These points will make you crazy!

Geometry Level 5

Let A 1 A_{1} , A 2 A_{2} , A 3 A_{3} , A 4 A_{4} and A 5 A_{5} be five points in a plane whose coordinates are ( 1 , − 1 ) (1,-1) , ( 2 , − 3 ) (2,-3) , ( 3 , − 2 ) (3,-2) , ( − 10 , − 4 ) (-10,-4) and ( 4 , 10 ) (4,10) respectively.

A 1 A 2 A_{1}A_{2} is bisected at B 1 B_{1} ; B 1 A 3 \ B_{1}A_{3} is divided at B 2 B_{2} in the ratio 1 : 2 1:2 ; B 2 A 4 \ B_{2}A_{4} is divided at B 3 B_{3} in the ratio 1 : 3 1:3 and B 3 A 5 B_{3}A_{5} is divided at B 4 B_{4} in the ratio 1 : 4 1:4 .

Given that B 4 B_{4} is the circumcenter of △ K L M \triangle KLM with coordinates of K = ( sin ⁡ θ , cos ⁡ θ ) K=(\sin \theta,\cos \theta) , L = ( a , b ) L=(a,b) and M = ( c , d ) M=(c,d) .

Find the value of a 2 + b 2 + c 2 + d 2 a^{2}+b^{2}+c^{2}+d^{2} .


The answer is 2.

Harsh Shrivastava by Harsh Shrivastava , 20, India

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1 solution

From the given proportions, we get the co-ordinates of B n . n = 1 , 2 , 3 , 4. A 1 ( 1 , − 1 ) . . . . . . . . . . . . . . A 2 ( 2 , − 3 ) , r a t i o 1 : 1 ∴ B 1 ( 3 / 2 , − 2 ) . B 1 ( 3 / 2 , − 2 ) . . . . . . . . . . . A 3 ( 3 , − 2 ) , r a t i o 1 : 2 ∴ B 2 ( 2 , − 2 ) . B 2 ( 2 , − 2 ) . . . . . . . . . . . . A 4 ( − 10 , − 4 ) , r a t i o 1 : 3 ∴ B 3 ( − 1 , − 5 / 2 ) . B 3 ( − 1 , − 5 / 2 ) . . . . . . . . . . A 5 ( 4 , 10 ) , r a t i o 1 : 4 ∴ B 4 ( 0 , 0 ) . ∴ c i r c u m r a d i u s R 2 = ( a − 0 ) 2 + ( b − 0 ) 2 = ( c − 0 ) 2 + ( d − 0 ) 2 = C o s 2 θ + s i n 2 θ = 1. ⟹ a 2 + b 2 + c 2 + d 2 = 1 + 1 = 2 \text{From the given proportions, we get the co-ordinates of } B_n. n=1, 2, 3, 4.\\ A_1(1,-1)..............A_2(2,-3), \ \ \ \ \ \ ratio\ \ 1:1 \ \ \ \ \ \therefore\ \ B_1(3/2,-2).\\ B_1(3/2,-2)...........A_3(3,-2), \ \ \ \ \ \ ratio\ \ 1:2\ \ \ \ \ \therefore\ \ B_2(2,-2).\\ B_2(2,-2)............A_4(-10,-4), \ \ \ ratio\ \ 1:3\ \ \ \ \ \therefore\ \ B_3(-1,-5/2).\\ B_3(-1,-5/2)..........A_5(4,10), \ \ \ \ \ \ ratio\ \ 1:4\ \ \ \ \ \therefore\ \ B_4(0,0).\\ \therefore\ circumradius\ R^2=(a-0)^2+(b-0)^2=(c-0)^2+(d-0)^2=Cos^2\theta+sin^2\theta=1.\\ \implies\ a^2+b^2+c^2+d^2=1+1=\Large\ \ \ \ \ \color{#D61F06}{2}

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