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Find the relationship of the length of triangle's sides.
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Denote the three sides of $triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 $$
Now determine what kind of triangle $triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
C.Right triangle
D.Isosceles Right triangle
The only information I got is from the number in the radical need to be greater than $0$. Then $bge4$ and $cge 1$. Also $10a+2sqrtb-4-22ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
inequality triangle
edited 2 hours ago
TheSimpliFire
11.1k62255
asked 3 hours ago
LOIS
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up vote
2
down vote
favorite
Denote the three sides of $triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 $$
Now determine what kind of triangle $triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
C.Right triangle
D.Isosceles Right triangle
The only information I got is from the number in the radical need to be greater than $0$. Then $bge4$ and $cge 1$. Also $10a+2sqrtb-4-22ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
inequality triangle
edited 2 hours ago
TheSimpliFire
11.1k62255
asked 3 hours ago
LOIS
2538
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Denote the three sides of $triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 $$
Now determine what kind of triangle $triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
C.Right triangle
D.Isosceles Right triangle
The only information I got is from the number in the radical need to be greater than $0$. Then $bge4$ and $cge 1$. Also $10a+2sqrtb-4-22ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
inequality triangle
edited 2 hours ago
TheSimpliFire
11.1k62255
asked 3 hours ago
LOIS
2538
Denote the three sides of $triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 $$
Now determine what kind of triangle $triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
C.Right triangle
D.Isosceles Right triangle
The only information I got is from the number in the radical need to be greater than $0$. Then $bge4$ and $cge 1$. Also $10a+2sqrtb-4-22ge0 $. But they are all inequalities. What we need is some equalities. It would be great to have some hints.
inequality triangle
inequality triangle
edited 2 hours ago
TheSimpliFire
11.1k62255
asked 3 hours ago
LOIS
2538
edited 2 hours ago
TheSimpliFire
11.1k62255
asked 3 hours ago
LOIS
2538
edited 2 hours ago
TheSimpliFire
11.1k62255
edited 2 hours ago
TheSimpliFire
11.1k62255
edited 2 hours ago
TheSimpliFire
11.1k62255
11.1k62255
asked 3 hours ago
LOIS
2538
asked 3 hours ago
LOIS
2538
asked 3 hours ago
LOIS
2538
2538
add a comment |Â
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
4
down vote
accepted
We have
$$|sqrtc-1-2|=10a+2sqrtb-4-22-a^2-btag1$$
from which
$$10a+2sqrtb-4-22-a^2-bge 0tag2$$
follows.
$(2)$ is equivalent to
$$a^2-10a+b-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2-25+(b-4)+4-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2+(sqrtb-4-1)^2le 0$$
from which
$$a-5=sqrtb-4-1=0$$
i.e.
$$a=b=5$$
follows.
Now you can get $c$ from $(1)$.
answered 2 hours ago
mathlove
87.9k877209
add a comment |Â
up vote
2
down vote
HINT:
Write the equality as $$a^2-10a+22+b+|sqrtc-1-2|-2sqrtb-4=0$$ and since we know that $a$ is real, $$22+b+|sqrtc-1-2|-2sqrtb-4le25\b-2sqrtb-4le3-|sqrtc-1-2|le3.$$ But if $f(b)=b-2sqrtb-4$, $f'(b)=1-dfrac1sqrtb-4=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.
Hence that is the only value for $b$, meaning that $c=cdots,,?$
Spoiler:
The triangle is equilateral.
answered 2 hours ago
TheSimpliFire
11.1k62255
add a comment |Â
up vote
1
down vote
Note that:
$$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 iff \
(a-5)^2+|sqrtc-1-2|+(sqrtb-4-1)^2=0 Rightarrow \
a=5, b=5, c=5.$$
answered 2 hours ago
farruhota
15.8k2734
add a comment |Â
up vote
0
down vote
$(a-5)^2+(sqrtb-4-1)^2+|sqrt-2|=0$
This lead $a=b=c=5$.
answered 2 hours ago
Takahiro Waki
1,973620
add a comment |Â
up vote
0
down vote
Noodling:
$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.
So I get
$a^2 - 10a +25+b +|sqrtc-1 - 2|=(a-5)^2 + b+ |sqrtc-1 -2|= 2sqrtb-4 + 3$
And, well this seems a bit weird but that is an even number in front of the $sqrtb-4$ and we have $sqrtb-4$ and $b$ variables to deal with so we can complete the square again with $v = sqrtb-4$ and $v^2 = b-4$.
(In the back of my mind I'm worrying about the $sqrtc-1$ which is just a single term; I'm not sure at this point what will happen.)
$(a -5)^2 + (b-4) - 2sqrtb-4 + 1 +|sqrtc-1 - 2|=3-4 + 1$
$(a-5)^2 + (sqrtb-4 - 1)^2 +|sqrtc-1 - 2| = 0$.
Oh......
