IPMAT Indore 2024 · Quantitative Ability · Question 19

Step 1 — place the figure. Let A\left( 0,0 \right), B\left( k,0 \right), C\left( k/2,k\sqrt{3}/2 \right). The circle has centre O\left( k/2,0 \right) and radius k/2.

Step 2 — find where the circle meets the sides. The circle on diameter AB passes through any point seeing AB at {90}^{∘}. Let M be the midpoint of AC. Then \angle AMB={90}^{∘} (the median from B in an equilateral triangle is also the altitude), so M lies on the circle. By symmetry the midpoint N of BC lies on it too.

Step 3 — decompose the region. The part of the triangle inside the circle is

\underset{\text{quadrilateral}AMNB}{\underset{⏟}{\triangle AMNB}}+ \underset{\text{bulge beyond}MN}{\underset{⏟}{\text{circular segment on}MN}}

It is cleaner to split as: triangle AOM + triangle BON + circular sector MON.

OA=OB=OM=ON=k/2.

\angle AOM: M is the midpoint of AC, so OM\parallel BC (midsegment), giving \angle AOM={60}^{∘}. Likewise \angle BON={60}^{∘}, leaving \angle MON={60}^{∘}.

Step 4 — compute each piece.

\left[ \triangle AOM \right]=\frac{1}{2}\cdot \frac{k}{2}\cdot \frac{k}{2}\cdot \sin{60}^{∘}=\frac{{k}^{2}\sqrt{3}}{16}

and the same for \left[ \triangle BON \right]. The sector:

\left[ \text{sector}MON \right]=\frac{60}{360}\pi{\left( \frac{k}{2}\right)}^{2}=\frac{\pi{k}^{2}}{24}

Step 5 — add.

\text{Area}=2\cdot \frac{{k}^{2}\sqrt{3}}{16}+\frac{\pi{k}^{2}}{24}=\frac{{k}^{2}\sqrt{3}}{8}+\frac{\pi{k}^{2}}{24}=\frac{3\sqrt{3}{k}^{2}+\pi{k}^{2}}{24}=\left( 3\sqrt{3}+\pi \right)\frac{{k}^{2}}{24}

Numerical check (k=1): \left( 5.196+3.142 \right)/24\approx 0.347. A grid count over the figure gives 0.3474 ✔

Answer: Option 3 — \left( 3\sqrt{3}+\pi \right)\frac{{k}^{2}}{24}

Shortcut. Eliminate, don’t compute. The region is a decent chunk of the triangle, whose area is \frac{\sqrt{3}}{4}{k}^{2}\approx 0.433{k}^{2}. Test each option at k=1:

\left( 3\sqrt{3}+\pi \right)/6\approx 1.39 — bigger than the whole triangle. Out.

\left( 3\sqrt{3}-\pi \right)/24\approx 0.086 — far too small. Out.

\left( 3\sqrt{3}-\pi \right)/6\approx 0.342 and \left( 3\sqrt{3}+\pi \right)/24\approx 0.347 — both plausible.

Now the clincher: the region is a sector plus two triangles, so the \pi term must be added, not subtracted. That leaves \left( 3\sqrt{3}+\pi \right){k}^{2}/24. Sign of the \pi term is the whole question.