Advanced 3 Rather hard Geometry Problems - XP Math - Forums

 XP Math - Forums Advanced 3 Rather hard Geometry Problems

 03-04-2014 #1 lolkidman     Join Date: Jun 2011 Posts: 53 3 Rather hard Geometry Problems ONLY GEOMETRY (no calc, trig, etc) 1. Prove that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices (not so hard) 2. Prove that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals (this one took me 20 min) 3. Prove that in any quadrilateral the sum of the squares of the sides is equal to the sum of the squares of the diagonals plus four times the square of the segment joining the midpoints of the diagonals (didnt get this one) __________________ Czech out my fb page at: https://www.facebook.com/Grandmaster.Dudedudedudeo
03-06-2014   #2
MAS1

Join Date: Dec 2008
Posts: 249
Problem 1

Quote:
 Originally Posted by lolkidman ONLY GEOMETRY (no calc, trig, etc) 1. Prove that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices (not so hard) 2. Prove that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals (this one took me 20 min) 3. Prove that in any quadrilateral the sum of the squares of the sides is equal to the sum of the squares of the diagonals plus four times the square of the segment joining the midpoints of the diagonals (didnt get this one)
I am going to use a coordinate proof for this problem.

Assume a right triangle with the right angle located at (0,0) and the other tow vertices located at (a,0) and (0,b).

The mid point of the hypotenuse is then located at (a/2, b/2) using the midpoint formula.

Distance from the right angle to the midpoint = sqrt((a/2 - 0)^2 + (0 - b/2)^2)
=sqrt((a/2)^2 + (b/2)^2)
= sqrt(a^2/4 + b^2/4)
=(sqrt(a^2 + b^2))/2

Distance from (a,0) to the midpoint = sqrt((a - a/2)^2 + (0 - b/2)^2)
=sqrt((a/2)^2 + (b/2)^2)
= sqrt(a^2/4 + b^2/4)
=(sqrt(a^2 + b^2))/2

Distance from (0,b) to the midpoint = sqrt((0 - a/2)^2 + (b - b/2)^2)
=sqrt((a/2)^2 + (b/2)^2)
= sqrt(a^2/4 + b^2/4)
=(sqrt(a^2 + b^2))/2

So all of the distances are the same.

03-06-2014   #3
MAS1

Join Date: Dec 2008
Posts: 249
Problem 2

Quote:
 Originally Posted by lolkidman ONLY GEOMETRY (no calc, trig, etc) 1. Prove that the midpoint of the hypotenuse of a right triangle is equidistant from the three vertices (not so hard) 2. Prove that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals (this one took me 20 min) 3. Prove that in any quadrilateral the sum of the squares of the sides is equal to the sum of the squares of the diagonals plus four times the square of the segment joining the midpoints of the diagonals (didnt get this one)
Another coordinate proof:

For parallelogram ABCD with vertices at A(0,0), B(b,0), C(a+b,c), and D(a,c):

(Length of side AD)^2 = (a^2 + c^2)
(Length of side BC)^2 = (a^2 + c^2)
(Length of side AB)^2 = (b^2)
(Length of side CD)^2 = (b^2)

(Length of diagonal AC)^2 = (a + b)^2 + c^2
(Length of diagonal BD)^2 = (b - a)^2 + c^2

Sum of the squares of the side lengths = 2(a^2 + b^2 + c^2)
Sum of the squares of the diagonals = a^2 + 2ab +b^2 + c^2 + b^2 - 2ab + a^2 + c^2 = 2(a^2 + b^2 + c^2)

So the sums are equal.

03-06-2014   #4
MAS1

Join Date: Dec 2008
Posts: 249
Problem 3

Quote:
 Originally Posted by lolkidman ONLY GEOMETRY (no calc, trig, etc) 3. Prove that in any quadrilateral the sum of the squares of the sides is equal to the sum of the squares of the diagonals plus four times the square of the segment joining the midpoints of the diagonals (didnt get this one)
Does this apply only to convex quadrilaterals or both convex and concave?

03-06-2014   #5
lolkidman

Join Date: Jun 2011
Posts: 53

Quote:
 Originally Posted by MAS1 Does this apply only to convex quadrilaterals or both convex and concave?
just convex

i also did a coordinate proof for the first problem
for the second one you solved, i didnt use a coordinate proog.
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