We have three things that can't be negative adding up to $0$. So they must each equal $0$.
So $(a-5)^2 = 0$ and $a =5$. And $(sqrtb-4-1)^2=0$ and $b=5$. And $|sqrtc-1 -2| = 0$ so $c=5$.
Well.... okay then......
answered 2 hours ago
fleablood
62.3k22678
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
We have
$$|sqrtc-1-2|=10a+2sqrtb-4-22-a^2-btag1$$
from which
$$10a+2sqrtb-4-22-a^2-bge 0tag2$$
follows.
$(2)$ is equivalent to
$$a^2-10a+b-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2-25+(b-4)+4-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2+(sqrtb-4-1)^2le 0$$
from which
$$a-5=sqrtb-4-1=0$$
i.e.
$$a=b=5$$
follows.
Now you can get $c$ from $(1)$.
answered 2 hours ago
mathlove
87.9k877209
add a comment |Â
up vote
4
down vote
accepted
We have
$$|sqrtc-1-2|=10a+2sqrtb-4-22-a^2-btag1$$
from which
$$10a+2sqrtb-4-22-a^2-bge 0tag2$$
follows.
$(2)$ is equivalent to
$$a^2-10a+b-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2-25+(b-4)+4-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2+(sqrtb-4-1)^2le 0$$
from which
$$a-5=sqrtb-4-1=0$$
i.e.
$$a=b=5$$
follows.
Now you can get $c$ from $(1)$.
answered 2 hours ago
mathlove
87.9k877209
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
We have
$$|sqrtc-1-2|=10a+2sqrtb-4-22-a^2-btag1$$
from which
$$10a+2sqrtb-4-22-a^2-bge 0tag2$$
follows.
$(2)$ is equivalent to
$$a^2-10a+b-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2-25+(b-4)+4-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2+(sqrtb-4-1)^2le 0$$
from which
$$a-5=sqrtb-4-1=0$$
i.e.
$$a=b=5$$
follows.
Now you can get $c$ from $(1)$.
answered 2 hours ago
mathlove
87.9k877209
We have
$$|sqrtc-1-2|=10a+2sqrtb-4-22-a^2-btag1$$
from which
$$10a+2sqrtb-4-22-a^2-bge 0tag2$$
follows.
$(2)$ is equivalent to
$$a^2-10a+b-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2-25+(b-4)+4-2sqrtb-4+22le 0,$$
i.e.
$$(a-5)^2+(sqrtb-4-1)^2le 0$$
from which
$$a-5=sqrtb-4-1=0$$
i.e.
$$a=b=5$$
follows.
Now you can get $c$ from $(1)$.
answered 2 hours ago
mathlove
87.9k877209
answered 2 hours ago
mathlove
87.9k877209
answered 2 hours ago
mathlove
87.9k877209
answered 2 hours ago
mathlove
87.9k877209
87.9k877209
add a comment |Â
add a comment |Â
up vote
2
down vote
HINT:
Write the equality as $$a^2-10a+22+b+|sqrtc-1-2|-2sqrtb-4=0$$ and since we know that $a$ is real, $$22+b+|sqrtc-1-2|-2sqrtb-4le25\b-2sqrtb-4le3-|sqrtc-1-2|le3.$$ But if $f(b)=b-2sqrtb-4$, $f'(b)=1-dfrac1sqrtb-4=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.
Hence that is the only value for $b$, meaning that $c=cdots,,?$
Spoiler:
The triangle is equilateral.
answered 2 hours ago
TheSimpliFire
11.1k62255
add a comment |Â
up vote
2
down vote
HINT:
Write the equality as $$a^2-10a+22+b+|sqrtc-1-2|-2sqrtb-4=0$$ and since we know that $a$ is real, $$22+b+|sqrtc-1-2|-2sqrtb-4le25\b-2sqrtb-4le3-|sqrtc-1-2|le3.$$ But if $f(b)=b-2sqrtb-4$, $f'(b)=1-dfrac1sqrtb-4=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.
Hence that is the only value for $b$, meaning that $c=cdots,,?$
Spoiler:
The triangle is equilateral.
answered 2 hours ago
TheSimpliFire
11.1k62255
add a comment |Â
up vote
2
down vote
up vote
2
down vote
HINT:
Write the equality as $$a^2-10a+22+b+|sqrtc-1-2|-2sqrtb-4=0$$ and since we know that $a$ is real, $$22+b+|sqrtc-1-2|-2sqrtb-4le25\b-2sqrtb-4le3-|sqrtc-1-2|le3.$$ But if $f(b)=b-2sqrtb-4$, $f'(b)=1-dfrac1sqrtb-4=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.
