Talk:Altitude (triangle)

Symphonic Theorem
An interesting theorem relating the altitudes of the right triangle might be added here; the reciprocals of their squares are harmonic, and their reciprocals are Pythagorean. (see: http://arxiv.org/abs/math.HO/0701554)

—The preceding unsigned comment was added by 74.69.93.70 (talk) 17:23, 26 January 2007 (UTC). orthocenter redirects here -- does it have a more general meaning (as Orthocentric system seems to imply)? -- Tarquin 11:53 Dec 24, 2002 (UTC)

According to my dictionary the line is called a perpendicular, the altitude would then be its length. Patrick 13:53 Dec 24, 2002 (UTC)

Triangle's height
Given the length of two sides of a triangle, the ttriangle being a right triangle, how would on figure out the triangl's height? Bypegorr 23:44, 28 November 2005 (UTC)

Given that it's a right angle triangle, just use c^2=a^2+b^2 where c is opposite your right angle. Uncle_Math

If you were given two sides of a right triangle, then one of those sides must be a leg. One of those legs can be your height. - Kevin (TALK) 20:52, 8 January 2007 (UTC)

Insightful
"In geometry, an altitude of a triangle is a straight line through a vertex and perpendicular to (i.e. forming a right angle with) the opposite side or an extension of the opposite side. The intersection between the (extended) side and the altitude is called the foot of the altitude. This opposite side is called the base of the altitude. The length of the altitude is the distance between the base and the vertex."

Oh, thanks for clearing that up! See, I was a little confused...... 209.40.210.222 (talk) 12:47, 18 November 2008 (UTC)

It is important to the article's discussion that the ratios of the distance from the orthocenter to a point (vertex) and the corresponding (opposite) line, in all three cases, have a ratio. If i remember correctly from high school geometry, it was either that one side was 2x bigger than the other in a special case, or in all cases. If someone would clarify, this would be helpful not for me (as i needed this information as a quick reference) then for people after me. This was part of the basic concepts of the Orthocentric discussion all the other points of which are in this article. — Preceding unsigned comment added by 174.227.7.157 (talk) 15:42, 2 April 2012 (UTC)

Orphaned references in Altitude (triangle)
I check pages listed in Category:Pages with incorrect ref formatting to try to fix reference errors. One of the things I do is look for content for orphaned references in wikilinked articles. I have found content for some of Altitude (triangle)'s orphans, the problem is that I found more than one version. I can't determine which (if any) is correct for this article, so I am asking for a sentient editor to look it over and copy the correct ref content into this article.

Reference named "ac": From Incircle and excircles of a triangle: Altshiller-Court, Nathan. College Geometry, Dover Publications, 1980. From Euler line: Altshiller-Court, Nathan, College Geometry, Dover Publications, 2007 (orig. Barnes & Noble 1952). From Incenter:. #84, p. 121. From Medial triangle: Altshiller-Court, Nathan. College Geometry. Dover Publications, 2007. 

I apologize if any of the above are effectively identical; I am just a simple computer program, so I can't determine whether minor differences are significant or not. AnomieBOT ⚡ 00:15, 16 January 2015 (UTC)

Orthocenter


The three (possibly extended) altitudes intersect in a single point, called the orthocenter of the triangle.



It is a point all of three altitudes pass. In fact, existence of such a point is not proven beforehand. It might be like a definition of creature which has human face and lion body. If point $$H_C$$ is on the altitude to line $$\overline{BC}$$ ($$\overline{AP}$$ in the figure), it must satisfy


 * $$(\vec{AH_C},\vec{BC}) = 0.$$ -- (1a)

Because $$\overline{BH_C}$$ (part of $$\overline{BQ}$$ in the figure) and $$\overline{CA}$$ are also vertical, following equation also stands
 * $$(\vec{BH_C},\vec{CA}) = 0.$$ -- (1b)

Such a point must apparently be unique, so, if some point happened to satisfy above two equations, it must be $$H_C$$. Further, although it is unlikely from the view of Fig 2, if following formula


 * $$(\vec{CH_C},\vec{AB}) = 0$$ -- (1c)

also simultaneously stands, it must be the sphinx $$H$$.

