## Friday, May 24, 2013

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

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

Click the figure below to see the complete problem 881.

1. 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
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

2. Join BM, BN.

BM^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

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

4. Problem 881: solution by Michael Tsourakakis from Greece at
www.gogeometry.com/school-college/p881-geometry-mixalis-greece.pdf

Thanks,
Mixalis

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

1. this is very easy.
Because 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

6. What is S?. Kindly clarify.
If 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.