constructing a triangle given length of 3 altitudes?

Some 4 years ago I answered a similar question regarding the construction of a triangle given the lengths of the 3 medians (the current question is also discussed partly there):

http://answers.yahoo.com/question/index;_ylt=AlgDU...

This is also a classic old problem and I have seen the following 'solution' in some books: let a, b, c are the unknown sides, h_a, h_b, h_c - the given altitudes. Since

2*Area = a*h_a = b*h_b = c*h_c, we'll have

a : b : c = (1/h_a) : (1/h_b) : (1/h_c), or h_a : h_b : h_c = (1/a) : (1/b) : (1/c) ,

hence a, b, c will be proportional to the altitudes of the triangle, whose sides are h_a, h_b, h_c. So we construct a triangle with sides h_a, h_b, h_c and the altitudes of the latter are sides of a triangle, similar to the required one. The rest is easy - it remains to adjust the latter triangle size homothetically (shrinking or expanding until one of its altitudes becomes equal to the corresponding given altitude).

Alas, the above is wrong, i.e. not universal - this way we can construct a triangle with altitudes 6, 7, 8 as in your example, be we can NOT with altitudes 1, 1, 8 (such 'very tall' isosceles triangle DOES exist!) - the flaw is that h_a, h_b, h_c in an arbitrary triangle not necessarily satisfy the Triangle Inequality what makes the construction of the helper triangle not always possible.

Here is one of many possible correct solutions of a).

The above proportion leads to a : b = h_b : h_a and

b : c = h_c : h_b = h_a : (h_a * h_b / h_c), hence

a : b : c = h_b : h_a : (h_a * h_b / h_c),

so we need to construct a triangle with sides h_b, h_a and h_a * h_b / h_c - Euclid's Construction, the necessary and sufficient condition is

| h_a - h_b | < h_a * h_b / h_c < h_a + h_b

and thus we have a triangle, similar to the requested one.

Your example leads to the construction of the triangle with sides 7, 6 and 21/4

(7 * 6 = 6 * 7 = 21/4 * 8)

Triangle with altitudes 1, 1, 8 needs to construct a triangle with sides 8, 8, 1 etc.

b) You can use the well-known formula for the area by 3 altitudes (2nd in the section below):

http://en.wikipedia.org/wiki/Triangle#Formulas_mim...

and then, of course a = 2*Area/h_a, b = 2*Area/h_b, c = 2*Area/h_c

P.S. In a) above we need of course the construction of the 4th proportional segment x, given 3 segments k, l, m, i.e. x = kl/m - a well-known construction (x/k = l/m).

P.S.(2 - after having read other answer updates and personal emails received)

To Madhukar Daftary: I followed the link to 'More Geometric Constructions' and read Answer #6. The author (Jim Loy) first explains the popular correct approach, but then, suggesting an alternative approach (see 'Note'), makes the same common mistake I have described above. He has written: '...This is an interesting alternative construction for the above. We construct a triangle from just the three given altitudes (far left)....' - THE LATTER CONSTRUCTION IS NOT ALWAYS POSSIBLE - take an isosceles triangle with a very short base and other 2 equal sides sufficiently long. The altitude to the base will be obviously much longer then the sum of the other 2 altitudes.

Yet it works with the current 6, 7, 8 example, but, remember, some sources in the Internet are not 100% trustworthy.

To Slowfinger: the approach in Your answer is equivalent to the described above, leading to construction of triangle with sides 7, 6 and 21/4: the triangle with sides 28, 24, 21 has simply 4 times longer sides.

All correct solutions of this classical construction problem I know use in one or other form a construction of a triangle, similar to the requested one - technically, there is little difference between them - I picked the most convenient in my opinion.

P.S.(3) I have received emails asking me to demonstrate how it can be done: here is an example where one of the altitudes is longer than the sum of other two, see the picture:

http://farm4.static.flickr.com/3501/5746007939_0e3...

Let the triangle have area A, and have side lengths a, b, c; let the altitudes intersecting these sides have lengths d, e, f, respectively. By the area formula,

2A = ad = be = cf.

Let the altitude intersecting side A divide side A into two segments of length g and h. (This still works if the triangle is obtuse, but one of g and h may have to be negative.) Note that

g + h = a.

By the Pythagorean theorem, we have

b^2 = g^2 + d^2,

c^2 = h^2 + d^2.

