IPMAT Indore 2023 · Quantitative Ability · Question 16

Step 1 — set the sides. Let AB=2k, BC=4k, CA=3k.

Step 2 — find BM with the projection formula. M is the foot of the altitude from A, so BM is the projection of AB onto BC:

BM=\frac{A{B}^{2}+B{C}^{2}-C{A}^{2}}{2\cdot BC}=\frac{4{k}^{2}+16{k}^{2}-9{k}^{2}}{8k}=\frac{11{k}^{2}}{8k}=\frac{11k}{8}

(This is just the cosine rule rearranged: BM=AB\cos B.)

Step 3 — find AD with the median formula. D is the midpoint of BC, so AD is the median from A:

A{D}^{2}=\frac{2A{B}^{2}+2C{A}^{2}-B{C}^{2}}{4}=\frac{8{k}^{2}+18{k}^{2}-16{k}^{2}}{4}=\frac{10{k}^{2}}{4}

AD=\frac{k\sqrt{10}}{2}

Step 4 — take the ratio.

\frac{BM}{AD}=\frac{11k/8}{k\sqrt{10}/2}=\frac{11}{8}\times \frac{2}{\sqrt{10}}=\frac{11}{4\sqrt{10}}

BM:AD=11:4\sqrt{10}

Sanity check: 4\sqrt{10}\approx 12.65, so the ratio is a little under 11:12 — consistent with options (b) and (c) being close, and (c) being the exact one.

Answer: Option 3 — 11:4\sqrt{10}

Shortcut. Two formulas, both worth memorising cold:

\text{projection:}BM=\frac{{c}^{2}+{a}^{2}-{b}^{2}}{2a}\text{median:}{m}_{a}^{2}=\frac{2{b}^{2}+2{c}^{2}-{a}^{2}}{4}

With sides 2,4,3 they give 11/8 and \sqrt{10}/2 directly.

Then use the options. A surd appears in AD but not in BM, so the answer must contain exactly one surd — which kills (a) and (b) instantly. Between (c) and (d), BM<AD numerically (1.375<1.58), so the smaller number comes first: 11:4\sqrt{10}.