Točno
17. travnja 2016. 12:19 (8 godine, 3 mjeseci)
Veoma dobro poznata je nejednakost trokuta, koja govori da u svakom trokutu sa stranicama
![a, b, c](/media/m/9/e/9/9e9dfe78930065fbe5a777e9b07c27c4.png)
vrijedi
![a+b\geq c](/media/m/5/1/2/51208bc4030911cfbdaf8328994201af.png)
,
![b+c\geq a](/media/m/2/1/b/21bb6feba240efb00f8771fe8cac0eca.png)
,
![a+c \geq b](/media/m/9/4/4/944e5b8464f1e357de7986775e37f71b.png)
pri kojoj se jednakost postiže u degeneriranim trokutima.
Dokažite jaču nejednakost
![a+b \geq \sqrt{c^2+4h_c^2}](/media/m/6/5/0/6504b5d85939e68d6cc77ab76f4aeb11.png)
,
![a+c \geq \sqrt{b^2+4h_b^2}](/media/m/3/f/0/3f016ca50b82820f9316d214f4be37f4.png)
,
![b+c \geq \sqrt{a^2+4h_a^2}](/media/m/8/7/0/87005d302ae77cee3e97fc5261913163.png)
gdje su
![h_a, h_b, h_c](/media/m/c/2/7/c2743927db9d707122c30003fb440356.png)
visine na stranice
![a, b, c](/media/m/9/e/9/9e9dfe78930065fbe5a777e9b07c27c4.png)
. Kada se postiže jednakost?
%V0
Veoma dobro poznata je nejednakost trokuta, koja govori da u svakom trokutu sa stranicama $a, b, c$ vrijedi $a+b\geq c$, $b+c\geq a$, $a+c \geq b$ pri kojoj se jednakost postiže u degeneriranim trokutima.
Dokažite jaču nejednakost $a+b \geq \sqrt{c^2+4h_c^2}$, $a+c \geq \sqrt{b^2+4h_b^2}$, $b+c \geq \sqrt{a^2+4h_a^2}$ gdje su $h_a, h_b, h_c$ visine na stranice $a, b, c$. Kada se postiže jednakost?
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![(1)](/media/m/2/5/e/25e5e167a3616378fc4ad422677ae0c4.png)
Primjetimo da je
![a = \sqrt{b^2-h_a^2} \pm \sqrt{c^2-h_a^2} \leq \sqrt{b^2-h_a^2} + \sqrt{c^2-h_a^2}](/media/m/5/6/c/56c83a182ab82db318bb0b1ccea6cfa3.png)
Prema tome,
![a^2 \leq b^2+c^2-2h_a^2 +2\sqrt{(b^2-h_a^2)(c^2-h_a^2)}](/media/m/3/b/a/3bafa62d7e669419d51fca86cc19b7a1.png)
Nadalje,
![b+c \geq \sqrt{a^2+4h_a^2} \iff (b+c)^2 \geq a^2 + 4h_a^2](/media/m/c/e/7/ce75b6935b4c19162c3a47b4aa3ece6d.png)
![b^2 + c^2 + 2bc \geq b^2 + c^2 - 2h_a^2 + 2\sqrt{(b^2 - h_a^2)(c^2 - h_a^2)} + 4h_a^2](/media/m/1/d/6/1d6a494958b3f56bfed5616f328b19e4.png)
![bc \geq h_a^2 + \sqrt{(b^2 - h_a^2)(c^2 - h_a^2)}](/media/m/e/d/a/eda843b65e758513e280b2157c4c3f07.png)
![(bc-h_a^2)^2 \geq (b^2 - h_a^2)(c^2 - h_a^2)](/media/m/e/5/8/e582f89bfe525c4908c41a424a032417.png)
![b^2c^2 + h_a^4 - 2bch_a^2 \geq b^2c^2 - (b^2+c^2)h_a^2 + h_a^4](/media/m/9/f/c/9fca71be1f58153ca81e5dfe271d8832.png)
![(b^2+c^2)h_a^2 \geq 2bch_a^2 \iff b^2 +c^2 \geq 2bc \iff (b-c)^2 \geq 0](/media/m/b/5/c/b5c75102db1469876ebfdd50693be879.png)
Što dokazuje originalnu nejednakost.
Jednakost može vrijediti samo ako
![b = c](/media/m/7/6/c/76c5156e45e6c5aee900e8a3c696e3ce.png)
ili
![h_a = 0](/media/m/e/f/2/ef2befa0a8a945d7b2e347d8b6ad412d.png)
te provjerom dobivamo da u jednakokračnim i degeneriranim trokutima jednakost uistinu vrijedi.
Analogno provodimo dokaz za druge dvije stranice.
