The two sides of a triangle (a and b) are given: b≤a≤2b. The medians to those sides are orthogonal.
Build this triangle.
ANALYSIS:
Let ABC be the desired triangle with standard
side lengths of a, b, and c. Let AA' abd BB'
be the medians intersecting orthogonally at
point I. Let |AA'| = 3m and |BB'| = 3n. Around
point I we have three right triangles AIB,
A'IB, and AIB'. Using the Pathagorean theorem:
AIB: c^2 = 4(m^2 + n^2) (1)
A'IB: a^2 = 4(m^2 + 4n^2) (2)
AIB': b^2 = 4(4m^2 + n^2) (3)
Combining (2) and (3) gives
a^2 + b^2 = 20(m^2 + n^2) (4)
Combining (1) and (4) gives
a^2 + b^2 = 5c^2 (5)
CONSTRUCTION:
Using lengths a and b construct a length c
satifying (5). Let C(P,r) denote a circle
with center P and radius r. WOLOG we will
assume that a <= b.
Construct a diameter CD of C(B,a). Construct
points E and F on the same side of CD such
that both line segments CE and DF are
perpendicular to CD with |CE| = b and
|DF| = 2a. Construct line segments BE, BF, and
EF. Line segment BF intersects C(B,a) at
point G. Construct point on line segment BE
such that line segments GH and EF are parallel.
Let A be the point of intersection of C(B,|BH|)
and C(C,b) on the opposite side of CD from
point G. The desired triangle is ABC.
Note: While checking out the above construction
with Geometer's SketchPad it was found
that b must be less than 2a. This is a
consequence of equation (5) and the
triangle inequality b < a + c.
PROOF:
We have to prove that |AB| = c satisfies (5).
|AB| = |BH| by construction
|BH| |BE|
------ = ------ triangles BHG and BEF
|BG| |BF| are similar by con-
struction.
Therefore,
|BE|^2
|BH|^2 = |BG|^2 --------
|BF|^2
a^2 + b^2
= a^2 -----------
5a^2
or
a^2 + b^2 = 5|BH|^2 = 5|AB|^2
QED
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Posted by Bractals
on 2012-03-15 12:45:48 |
Solution