By the way, using inner products of sides defined as following,
 * $$i_A\equiv(\vec{AB},\vec{AC}),$$ -- (2a)
 * $$i_B\equiv(\vec{BC},\vec{BA}),$$ -- (2b)
 * $$i_C\equiv(\vec{CA},\vec{CB}),$$ -- (2c)

we can here define a point
 * $$G=\frac{i_B\cdot i_C\cdot A+i_C\cdot i_A\cdot B+i_A\cdot i_B\cdot C}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B}.$$ -- (3)

In fact, it satisfies above equations in the place of $$H_C$$. For example,
 * $$(\vec{AG},\vec{BC}) =(\frac{i_B\cdot i_C\cdot A+i_C\cdot i_A\cdot B+i_A\cdot i_B\cdot C-(i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B)A}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B},\vec{BC})$$
 * $$=(\frac{i_C\cdot i_A\cdot (B-A)+i_A\cdot i_B\cdot (C-A)}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B},\vec{BC})$$
 * $$=\frac{i_C\cdot i_A\cdot (\vec{AB},\vec{BC})+i_A\cdot i_B\cdot (\vec{AC},\vec{BC})}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B}$$
 * $$=\frac{-i_C\cdot i_A\cdot i_B+i_A\cdot i_B\cdot i_C}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B}$$
 * $$=0.$$ -- (4a)

In the same manner,
 * $$(\vec{BG},\vec{CA}) =0,$$ -- (4b)
 * $$(\vec{CG},\vec{AB}) =0.$$ -- (4c)

Thus, we found the orthocenter $$H$$ at last.

In the process above, divisor used in formula (3) was $$4s^2$$, where $$s$$ is the area of the triangle. In fact, if rewrite $$\vec{BC}$$, $$\vec{CA}$$, $$\vec{AB}$$, as following,
 * $$\vec{BC}=\vec{a}$$, -- (5a)
 * $$\vec{CA}=\vec{b}$$, -- (5b)
 * $$\vec{AB}=\vec{c}$$, -- (5c)

and, using $$\vec{a}+\vec{b}+\vec{c}=\vec{0}$$, rewrite $$\vec{i_A}$$, $$\vec{i_B}$$, $$\vec{i_C}$$ as,


 * $$i_A\equiv(\vec{AB},\vec{AC})=-(\vec{c},\vec{b})=(\vec{a}+\vec{b},\vec{b}),$$ -- (6a)
 * $$i_B\equiv(\vec{BC},\vec{BA})=-(\vec{a},\vec{c})=(\vec{a},\vec{a}+\vec{b}),$$ -- (6b)
 * $$i_C\equiv(\vec{CA},\vec{CB})=-(\vec{b},\vec{a}),$$ -- (6c)

the divisor will be


 * $$i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B=(\vec{a}_x\cdot \vec{b}_y-\vec{a}_y\cdot \vec{b}_x)^2=(|\vec{a}||\vec{b}|sin C)^2 =4s^2.$$

Tsukitakemochi (talk) 01:22, 14 May 2017 (UTC) Tsukitakemochi (talk) 01:45, 14 May 2017 (UTC) Tsukitakemochi (talk) 02:03, 14 May 2017 (UTC)


 * Thank you for placing this here. I can be more explicit about my objections, as I will do below. First of all, I believe that this is your own work (the nature of the errors leads me to believe that) and Wikipedia does not publish such things (see WP:NOR). If I am wrong about that, you should be able to provide a reference to a reliable secondary source (see WP:RSS) for the proof. Giving you the benefit of the doubt, I will now edit this contribution for style and correctness of the English with strike-outs, additions (in red) and comments (in green).--Bill Cherowitzo (talk) 05:41, 14 May 2017 (UTC)

It is a point through which all three altitudes pass. In fact, the existence of such a point must be established. It might be like a definition of creature which has human face and lion body. This is repetitive and unencyclopedic.

If the point $$H_C$$ (poor choice of notation - how is this point related to C?) is on the altitude to line $$\overline{BC}$$ ($$\overline{AP}$$ in the figure), it must satisfy


 * $$(\vec{AH_C},\vec{BC}) = 0.$$ (since these vectors are orthogonal) -- (1a)

Because $$\overline{BH_C}$$ (part of $$\overline{BQ}$$ in the figure) and $$\overline{CA}$$ are also vertical perpendicular, the following equation also stands holds
 * $$(\vec{BH_C},\vec{CA}) = 0.$$ -- (1b)

Such a point must apparently be unique (why?? - this needs to be rigorously established as it is the heart of the proof), so, if some point happens to satisfy the above two equations, it must be $$H_C$$. Furthermore, although it is unlikely from the view of Fig 2, if the following formula condition


 * $$(\vec{CH_C},\vec{AB}) = 0$$ -- (1c)

also simultaneously stands holds, it (what does it refer to?) must be the sphinx $$H$$.