![(2)](/media/m/6/6/a/66ab623e63546b6c830c0a02c99d5444.png)
Neka su stranice
![a, b, c](/media/m/9/e/9/9e9dfe78930065fbe5a777e9b07c27c4.png)
nasuprotne vrhovima
![A, B, C](/media/m/5/2/5/5251ced8c37ecf5247e7f644e571612f.png)
i neka je
![N](/media/m/f/1/9/f19700f291b1f2255b011c11d686a4cd.png)
nožište visine iz vrha
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
na stranicu
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
.
![(|AN| = h_a)](/media/m/e/5/c/e5cd952a9c5b6cca8e8cce66837b0e2e.png)
Neka je
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
točka takva da je
![NASB](/media/m/c/7/d/c7dd17c36c4d6129223160a0c693ba4d.png)
pravokutnik. Neka je
![Q](/media/m/4/5/c/45ce8d14aa1eb54f755fd8e332280abd.png)
točka na na polupravcu
![BS](/media/m/3/0/f/30f1debf47f15e4cdb55f1f18a034c23.png)
takva da
![\triangle ASB \cong \triangle ASQ \implies |AQ| = |AB| = c](/media/m/5/e/7/5e7a4ad29d554f06f7903b77e13ed3c0.png)
pa po nejednakosti trokuta na
![\triangle QAC](/media/m/d/0/f/d0f32162e3c35af1ab365b0e613e4b89.png)
mora vrijediti
![|AQ| + |AC| \geq |CQ| \iff b + c \geq \sqrt{a^2 +4h_a^2}](/media/m/1/6/3/163d231b43e50a31e3773cca77f83321.png)
Jednakost se postiže kada
![A \in QC](/media/m/e/b/a/eba9bb86f1618e9b42f5dc5d9319a113.png)
a tada
![\triangle CAN \sim \triangle CQB, k = 2 \implies |CQ| = 2|CA| \implies |CA| = |AQ| = |AB|](/media/m/9/9/b/99b45747e03d228753327739e4ccc552.png)
pa je
![\triangle ABC](/media/m/1/f/3/1f3c3c0f3e134a169655f9511ba6ea82.png)
jednakokračan.
Lako je vidjeti da se jednakost postiže i kada je
![\triangle ABC](/media/m/1/f/3/1f3c3c0f3e134a169655f9511ba6ea82.png)
degeneriran.
Analogno provodimo dokaz za druge dvije stranice.
%V0
$(1)$
Primjetimo da je $a = \sqrt{b^2-h_a^2} \pm \sqrt{c^2-h_a^2} \leq \sqrt{b^2-h_a^2} + \sqrt{c^2-h_a^2}$
Prema tome, $a^2 \leq b^2+c^2-2h_a^2 +2\sqrt{(b^2-h_a^2)(c^2-h_a^2)}$
Nadalje,
$$b+c \geq \sqrt{a^2+4h_a^2} \iff (b+c)^2 \geq a^2 + 4h_a^2 $$
$$b^2 + c^2 + 2bc \geq b^2 + c^2 - 2h_a^2 + 2\sqrt{(b^2 - h_a^2)(c^2 - h_a^2)} + 4h_a^2$$
$$bc \geq h_a^2 + \sqrt{(b^2 - h_a^2)(c^2 - h_a^2)}$$
$$(bc-h_a^2)^2 \geq (b^2 - h_a^2)(c^2 - h_a^2)$$
$$b^2c^2 + h_a^4 - 2bch_a^2 \geq b^2c^2 - (b^2+c^2)h_a^2 + h_a^4$$
$$(b^2+c^2)h_a^2 \geq 2bch_a^2 \iff b^2 +c^2 \geq 2bc \iff (b-c)^2 \geq 0$$
Što dokazuje originalnu nejednakost.
Jednakost može vrijediti samo ako $b = c$ ili $h_a = 0$ te provjerom dobivamo da u jednakokračnim i degeneriranim trokutima jednakost uistinu vrijedi.
Analogno provodimo dokaz za druge dvije stranice.
$(2)$
Neka su stranice $a, b, c$ nasuprotne vrhovima $A, B, C$ i neka je $N$ nožište visine iz vrha $A$ na stranicu $a$. $(|AN| = h_a)$
Neka je $S$ točka takva da je $NASB$ pravokutnik. Neka je $Q$ točka na na polupravcu $BS$ takva da $|BQ| = 2|BS| = 2h_a$
$\triangle ASB \cong \triangle ASQ \implies |AQ| = |AB| = c$ pa po nejednakosti trokuta na $\triangle QAC$ mora vrijediti $|AQ| + |AC| \geq |CQ| \iff b + c \geq \sqrt{a^2 +4h_a^2}$
Jednakost se postiže kada $A \in QC$ a tada $\triangle CAN \sim \triangle CQB, k = 2 \implies |CQ| = 2|CA| \implies |CA| = |AQ| = |AB| $ pa je $\triangle ABC$ jednakokračan.
Lako je vidjeti da se jednakost postiže i kada je $\triangle ABC$ degeneriran.
Analogno provodimo dokaz za druge dvije stranice.
4. svibnja 2016. 14:24 | grga | Točno |