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 881.

## Friday, May 24, 2013

### Problem 881: Triangle, three Squares, Centers, Areas

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BM=4 , BN= SQRT(23), MN=SQRT(67)

ReplyDeleteIn triangle MBN we have MN^2=MB^2+NB^2- 2.MB.NB.cos(B+90)

But cos(B+90)=- sin(B)

So sin(B)=(MN^2-MB^2-NB^2)/2.MB.NB

=(67-16-23)/(2 * 4 .*SQRT(23)= 7/(2 . SQRT(23))

Area (ABC)=1/2 . AB* BC* Sin(B) = 14

BM=4 , BN= SQRT(23), MN=SQRT(67)

In triangle MBN we have MN^2=MB^2+NB^2- 2.MB.NB.cos(B+90)

But cos(B+90)=- sin(B)

So sin(B)=(MN^2-MB^2-NB^2)/2.MB.NB

=(67-16-23)/(2 * 4 .*SQRT(23)= 7/(2 . SQRT(23))

Area (ABC)=1/2 . AB* BC* Sin(B) = 14

Join BM, BN.

ReplyDeleteBM^2 = BD^2 / 2 = 16

BN^2 = BF^2 / 2 = 23

MN^2 = 67

MN^2 = BM^2 + BN^2 − 2×BM×BN×cos(∠B+90°)

67 = 16 + 23 + 2×BM×BN×sin∠B

BM×BN×sin∠B = 14

S(ABC) = 1/2×BA×BC×sin∠B

= 1/2×(√2×BM)×(√2×BN)×sin∠B

= BM×BN×sin∠B

= 14

BD=AB=sqrt(32), BF=BC=sqrt(46), MN=sqrt(67)

ReplyDeleteBM=sqrt(16)=4, BN=sqrt(23)

By cosine law,

MN^2 = BM^2 + BN^2 - 2*BM*BN*cosMBN

MN^2 = BM^2 + BN^2 - 2*BM*BN*cos(90+ABC)

MN^2 = BM^2 + BN^2 + 2*BM*BN*sinABC

MN^2 = BM^2 + BN^2 + 2*1/sqrt2*1/sqrt2*(AB*BC*sinABC)

67 - 16 - 23 = AB*BC*sinABC

28 = AB*BC*sinABC

14 = Area

Problem 881: solution by Michael Tsourakakis from Greece at

ReplyDeletewww.gogeometry.com/school-college/p881-geometry-mixalis-greece.pdf

Thanks,

Mixalis

Prove that the center of the square MNPQ coincides with the midpoint of AC.

ReplyDeletethis is very easy.

DeleteBecause MSP, is, diagonal of the square MNPQ ,and angle MSN=90 ,angle SMN=angle MNS = 45 ,angle MNP=90 ,then, angle SNP=45 ,so, angle NPS =45 .Therefore SP=SN=SM so,,S midpoint of the MP

What is S?. Kindly clarify.

ReplyDeleteIf you mean S is the point where diagonals of the square MNPQ meet, it is trivial to see that SP = SN = SM.

If you mean S is the point where a diagonal, say MP, meets AC, you need to prove that S is the midpoint of the side AC of the triangle ABC.

Please read my solution.S, is midpoint of the AC

ReplyDelete