IPMAT Indore 2023 · Quantitative Ability · Question 14

Step 1 — write the area in terms of y alone. AB lies along the x-axis with length 4, so for C=\left( x,y \right) the height is \left| y \right|:

\left[ \triangle ABC \right]=\frac{1}{2}\times 4\times \left| y \right|=2\left| y \right|

Step 2 — find the range of the area. C lies on the circle of radius 2, so \left| y \right|\le 2 and

0\le \text{area}\le 4

Step 3 — list the integer areas. The candidates are 0,1,2,3,4. But area 0 means y=0, i.e. C=\left( \pm 2,0 \right)=A or B — then A, B, C are collinear and there is no triangle. So discard it.

\text{area}\in \{1,2,3,4\}

Step 4 — count the points for each area.

The last row is the special case: \left| y \right|=2 forces x=0, giving the two poles \left( 0,\pm 2 \right) rather than four points.

Step 5 — total.

4+4+4+2=14

Answer: 14

Shortcut. Area =2\left| y \right|, so the question is really “how many points have \left| y \right|=1/2,1,3/2,2?”

Each horizontal line y=\pm h with 0<h<2 cuts the circle in two points, giving 4 points per area value — except at the top and bottom, where the line is tangent and gives one point each.

3\times 4 + 1\times 2=14

Two things decide the answer, and both are easy to miss: the maximum area is 4 (not 2 — don’t forget the factor 1/2\times 4=2), and the extreme case \left| y \right|=2 yields two points rather than four. Answering 16 means you treated the poles like the rest; answering 8 means you capped the area at 2.