Try to use elementary geometry (Euclid's Elements).

See complete Problem 239 at:

gogeometry.com/problem/p239_square_midpoints_congruence_pythagoras.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Sunday, February 1, 2009

### Elearn Geometry Problem 239: Square, Midpoints, Congruence, Pythagoras

Labels:
congruence,
midpoint,
Pythagoras,
square

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Let A be (0,0) then we've B;(0,a),C:(a,a) & D:(a,0). With routine analytical geometry one can determine DH:x+2y=a and CG:2x-y=a which gives us N:(3a/5,a/5). Likewise, we find M:(2a/5,4a/5).

ReplyDeleteHence, MN^2 = b^2 = (3a/5-2a/5)^2+(a/5-4a/5)^2

= 2a^2/5 = 10a^2/5^2

Hence, b=(a/5)*(10)^(1/2)

Ajit: ajitathle@gmail.com

Let MA and HN intersect at P.

ReplyDeleteLet NC and MF intersect at Q.

MPNQ is a square.

We can show that MP = AM - BM

So,

MP*MP = AM*AM + BM*BM - 2*AM*BM

= a*a - 2*AM*BM

In triangle ABE,

1/[AB*AB] + 1/[BE*BE] = 1/[BM*BM]

So,

5/ [a*a] = 1/[BM*BM]

BM = a/ [square root of 5]

AM * AE = AB*AB

so, AM * [square root of 5]*a/2 = a*a

AM = 2*a/[square root of 5]

So,

MP*MP = a*a - 4*a*a/5

= a*a/5

So,

MN*MN = 2*a*a/5

= 10*a*a/25

So,

MN = [square root of 10]*a /5

The area of the newly formed inner square is 1/5 the area of the original square because:

ReplyDelete1-there are 4 copies of the same triange (AMB), 3 have been rotated by 90degrees, 180 degrees and 270degrees respectively

2 - each of those has an area of (AM)(MB)/2

3 - those triangles can be arranged so that the hypotenuse (a) becomes the side of the square

4 - This leaves the original square with a square "hole" in the center with an area of (AM-MB)^2

5 - The area of four triangles is 4*[(AM)(MB)/2] or 2(AM)(MB)

6 - The area of the entire square is equal to the area of the small square plus the area of the four triangles, so a^2=2(AM)(MB)+(AM-MB)^2

7- we also know that (AM)^2 + (MB)^2 = a^2 because the triangles are right triangles

What I can't figure out is how to get to my final answer which should be (AM-MB)^2 = 1/5a^2??

Any ideas or help?

It is so simple Anonymous

ReplyDeleteSince you said that the inner square is (1/5) the are of the outer square which has an area of a*a

This means that the inner square has an area of

(a*a)/5

Assuming that the left lower corner of the inner square is Q we have

MQ = NQ = a/sqrt(5)

So b = sqrt(MQ^2 + NQ^2) = sqrt(a^2/5 + a^2/5)

= sqrt(2a^2/5).

Multiplying both numerator and denominator within the sqrt by 5, yields a*sqrt(10)/5

which is the answer.

Joining HF, it is easy to see that

ReplyDeleteArea of parallelogram BHDF

= sum of areas of triangles AHD & BCF

= (a^2)/2

For the parallelogram

Base = HD = a (sqrt 5)/2 ,

Height = b/(sqrt 2) &

Area = [a(sqrt 5)/ 2] x (b / sqrt 2)

So ab(sqrt 5)/2(sqrt 2) =(a^2)/2,

b = a(sqrt2)/(sqrt5) = a(sqrt 10)/5

This comment has been removed by the author.

ReplyDeleteSee the

DeletedrawingDefine O intersection of AE and HD and P intersection of BF and GC

ABE is congruent to CBF by rotation of pi/2

=> x = 2 y

S= a^2 = 2 𝑥^2+ 3 𝑥^2 = 5 𝑥^2

=>𝑥^2 = a^2/5

b hypotenuse =>b^2= 2𝑥^2 = 2a^2/5

Therefore b^2 = 10 a^2/25

GN=(1/2)ND=>PN=(2/5)ND (H,P,N,D).=>b=((2v2)/5)HD

ReplyDelete2 proofs by area method :

ReplyDeleteP the intersection of BF and CG, Q that of AE and DH.

1/ Rotating MBE clockwise around E by an angle of π, M->M’, B->C, E->E, yields a square MM’CP the same size as the central square MPNQ.

Doing the same with PCF around F, NDG around G and QAH around H yields a total of 5 identical squares for ABCD.

Therefore, a.a = 5.b.b/2 or b = a.sqrt(10)/5 QED

2/ ABE = ABCD/4 = 5*EMB (since EMB = HQA = ABM/4)

So square MPNQ = 4*EMB = ABCD/5 = a*a/5, has a side equal to a*sqrt(5)/5 and its diagonal b = a*sqrt(10)/5 QED

Tr.s BCF & ABE are congruent SAS so AE is perpendicular to BM and similarly to HD

ReplyDeleteHence AM = a^2/ (sqrt5. a/2) = 2a/(sqrt5)

Since HD also bisects AM, b = (AM/2). (sqrt2) = sqrt(2/5).a = a.(sqrt10)/5

Sumith Peiris

Moratuwa

Sri Lanka

Read V as Sqrt

ReplyDeleteBF=aV5/2

Area of //gm BHFD=BF.MN/V2=a^2/2

=>MN.aV5/2V2=a^2/2

=>MN=aV2/V5=aV10/5

Triangle BME ~ Triangle BCF

ReplyDeleteBM:ME=2:1

ME=c, BM=2c

----------------------------------------

By Pyth. Theorem (Triangle BME)

BE^2=BM^2+ME^2=5c^2

BE^2=(a/2)^2

=>a^2=20c^2

----------------------------------------

By Pyth. Theorem (Triangle BAE)

BE^2+AB^2=AE^2

AE^2=1.25a^2=25c^2

AE=5c

AG=AE-GE=4c

----------------------------------------

Let AM & HN meet at X and by mid-pt. theorem AX=XM=2c

Similarly, NX=2c

So by Pyth. Theorem on triangle MXN, b^2=8c^2

b^2/a^2=2/5

b/a=sqrt(2/5)

b=[sqrt(10)/5]*a