By the way, Using the inner products of sides defined as follows,
 * $$i_A\equiv(\vec{AB},\vec{AC}),$$ -- (2a)
 * $$i_B\equiv(\vec{BC},\vec{BA}),$$ -- (2b)
 * $$i_C\equiv(\vec{CA},\vec{CB}),$$ -- (2c)

we can here define a point G that is a weighted average of the three vertices, by
 * $$G=\frac{i_B\cdot i_C\cdot A+i_C\cdot i_A\cdot B+i_A\cdot i_B\cdot C}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B}.$$ -- (3)

In fact, it satisfies above equations With G in the place of $$H_C$$, the equations 1a,b and c are satisfied. For example, In particular,
 * $$(\vec{AG},\vec{BC}) =(\frac{i_B\cdot i_C\cdot A+i_C\cdot i_A\cdot B+i_A\cdot i_B\cdot C-(i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B)A}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B},\vec{BC})$$
 * $$=(\frac{i_C\cdot i_A\cdot (B-A)+i_A\cdot i_B\cdot (C-A)}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B},\vec{BC})$$
 * $$=\frac{i_C\cdot i_A\cdot (\vec{AB},\vec{BC})+i_A\cdot i_B\cdot (\vec{AC},\vec{BC})}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B}$$
 * $$=\frac{-i_C\cdot i_A\cdot i_B+i_A\cdot i_B\cdot i_C}{i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B}$$
 * $$=0.$$ -- (4a)

In the same manner,
 * $$(\vec{BG},\vec{CA}) =0,$$ and -- (4b)
 * $$(\vec{CG},\vec{AB}) =0.$$ -- (4c)

Thus, we have found the orthocenter $$H$$ at last.

(The material below is not relevant to the topic and should be omitted)

In the process above, divisor used in formula (3) was $$4s^2$$, where $$s$$ is the area of the triangle. In fact, if rewrite $$\vec{BC}$$, $$\vec{CA}$$, $$\vec{AB}$$, as following,
 * $$\vec{BC}=\vec{a}$$, -- (5a)
 * $$\vec{CA}=\vec{b}$$, -- (5b)
 * $$\vec{AB}=\vec{c}$$, -- (5c)

and, using $$\vec{a}+\vec{b}+\vec{c}=\vec{0}$$, rewrite $$\vec{i_A}$$, $$\vec{i_B}$$, $$\vec{i_C}$$ as,


 * $$i_A\equiv(\vec{AB},\vec{AC})=-(\vec{c},\vec{b})=(\vec{a}+\vec{b},\vec{b}),$$ -- (6a)
 * $$i_B\equiv(\vec{BC},\vec{BA})=-(\vec{a},\vec{c})=(\vec{a},\vec{a}+\vec{b}),$$ -- (6b)
 * $$i_C\equiv(\vec{CA},\vec{CB})=-(\vec{b},\vec{a}),$$ -- (6c)

the divisor will be


 * $$i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B=(\vec{a}_x\cdot \vec{b}_y-\vec{a}_y\cdot \vec{b}_x)^2=(|\vec{a}||\vec{b}|sin C)^2 =4s^2.$$

To Bill Cherowitzo:

Thank you for your suggestion and precise correction. I really appreciate. It would be my precious resource in studying English.

As you pointed, main equation (3) itself is my own work prepared on this occasion. An equation to get circumcenter which also is on a Wikipedia's page is also by my past proposal.

In spite of your suggestion, fact that two altitudes always cross is not a mysterious one for me, because two lines always cross except they are parallel. As two sides of a triangle cannot be parallel, altitudes that are vertical to them cannot be parallel.

Three altitudes, however, do not meet at a single point except there is some good fortune. For one thing, I wanted to point it out.

Orthocenter can be either defined as "point two altitudes cross" or "point all three altitudes pass". Difference is not a matter of importance, because they are the same thing as long as "point on two altitudes is necessarily on three altitudes" is somehow testified, which, I believe, had not been done until my proposal at least on the page.

Tsukitakemochi (talk) 13:52, 14 May 2017 (UTC)

An excellent expression for the orthocentre and proof of its existence, if oddly worded. Can you prove the following equality that you have cited though: $$i_B\cdot i_C+i_C\cdot i_A+i_A\cdot i_B=(\vec{a}_x\cdot \vec{b}_y-\vec{a}_y\cdot \vec{b}_x)^2=(|\vec{a}||\vec{b}|sin C)^2 =4s^2.$$ ? Also how have you defined $$\vec{a}_x$$ and $$\vec{b}_x$$ ?Overlordnat1 (talk) 17:29, 19 April 2021 (UTC)

I have demonstrated that the orthocentre has the following coordinates, which tallies with your calculation once you convert the dot products to side lengths. If we call the orthocentre D, we have:-

$$D=D_a\vec{a}+D_b\vec{b}+D_c\vec{c}, where D_a=[2b^2c^2+(a^4-b^4-c^4)]/[2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)], D_b=[2c^2a^2+(b^4-c^4-a^4)]/[2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)] and D_c=[2a^2b^2+(c^4-a^4-b^4)]/ [2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)]$$ Overlordnat1 (talk) 16:58, 17 May 2021 (UTC)

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altitude using Cosine and Sine with sides a,b,c
>For Cosine:

For a triangle with  angles$$\angle ABC$$ there exist  lengths  $$a,b,c$$ opposite to the angles. By using the trigonometric function for cosines, I multiplied the length of $$AC\times\cos\angle B$$ for example: $$5+5+4=14$$ where $$\cos\angle B=0.4,\cos\angle A=0.4,\cos\angle C=0.68$$ and there are $$3$$ $$\theta$$,


 * $$AC\times\cos\angle B=2$$ or $$BC\times\cos\angle A=2$$ which is the place of the altitude and $$2+2=4$$,

For a triangle of sides a,b,c there exist altitudes from three sides of all triangles such that sides of triangles give the altitude in term of Cosine,which gives you the place of the altitude, and with the term of Sine you get directly the altitude itself.


