# Sides of triangle and an altitude

Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Let $h$ be the altitude drawn on the side of length $a$ Then is $a^2 + 4h^2 - (b+c)^2$ always negative ?

-

For acute/right angled triangle $h=b\sin C=c\sin B$ and

for obtuse triangle (let $\angle B>\frac \pi 2$), $h=c\sin(\pi-B)=c\sin B$

$4h^2=2h\cdot 2h=(2b\sin C)(2c\sin B)$

$=2bc(\cos(B-C)-\cos(B+C))$

$=2bc(\cos(B-C)+\cos A)$ as $\cos(B+C)=\cos(\pi-A)=-\cos A$

We know, $$\cos A=\frac{b^2+c^2-a^2}{2bc}$$

$$\implies 1+\cos A=\frac{b^2+c^2-a^2}{2bc}+1=\frac{(b+c)^2-a^2}{2bc}$$

$\implies (b+c)^2-a^2=2bc (1+\cos A)$

$4h^2-((b+c)^2-a^2)$

$=2bc(\cos(B-C)+\cos A)-2bc (1+\cos A)$

$=2bc(\cos(B-C)-1)\le0$ as $\cos(B-C)\le 1$ and $bc>0$

-
That's a wonderful proof ! I have also given a proof , don't forget to see it – Souvik Dey Oct 19 '12 at 10:51

I also have been able to come up with a proof that the expression indeed is negative ;
$-(b-c)^2 ≤ 0$

$\implies a^2 - (b-c)^2 ≤ a^2$

$\implies (a+b-c)(a-b+c) ≤ a^2$

$\implies (a+b+c)(b+c-a)(a+b-c)(a-b+c) ≤ a^2 ( a+b+c)(b+c-a)$ {since $a+b+c >0$ and for a triangle $b+c > a$ , the multiplication in the above line doesn't change sign of inequality}

$\implies 16 A^2 ≤ a^2 \{ (b+c)^2 - a^2 \}$ [ A is the area of the triangle , by Heron's formula ]

$\implies 16 ( \frac 12 ah )^2 ≤ a^2 \{(b+c)^2 - a^2\}$ [ since h is altitude on a , A= ah/2 ]
$\implies 4 (ah)^2 ≤ a^2 \{ (b+c)^2 - a^2 \}$

$\implies 4 h^2 ≤ (b+c)^2 - a^2$

-

Let's suppose the triangle is acute

We have $$a_1^2 + h^2 = b^2\\ a_2^2 + h^2 = c^2$$ and by Cauchy-Schwarz inequality $$a_1\cdot a_2 + h\cdot h \leq \sqrt{a_1^2 + h^2}\sqrt{a_2^2 + h^2} = bc$$ Summing the above relations we get $$a^2 + 4h^2 = a_1^2 + a_2^2 + 2a_1 a_2 + 4h^2 \leq b^2 + c^2 +2bc = (b + c)^2$$ A similar reasoning shows the inequality is true even for obtuse triangles.

-
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "url": "http://math.stackexchange.com/questions/216804/sides-of-triangle-and-an-altitude/216829", "fetch_time": 1469279942000000000, "content_mime_type": "text/html", "warc_filename": "crawl-data/CC-MAIN-2016-30/segments/1469257822598.11/warc/CC-MAIN-20160723071022-00274-ip-10-185-27-174.ec2.internal.warc.gz", "warc_record_offset": 155304879, "warc_record_length": 18292, "token_count": 816, "char_count": 1880, "score": 4.4375, "int_score": 4, "crawl": "CC-MAIN-2016-30", "snapshot_type": "latest", "language": "en", "language_score": 0.7151437401771545, "file_path": "/sailhome/yjruan/project/latent-lm/data/processed/finemath-4plus/finemath-4plus/train-00041-of-00064.parquet"}