See complete Problem 65

Right Triangle, Midpoints. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 65

Labels:
midpoint,
right triangle

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Let B be (0,0) BE the x-axis & BA the y-axis. Further, let BE=p and BD=q. This gives, E:(p,0), C:(p+e,0),D:(0,q), A:(0,d+q). Thus N:((p+e)/2,(q+d)/2)and M;(p/2,q/2). By the distance formula, MN^2 = x^2 = ((p+e)/2 -p/2)^2+((q+d)/2 -q/2)^2 = e^2/4 + d&2/4 0r x^2 =(1/4)*(d^2+e^2)

ReplyDeleteor x=(1/2)*(d^2+e^2)^(1/2)

Ajit: ajitathle@gmail.com

I like the geometric solution Joe so I got this solution : Let K and L be the orthogonal projections of A and C on (DE), take S the midpoint of [KL] such that SM=p now KBEA and DBLC are concyclic so the two triangles AKD and CLE are similar then LC*KA=LE*KD-------(1)

ReplyDeletenow p=(KM-LM)/2 and the median NS=(CL+AK)/2

Now x^2=(NS)^2+P^2

=(KM^2-2KM*LM+LM^2+CL^2+2CL*AK+AK^2)/4

Let DM=ME=y

so x^2=((KD+y)^2-2(KD+y)(LE+y)+(LE+y)^2+CL^2+2CL*AK+AK^2)/4

= (CL^2+AK^2+KD^2+2yKD+y^2+LE^2+2yLE+y^2-2KD*LE-2yKD-2yLE-2y^2+2CL*AK)/4

=(CL^2+AK^2+KD^2+LE^2-2KD*LE+CL*AK)/4

=(e^2+d^2)/4 (from (1)

but this solution is long Is there shorter Antonio????? why there is no post for problem 358 the solution is already prepared and I have suggest for you

make an area in your site to our problems if we have because I have many problems didn't mention in this with these problems thx a lot

http://geometri-problemleri.blogspot.com/2009/11/problem-48-ve-cozumu.html

ReplyDeleteExtend AM to P such that AM = MP so ADPE is a parellelogram and Tr. PEC is a right Tr. with EP = d

ReplyDeleteNow extend MN to Q such that MN = NQ = x. So AMCQ is a //ogram with QS = AM = MP. Hence MPCQ is a //ogram and PC = MQ = 2x.

Now apply Pythagoras to right Tr. CEP and d^2 + e^2 = 4x^2 and so the result follows.

Sumith Peiris

Moratuwa

Sri Lanka

Variation of this problem

ReplyDeleteIf B = 60 and d = e then d = 2x