Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 878.

## Wednesday, May 22, 2013

### Problem 878: Square, Inscribed Circle, Quadrant, Center, Perpendicular, Metric Relations

Labels:
circle,
inscribed,
metric relations,
perpendicular,
quadrant,
square

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since circle O inscribed in the circle then OE = OF = 2 and AE = AF = 4.

ReplyDeleteAC = 4√2 then AO = 2√2. if HO = x and FH = a then

4^2 = (2√2 + x)^2 + a^2

OH = x = 0.5√2.

how did u get from the third line to the last line?

DeleteInsert Complex Coordinate plane with O(0,0)

ReplyDeleteThe inscribed circle : |z|=2,

The circle BEFD : |z + (2+2i)| = 4

By solving and get the mid point, point H = 1/2 + 1/2i

OH = 1/sqrt2

EF is the radical axis of the two circles so the power of circle at H is the same.

ReplyDeletex^2 - 2^2 = (OA + x)^2 - 4^2, where OA = sqrt(8)

By solving, x = 1/sqrt2

AE=4, AO=2*sqrt2, OE=2

ReplyDeleteBy Law Of Cosines, angle AOE=110°41'17"

Angle EOH=69°17'43"

OH=2*cosEOH=0.707107

AE=4, EO=2, and AO=2*sqrt(2). Then 4^2+(2*sqrt(2))^2-2(4)(2*sqrt(2))cos<EAO=2^2, so cos<EAO=5/(4*sqrt(2)), making AH=4*cos<EAO=5/sqrt(2). AH-AO=1/sqrt(2)=x.

ReplyDeleteEH^2 = 4^2 - (2sqrt2+x)^2 = 2^2 - x^2

ReplyDeleteSolving x = 1/sqrt2

Sumith Peiris

Moratuwa

Sri Lanka