Subtracting these equations, we get

b^2 - c^2 = g^2 - h^2 = (g+h) (g-h) = a(g-h).

We can also plug in b = ad/e and c = ad/f, so this becomes

a^2 d^2 (1/e^2 - 1/f^2) = a(g-h),

a d^2 (1/e^2 - 1/f^2) = g - h.

Adding a = g+h, we get

a d^2 (1/d^2 + 1/e^2 - 1/f^2) = 2g,

g/a = d^2 (1/d^2 + 1/e^2 - 1/f^2)/2.

Note that g/a has a value which can be determined in terms of the given information. Call this number X. Now note that

d^2/e^2 = b^2/a^2 = (g^2 + d^2)/a^2 = (a^2 X^2 + d^2)/a^2 = X^2 + d^2/a^2.

Solving for a, we find that

a = 2e[4 - 4e^2 X^2 /d^2]^(-1/2)

= 2e[4 - d^2 e^2 (1/d^2 + 1/e^2 - 1/f^2)^2]^(-1/2).

Similar formulas apply to b and c. This proves that the triangle is uniquely determined. Once we know the side lengths, they can be used to construct the triangle; note that the side lengths are actually constructible, since it is possible to construct sums, differences, products, ratios, and square roots of given lengths.

Constructing A Triangle

This Site Might Help You.

RE:

constructing a triangle given length of 3 altitudes?

If I am given the lengths of the 3 altitudes of a triangle (say I take at random altitude 6cm, 7cm, 8 cm)

a) can i construct a triangle using pure construction methods (w/o calculating sides)

b) how would I do it by calculating sides

given 3 altitudes is the traingle formed unique

answers or...

This is a trial , i am not sure about!

Area of given triangle * 2= a * h_a= b * h_b =c * h_c

considering the last three equalities,we can apply the theorem of circle which says two chords say AB & CD & EFintersect at a point inside the circle at point P then AP* PB= PC* PD.=EP*PF

Take AP, PC ,EP as altitudes of triangleand extend those partial chords to intersect the circle and so we can find the sides of a triangle a` ,b` ,c` as PB,PD,PF respectively, which is in turn SIMILAR to the required triangle having sides a, b, c.

{Nomenclature; ABC is required triangle, A`B`C` is the similar triangle that we obtain

AD & A`D are the respective altitudes. a,b,c are sidelengths of ABC also a`,b`,c`are sidelengths of A`B`C`}

And thus after getting the sides of the above similar triangle,we construct the triangle.{name it A`B`C`}

then we draw their altitudes .

Now these altitudes drawn are also similar to given ones.

consider the altitude A`D` [where D` is the foot of altitude of triangle A`B`C`]

and mark an arc {keeping the pointer of the rounder on A`} of length AD{altitude of the required triangle} .

One may get A` - D`-D or A`- D -D` depending upon the radius of the circle taken initially

Now draw a perpendicular to that altitude at D

this perpendicular will intersect the sides A`B` & A`C` name them B&C respectively & of course replace A` by A.

and we get required triangle ABC

Triangle ΔABC.

Let the altitudes are h₁, h₂ and h₃, normal to sides a=BC, b=AC and c=AB, respectively.

Since the area of the triangle is

A = a h₁ / 2 = b h₂ / 2 = c h₃ / 2,

we have

a : b : c = 1/h₁ : 1/h₂ : 1/h₃.

Since the heights are given, h₁=6, h₂=7, h₃=8, we may construct a triangle ΔDEF that is similar to the sought one. Let d=EF, e=DF, f=DE

d : e : f = 1/6 : 1/7 : 1/8 ............ (*)

d : e : f = 28/168 : 24/168 : 21/168

d : e : f = 28 : 24 : 21

Because any triangle that is similar to the original one satisfies (*), we can choose the sides of triangle ΔDEF arbitrarily, so we'll take

d=EF=28, e=DF=24, f=DE=21

Using compass and ruler, construct ΔDEF:

http://img5.imageshack.us/img5/1218/triangle01.gif

Draw altitude FF' from F.

Draw an arbitrary line through F, and on that line draw arbitrary point C.

Draw a line through C, parallel to FF'

On that line, starting from point C mark a segment CC' of length h₃=8 units

http://img197.imageshack.us/img197/6277/triangle02...

Draw a line through F' and C'. It intercepts the line through F and C in point P.

Draw lines DP and EP.

http://img824.imageshack.us/img824/3310/triangle03...

