to 12 - with material worth about $0.57$ of a tile left unused. Rounding to 13 here is the trap: that would require metal that does not exist.

Final Answer: 12

The Setup: A Diophantine equation dressed as geometry. Convert the surface-area condition into a sum of three squares, find every integer solution, then convert back to volumes.
Step 1: Reduce the surface-area equation. A cube of edge $a$ has surface area $6a^2$, so with edges $a,b,c$:

$$6a^2+6b^2+6c^2=564 \implies a^2+b^2+c^2=94$$

Step 2: Bound the search before starting it. Order the edges $a\leq b\leq c$, with each at least 1 (an edge of 0 is not a cube). Then $c$ is the largest, so $3c^2\geq 94$ and $c^2\leq 94$:

$$\frac{94}{3}\leq c^2\leq 94 \implies 31.3\leq c^2\leq 94 \implies 6\leq c\leq 9$$

That leaves exactly four cases to test, which is what makes the search genuinely exhaustive rather than a lucky hunt.
Step 3: Test all four.

So there are exactly two admissible triples, and no more.
Step 4: Convert to volumes.

$$\{2,3,9\}: \quad 2^3+3^3+9^3=8+27+729=764$$

$$\{3,6,7\}: \quad 3^3+6^3+7^3=27+216+343=586$$

Both triples do give surface area $6(94)=564$ ✓, so both are genuine and the answer must list both values.

Final Answer: $764 \ \text{cm}^3$ and $586 \ \text{cm}^3$