 * $$AC\times\cos\angle A=2$$
 * $$BC\times\cos\angle B=2$$
 * $$AB\times\cos\angle B=1.6$$
 * $$AB\times\cos\angle A=1.6$$
 * $$AC\times\cos\angle C=3.4$$
 * $$BC\times\cos\angle C=3.4$$
 * $$2+2=4,1.6+3.4=5,1.6+3.4=5$$

199.7.157.73 (talk) 13:47, 29 November 2017 (UTC)
 * $$AC\times\cos\angle B=2$$
 * $$BC\times\cos\angle A=2$$
 * $$AB=2+2=4$$
 * $$1.6+3.4=5$$


 * $$\frac{\sin A}{\sin C}=a$$


 * $$\frac{\sin B}{\sin C}=b$$


 * $$\frac{\sin C}{\sin C}=c$$

>for Sine:
 * $$AC\times\sin\angle B=$$


 * $$BC\times\sin\angle A=$$


 * $$AC\times\sin\angle C=$$


 * $$BC\times\sin\angle C=$$


 * $$AB\times\sin\angle A=$$


 * $$AB\times\sin\angle B=$$

And to find Sine I subtract 1 from all three trigonometric function $$\cos ^2 A,\cos^2 B,\cos^2 C $$

And in general the sum of all angles is:


 * $$\sin^2 A+\sin^2 B+\sin^2 C+\cos^2 A+\cos^2 B+\cos^2 C=3$$

199.7.157.73 (talk) 13:47, 29 November 2017 (UTC) 199.119.233.236 (talk) 18:01, 27 November 2017 (UTC)
 * Fortunately we only include material that has appeared in reliable sources (WP:NOR). There is little chance of the above material appearing in such a source since, as written, it is pretty much unintelligible gibberish. --Bill Cherowitzo (talk) 18:39, 27 November 2017 (UTC)

Request to link articles from different languages
It seems there are at least 2 distinct groups of pages (each page in a different language) for this concept. One page in the first group is this one (in English), one page from the other group is this one: https://ro.wikipedia.org/wiki/%C3%8En%C4%83l%C8%9Bime_(geometrie) Can these 2 groups be joined? Thank you. — Preceding unsigned comment added by Cipak (talk • contribs) 16:46, 13 January 2018 (UTC)


 * Why? What is contained in the other page that you feel should be included in this one? --Bill Cherowitzo (talk) 19:14, 13 January 2018 (UTC)

Orthocentre
We could mention the following formula, true when one vertex (x3,y3) = (0,0):-

Orthocentre = (k(y2-y1),k(x1-x2)) where k = (x1x2+y1y2)/(x1y2-x2y1).

This can obviously be generalised by the translation of the point (0,0) to (x3,y3) where x3 and y3 are not equal to 0. (See the video ‘how to find orthocentre shortcut trick’ on YT, which I am unable to post as a link here).

Also I’ve personally proved that the orthocentre has the form (Y1,Y2), where Y1 = a1+X1(b1-a1)/c-X2(b2-a2)/c and Y2 = a2+X1(b2-a2)/c+X2(b1-a1)/c given that we define X1 = (c^2+b^2-a^2)/2c and X2 = (c^4-(a^2-b^2)^2)/(2c*sq rt (4b^2c^2-(c^2+b^2-a^2)^2))

We can easily see the above if we first prove (X1,X2) is the orthocentre of a triangle ((0,0),(b1,0),(c1,c2)), in other words line c runs along the x-axis. Then it’s just a case of rotating the triangle and translating it from (0,0) to (a1,a2)Overlordnat1 (talk) 17:14, 19 April 2021 (UTC)

Should orthocenter be a separate article?
While these topics overlap substantially, it seems like orthocenter is enough of a separate topic to warrant its own article. For example, I think we could include several proofs that the altitudes of a triangle concur (in the orthocenter), but those may seem a bit out of scope here at altitude (triangle). We could also talk add a section "orthocentric simplex" (a specific type of higher-dimensional simplex whose altitudes concur). Etc.

Possibly orthocentric system could be moved to orthocenter with "orthocentric system" made into one section.

Thoughts? –jacobolus (t) 02:49, 15 December 2022 (UTC)