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Olympiad workbook
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IOQM
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Vector Algebra

Olympiad workbook · IOQM · Vector Algebra

In parallelogram A B C D, A C=10 and B D=28 . The points K and L in...

In parallelogram ABCD,AC=10A B C D, A C=10 and BD=28B D=28. The points KK and LL in the plane of ABCDA B C D move in such a way that AK=BDA K=B D and BL=ACB L=A C. Let MM and NN be the midpoints of CKC K and DLD L, respectively. What is the maximum value of cot⁡2(∠BMD/2)+tan⁡2(∠ANC/2)?\cot ^2(\angle \mathrm{BMD} / 2)+\tan ^2(\angle \mathrm{ANC} / 2) ?In parallelogram A B C D, A C=10 and B D=28 . The points K and L in the plane of A B C D move in such a way that A K=BIn parallelogram A B C D, A C=10 and B D=28 . The points K and L in the plane of A B C D move in such a way that A K=B

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02

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In parallelogram A B C D, A C=10 and B D=28 . The points K and L in the plane of A B C D move in such a way that A K=B
CK1=18\mathrm{CK}_1=18 OM1=5+9=14∴∠BM1D=90∘(∠ in semicircle ) Similarly, CK2=10+28=38∴CM2=19AM2=9OM2=5+9=14∴∠BM2D=90∘(∠ in semicircle )\begin{aligned} & \mathrm{OM}_1=5+9=14 \\ & \therefore \quad \angle \mathrm{BM}_1 \mathrm{D}=90^{\circ}(\angle \text { in semicircle }) \\ & \text { Similarly, } \mathrm{CK}_2=10+28=38 \\ & \therefore \quad \mathrm{CM}_2=19 \\ & \quad \mathrm{AM}_2=9 \\ & \quad \mathrm{OM}_2=5+9=14 \\ & \therefore \quad \angle \mathrm{BM}_2 \mathrm{D}=90^{\circ}(\angle \text { in semicircle }) \end{aligned} If K1\mathrm{K}_1 or K2\mathrm{K}_2 moves ∠BMD\angle \mathrm{BMD} will increase By observation : ∠BMD2↑∴cot⁡(∠BMD2)↓\frac{\angle \mathrm{BMD}}{2} \uparrow \quad \therefore \cot \left(\frac{\angle \mathrm{BMD}}{2}\right) \downarrow ∴\therefore \quad For maximum ∠BMD=90∘\quad \angle \mathrm{BMD}=90^{\circ} cot⁡2(∠BMD2)=1\cot ^2\left(\frac{\angle \mathrm{BMD}}{2}\right)=1In parallelogram A B C D, A C=10 and B D=28 . The points K and L in the plane of A B C D move in such a way that A K=B
BL1=10,OL1=4,DL1=18,DN1=9,ON1=5∴∠AN1C=π2(∠ in semicircle )\begin{aligned} & \mathrm{BL}_1=10, \mathrm{OL}_1=4, \mathrm{DL}_1=18, \mathrm{DN}_1=9, \mathrm{ON}_1=5 \\ & \therefore \angle \mathrm{AN}_1 \mathrm{C}=\frac{\pi}{2}(\angle \text { in semicircle }) \end{aligned} For L2 point, DL2=14+14+10DL2=38DN2=19ON2=5\begin{aligned} & \mathrm{DL}_2=14+14+10 \\ & \mathrm{DL}_2=38 \\ & \mathrm{DN}_2=19 \\ & \mathrm{ON}_2=5 \end{aligned} ∠AN2C=π2(∠ in semicircle )\angle \mathrm{AN}_2 \mathrm{C}=\frac{\pi}{2}(\angle \text { in semicircle }) Now if L1L_1 or L2L_2 moves, ∠ANC↓∠ANC2↓tan⁡∠ANC2↓ (Observation) \angle \mathrm{ANC} \downarrow \quad \frac{\angle \mathrm{ANC}}{2} \downarrow \quad \tan \frac{\angle \mathrm{ANC}}{2} \downarrow \text { (Observation) } Hence for maximum, ∠ANC=π2∴tan⁡2(π4)=1∴ Maximum =1+1=2\begin{aligned} & \angle A N C=\frac{\pi}{2} \\ & \therefore \quad \tan ^2\left(\frac{\pi}{4}\right)=1 \\ & \therefore \quad \text { Maximum }=1+1=2 \end{aligned}
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