원 O O O A B C D ABCD A B C D P P P P P P E E E F F F G G G H H H A E P H AEPH A E P H E B F P EBFP E B F P P F C G PFCG P F C G H P G D HPGD H P G D S 1 S_1 S 1 S 2 S_2 S 2 S 3 S_3 S 3 S 4 S_4 S 4 S 1 + S 2 + S 3 + S 4 S_1+S_2+S_3+S_4 S 1 + S 2 + S 3 + S 4 O O O
A H ‾ = x \overline{AH}=x A H = x H D ‾ = y \overline{HD}=y H D = y D G ‾ = u \overline{DG}=u D G = u G C ‾ = v \overline{GC}=v GC = v x + y = u + v x+y=u+v x + y = u + v
또, S 1 + S 2 + S 3 + S 4 = ( 가 ) ( x 2 + y 2 + u 2 + v 2 ) S_1+S_2+S_3+S_4=\boxed{(\char"AC00 )}(x^2+y^2+u^2+v^2) S 1 + S 2 + S 3 + S 4 = ( 가 ) ( x 2 + y 2 + u 2 + v 2 )
원 O O O ( 나 ) ( x + y ) 2 \boxed{(\char"B098 )}(x+y)^2 ( 나 ) ( x + y ) 2
한편 2 ( x 2 + y 2 + u 2 + v 2 ) − ( 다 ) ( x + y ) 2 2(x^2+y^2+u^2+v^2)-\boxed{(\char"B2E4 )}(x+y)^2 2 ( x 2 + y 2 + u 2 + v 2 ) − ( 다 ) ( x + y ) 2
= ( x − u ) 2 + ( x − v ) 2 + ( y − u ) 2 + ( y − v ) 2 ≥ 0 =(x-u)^2+(x-v)^2+(y-u)^2+(y-v)^2\ge0 = ( x − u ) 2 + ( x − v ) 2 + ( y − u ) 2 + ( y − v ) 2 ≥ 0
이므로 S 1 + S 2 + S 3 + S 4 S_1+S_2+S_3+S_4 S 1 + S 2 + S 3 + S 4 O O O
위의 증명에서 (가), (나), (다)에 알맞은 값을 모두 더하면? [4점]