Draw a line through C' parallel to DE. The intersection to DP is point A and the intersection to EP is point B.

Connect the points A, B and C. That's the triangle we sought for.

http://img580.imageshack.us/img580/5664/triangle04...

Given three medians, you can construct a triangle, but it is not unique. -------- Ideas to construct the triangle: Start at 2/3 of the three segments of the medians, turn them around until you find a triangle.

Edit 7: (Explanation by Mr. Duke)

Mr. Duke asked me to try to construct a triangle whose altitudes are 1 , 2 and 3. I could understand from his question that my understanding that using the construction methods given him and by Kruti, triangle can be constructed even when the altitudes do not follow triangular inequality is incorrect. What is important is for the inverses of the altitudes to follow the triangular inequality. The advantage of their construction methods is the construction of triangles where the altitudes do not follow the triangular inequality, but the inverses follow the inequality which is not possible by the method given in the website link.

Edit 6: (about the construction given by Mr. Duke)

Initially, I had difficulty understanding the construction of Mr. Duke which he has posted on my request, but now it is clear to me and I find that also very interesting. Let me explain it for the benefit of those who had the same difficulty as I had in understanding the construction.

We need to construct a triangle whose sides are in the ratio

1/ha : 1/hb : 1/hc, (where ha, hb and hc are the altitudes), i.e.,

hbhc : hcha : hahb, i.e.,

hbhc/ha : hc : hb, i.e.,

h : hc : hb.

Mr. Duke has demonstrated in his construction how to draw h = hbhc/ha.

The rest of the steps are self-explanatory.

I understand that this method is also unique in that it is applicable in all cases and involves pure construction without the need of a scale.

Edit 5: (about the construction suggested by Kruti)

For the construction of the traingle given the altitudes based on the method outlined by Kruti with only last few steps changed slightly is described with the figure in the link posted as under.

http://www.flickr.com/photos/52771834@N00/57519392...

Edit 4:

New answer by Kruti is thought provoking.

I was thinking it over and felt that if the three altitudes are drawn with a common meeting point in three directions, say

PA = 6, PB = 7 and PC = 8 and a circle drawn passing through ABC.

PA, PB, PC extended to meet the circle in D, E, F, then

PD, PE and PF are proportional to the sides of the required triangle.

Taking these lengths, a triangle similar to the required triangle can be drawn and using the construction suggested by me in Edit 1, the required triangle can be drawn.

Just think it over. It seems to be the best of all.

The advantage of this method is that it can take care of altitudes that do not satisfy the triagular inequality and is thus a more general and simpler method. The construction is perfect and altitudes given in the form of line-segments will do. No scale is needed.

Edit 2:

The best method based on the link given in source list is explained as under.

Draw a triangle with sides equal to the given altitudes. Draw its altitudes. Draw another triangle DEF with sides equal to the altitudes of the first triangle. Draw an altitude DN on EF, the longest side. From N, take length AN = smallest given altitude. From A, draw lines, AB and AC parallel to DE and DF respectively meeting EF in B and C. ABC is the required triangle.

All my previous postings are pushed down below the line. Some postings which had become redundant have been deleted.

EDIT 3: IMPORTANT

Mr. Duke, who has mastery on the subject, drew my attention to the following limitations of the above construction method in e-mail communication with me.

"the altitudes of any triangle will NOT NECESSARILY satisfy the triangular inequality. Try an isosceles triangle with a short base (say 1) and 2 sufficiently long other equal sides (say length 10). The altitude to the base is much longer then the sum of other 2 altitudes. The author in the link, provided in Your answer, has made the same common mistake: "...This is an interesting alternative construction for the above. We construct a triangle from just the three given altitudes (far left)...." - THE LATTER CONSTRUCTION IS NOT ALWAYS POSSIBLE!"

Thus, if the given altitudes do not satisfy triangular inequality, a triangle cannot be drawn and the above method will fail in such a case. Quite interesting observation by Mr. Duke.

In such cases, the method given in Edit 1 can be used. However, in that method, the lengths of altitudes should be given in measurement and not in the form of drawn line segments and the construction will need the use of a scale.

======================================================================

I deleted all my previous postings which had become reducndant and were not allowing me to post more useful edits.

Draw a line 7 cm. ( AB)

at A , draw an arc of 6 cm (AC)

at Draw BC ( 8 cm) both cut at C.

Triangle is reday.