Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Thursday, May 26, 2016
Geometry Problem 1218: Scalene Triangle, Equilateral Triangles, Midpoints, 60 Degrees, Congruence
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http://s33.postimg.org/53zo0aqen/pro_1218.png
ReplyDeleteLet D is the midpoint of BC
CBC2 is 30-60-90 triangle
So BC2=1/2. BC= BD , BDC2 is equilateral and ∡ (CDC2)= 120 degrees.
Note that DB1= ½. AB= BA2
∡ (BDC2)= ∡ (B1DC)+ 120 = ∡ (ABC)+120= ∡ (A2BC2)
Triangle B1DC2 congruent to tri. A2BC2 ( case SAS)
So B1C2= A2C2 and ∡ (B1C2A2)= ∡ (DC2B)= 60 degrees
So B1C2A2 is a equilateral triangle.
Problem 1218
ReplyDeleteLet K, L medpoints the sides AB and AC.Then KA2=AB/2=A2B=KB=LB1, KB1=BL=LC2, <B1KA2=<ΑΚΒ1+120=<ABC+20=<B1LC+120=B1LC2=<ABC+60+60=120. Therefore
Triangles B1KA2, A2BC2 and B1LC2 are equals.Then triangle A2B1C2 is equilateral.
MANOLOUDIS APOSTOLIS
4th HIGH SCHOOL OF KORYDALLOS -PIRAEUS-GREECE
Each of the sides of the triangle are = to a/2 + c/2 using the mid point theorem and so the result follows
ReplyDeleteSumith Peiris
Moratuwa
Sri Lanka
Refer to your comment " Each of the sides of the triangle are = to a/2 + c/2".
ReplyDeletePer sketch shown on the link
http://s33.postimg.org/53zo0aqen/pro_1218.png
each side of triangle B1C2A2 is less than a/2 + c/2 .
Please explain.
Correction
ReplyDelete3 congruent triangles are formed within A2B2C2 viz A2BC2, A2B1X and B1C2Y where X and Y are the midpoints of AB and BC respectively, each having adjacent sides = to a/2 and c/2 (using the mid point theorem) and included angle = B + 120 and so the result follows
We use complex numbers; one complex number represents a point.
ReplyDeleteBecause A2 is the midpoint of B and C1, (c1 + b)/2 = a2.
Similarly, c2 = (a1 + b)/2, b1 = (a + c)/2.
As triangle AB(C1) and triangle (A1)BC are equilateral triangles, we use De Moivre’s theorem to find: (c1 - b) cis(pi/3) = (a – b), (b – a1) cis(pi/3) = c – a1.
It’s obvious that a2 – c2 = (c1 – b)/2 + (b – a1)/2. Using the equations found from De Moivre’s theorem, we conclude that (a2 – c2) cis(pi/3) = (a – b)/2 + (c – a1)/2 = (b1 – c2)
In conclusion, (a2 – c2) cis(pi/3) = (b1 – c2) which indicates two things: (A2)(C2) has equal length with (B1)(C2), and that these two lines make an angle of 60 degrees with each other. This indicates that Triangle (A2)(C2)(B1) is equilateral.