## Wednesday, March 2, 2016

### Geometry Problem 1194: Equilateral Triangle, Perpendicular, Distance, Side, Metric Relations

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1194.

1. Draw MN middle line of ADEC (= 12)Apply theorem of pythagore MNB, ADB, BEC
we get x=14

2. Draw CF ⊥ DA ..( F on DA)
Let z= x^2
Applying Pythagoras’s theorem in triangles ABD, BCE and ACF we get
BE^2=x^2-121=z-121
BD^2=z-169
AC^2=CF^2+AF^2 => (BE+BD)^2+4=z
Or 2.BE.BD=z-BE^2-BD^2-4= 286-z
4.BE^2.BD^2=4(z-121)(z-169)=(286-z)^2
Simplify this equation we get 3.z^2-588.z= 0
We get 2 solutions z= 0 and z= 196
Only z=196 is acceptable so x= 14

1. See below for geometry solution of problem 1194

Draw circles diameters BA and BC
Draw altitudes BG of tri. ABC and GH of triangle DGE
Observe that H is the midpoint of DE => DGE is isosceles tri.
∡ (DGE)= ∡ (DGB)+ ∡ (BGE)= ∡ (DAB)+ ∡ (BCE)= 180-120= 60 degrees
So DGE is an equilateral tri.
HG= ½( DA+CE)= 12= DE. √ (3)/2
So DE= 8. √3=CF
In right triangle ACF we have x^2= AF^2+CF^2= 4+(8√3)^2=196 => x= 14

3. DB + BE = √(x²-（13-11）²)
√(x²-13²) + √(x²-11²) = √(x²-（13-11）²)
x = 14

4. Sumith Peiris has left a new comment on your post "Geometry Problem 1194:
Equilateral Triangle, Perpe...":

Applying Pythagoras 3 times we see

DE = sqrt(x^2-13^2) + sqrt(x^2-11^2)=sqrt(x^2-2^2)

One can solve this equation by algebra but the equation becomes somewhat
unwieldy

By trial and error x = 14 (check: sqrt27 + sqrt75 = sqrt(27 + 75 +
2Xsqrt(27X75)= sqrt(102 + 2 X 45)=sqrt192

5. Let F be the mid point of AC.
Easily ADBF and BECF are concyclic and DFE is equilateral

Use Ptolemy in these 2 concyclic quadrilaterals and if DB = a and BE = b derive that a+b = 8sqrt3.

So x^2 = (13-11)^2 + (8sqrt3)^2 = 196
and so x = 14

Sumith Peiris
Moratuwa
Sri Lanka

6. MN middle line of ADEC, CP perpend AD. ∆MBN ~ ∆CPN => BN = √3
Pyth Th for BNM => 3+144=(3x²)/4 => x=14