Construction of a Triangle
Is a triangle only defined by its heights 10, 15 and 20 constructible?
Hint: You can start with side lengths beeing obvious (means no calculation!)!
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1 solution
First I don't agree with your system of equations, for example c/a must be equal 1/2. You understood this wrong because for "Yes" the task required the construction without pre-calculations. A start could be a=3/b=2/c=1.5 simply following from the relations h j h i = j i . But now you have to construct the real triangle varying this 'help triangle'!
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In fact, given arbitrary lengths marked on any line for the 3 altitudes, with a ruler and compass, one can construct a triangle with those altitudes, if such a triangle exists. Instead of offering proof of this, I simply calculated the sides of the triangle for the specific example of altitudes 10, 15, 20, and showed them to be constructible quantities.
There was some vagueness about this problem whether or not it's asking if it's constructible or such a triangle can exist. I thought I'd cover both bases.
As for c/a "must be equal to 1/2", that was a typo, and I just fixed it. Thank you for pointing that out.
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The question is if it is possible to construct the triangle given with 3 heights going out from 3 appropriate lengths of it!
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@Andreas Wendler – Yeah, given altitudes in the ratio of h a : h b : h c , first construct lengths in the ratios of h b : h a and then h c : h b , and use new lengths to create a triangle. Then scale so that one of its altitudes matches the corresponding altitude h i .
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@Michael Mendrin – But how to scale?
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@Andreas Wendler – Once you've constructed a triangle similar to the final triangle, draw a line parallel to one of the sides at a distance equal to the altitude associated with that side. Then extend one of the sides of the similar triangle to intersection with that line. Finish rest of final triangle.
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@Michael Mendrin – Yes, correct. But it's more elegant to determine graphically the relation of the heights to have the factor: So all is a question of central stretching!
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@Andreas Wendler – A problem like this is probably more properly handled with a Note instead of with a Problem. Did I just now utter a tongue-twister? But, anyway, yes, the subject of how to construct a triangle with a compass and ruler, given its three altitudes, is an interesting and fun exercise. Maybe later I'll post a graphic about this here---but right now I have a busy weekend coming up.
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@Michael Mendrin – Thanks for your attention! Till now you and another person were the only ones dealing with my problem!
With notation p, q, r as reciprocal of altitudes and 2H= p+q+r, and reciprocal of area U,
we have the relation
U
=
4
∗
H
∗
(
H
−
p
)
∗
(
H
−
q
)
∗
(
H
−
r
)
.
p
=
1
2
0
1
2
,
q
=
1
2
0
8
,
r
=
1
2
0
6
,
H
=
1
2
0
1
3
.
∴
a
r
e
a
o
f
t
h
e
Δ
=
5
∗
9
1
.
3
6
0
0
∴
t
h
e
s
i
d
e
s
a
r
e
,
.
5
∗
9
1
1
8
0
,
5
∗
9
1
1
3
5
,
5
∗
9
1
9
0
S
a
m
e
a
s
a
b
o
v
e
.
A system of 3 equations
2 1 1 0 a = 2 1 1 5 b
2 1 1 0 a = 2 1 1 2 c
b 2 − 1 0 2 + c 2 − 1 0 2 = a
provides the unique solution
a = 1 4 4 9 1 5
b = 9 6 9 1 5
c = 7 2 9 1 5
Since these values are constructible with a compass and ruler (relative to a given unit length), the triangle itself is constructible.