Points of Authority

Geometry Level pending

Consider the rectangle A B C D ABCD and points M , N , O , P M, N, O, P on the segments A B ‾ \overline{AB} , B C ‾ \overline{BC} , C D ‾ \overline{CD} and A D ‾ \overline{AD} , respectively.

M O MO and N P NP are perpendicular to each, intersecting at the point X X .

If [ A M X P ] = 6 , [ P X O D ] = 8 \left [ AMXP \right ] = 6, \left [ PXOD \right ]= 8 and [ M B N X ] = 18 \left [ MBNX \right ] = 18 , evaluate [ M N O P ] + [ A B C D ] \left [ MNOP \right ] + \left [ ABCD \right ] .


The answer is 84.

Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Let A M ‾ = a , M B ‾ = b , B N ‾ = c , N C ‾ = d \overline{AM} = a, \; \overline{MB} = b, \; \overline{BN} = c, \; \overline{NC} = d . We have a c = 6 , a d = 8 ⇔ d = 8 a , b c = 18 ⇔ b = 18 c ac = 6, \; ad = 8 \Leftrightarrow d = \dfrac{8}{a}, \; bc = 18 \Leftrightarrow b = \dfrac{18}{c} , and thus b d = 144 a c = 144 6 = 24 bd = \dfrac{144}{ac} = \dfrac{144}{6} = 24 . Then [ A B C D ] = 6 + 8 + 18 + 24 = 56 \left [ ABCD \right ] = 6 + 8 + 18 + 24 = 56 , and because [ A B C D ] = 2 [ M N O P ] \left [ ABCD \right ] = 2 \left [ MNOP \right ] , finally [ A B C D ] + [ M N O P ] = 84. \left [ ABCD \right ] + \left [ MNOP \right ] = \boxed{84.} .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...