Hence that is the only value for $b$, meaning that $c=cdots,,?$
Spoiler:
The triangle is equilateral.
answered 2 hours ago
TheSimpliFire
11.1k62255
HINT:
Write the equality as $$a^2-10a+22+b+|sqrtc-1-2|-2sqrtb-4=0$$ and since we know that $a$ is real, $$22+b+|sqrtc-1-2|-2sqrtb-4le25\b-2sqrtb-4le3-|sqrtc-1-2|le3.$$ But if $f(b)=b-2sqrtb-4$, $f'(b)=1-dfrac1sqrtb-4=0$ for stationary points, resulting in $b=5$ as a minimum, and $f(b)=3$.
Hence that is the only value for $b$, meaning that $c=cdots,,?$
Spoiler:
The triangle is equilateral.
answered 2 hours ago
TheSimpliFire
11.1k62255
edited 2 hours ago
answered 2 hours ago
TheSimpliFire
11.1k62255
answered 2 hours ago
TheSimpliFire
11.1k62255
answered 2 hours ago
TheSimpliFire
11.1k62255
11.1k62255
add a comment |Â
add a comment |Â
up vote
1
down vote
Note that:
$$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 iff \
(a-5)^2+|sqrtc-1-2|+(sqrtb-4-1)^2=0 Rightarrow \
a=5, b=5, c=5.$$
answered 2 hours ago
farruhota
15.8k2734
add a comment |Â
up vote
1
down vote
Note that:
$$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 iff \
(a-5)^2+|sqrtc-1-2|+(sqrtb-4-1)^2=0 Rightarrow \
a=5, b=5, c=5.$$
answered 2 hours ago
farruhota
15.8k2734
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that:
$$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 iff \
(a-5)^2+|sqrtc-1-2|+(sqrtb-4-1)^2=0 Rightarrow \
a=5, b=5, c=5.$$
answered 2 hours ago
farruhota
15.8k2734
Note that:
$$a^2+b+|sqrtc-1-2|=10a+2sqrtb-4-22 iff \
(a-5)^2+|sqrtc-1-2|+(sqrtb-4-1)^2=0 Rightarrow \
a=5, b=5, c=5.$$
answered 2 hours ago
farruhota
15.8k2734
answered 2 hours ago
farruhota
15.8k2734
answered 2 hours ago
farruhota
15.8k2734
answered 2 hours ago
farruhota
15.8k2734
15.8k2734
add a comment |Â
add a comment |Â
up vote
0
down vote
$(a-5)^2+(sqrtb-4-1)^2+|sqrt-2|=0$
This lead $a=b=c=5$.
answered 2 hours ago
Takahiro Waki
1,973620
add a comment |Â
up vote
0
down vote
$(a-5)^2+(sqrtb-4-1)^2+|sqrt-2|=0$
This lead $a=b=c=5$.
answered 2 hours ago
Takahiro Waki
1,973620
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$(a-5)^2+(sqrtb-4-1)^2+|sqrt-2|=0$
This lead $a=b=c=5$.
answered 2 hours ago
Takahiro Waki
1,973620
$(a-5)^2+(sqrtb-4-1)^2+|sqrt-2|=0$
This lead $a=b=c=5$.
answered 2 hours ago
Takahiro Waki
1,973620
answered 2 hours ago
Takahiro Waki
1,973620
answered 2 hours ago
Takahiro Waki
1,973620
answered 2 hours ago
Takahiro Waki
1,973620
1,973620
add a comment |Â
add a comment |Â
up vote
0
down vote
Noodling:
$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.
So I get
$a^2 - 10a +25+b +|sqrtc-1 - 2|=(a-5)^2 + b+ |sqrtc-1 -2|= 2sqrtb-4 + 3$
And, well this seems a bit weird but that is an even number in front of the $sqrtb-4$ and we have $sqrtb-4$ and $b$ variables to deal with so we can complete the square again with $v = sqrtb-4$ and $v^2 = b-4$.
(In the back of my mind I'm worrying about the $sqrtc-1$ which is just a single term; I'm not sure at this point what will happen.)
$(a -5)^2 + (b-4) - 2sqrtb-4 + 1 +|sqrtc-1 - 2|=3-4 + 1$
$(a-5)^2 + (sqrtb-4 - 1)^2 +|sqrtc-1 - 2| = 0$.
Oh......
We have three things that can't be negative adding up to $0$. So they must each equal $0$.
So $(a-5)^2 = 0$ and $a =5$. And $(sqrtb-4-1)^2=0$ and $b=5$. And $|sqrtc-1 -2| = 0$ so $c=5$.
Well.... okay then......
answered 2 hours ago
fleablood
62.3k22678
add a comment |Â
up vote
0
down vote
Noodling:
$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.
So I get
$a^2 - 10a +25+b +|sqrtc-1 - 2|=(a-5)^2 + b+ |sqrtc-1 -2|= 2sqrtb-4 + 3$
And, well this seems a bit weird but that is an even number in front of the $sqrtb-4$ and we have $sqrtb-4$ and $b$ variables to deal with so we can complete the square again with $v = sqrtb-4$ and $v^2 = b-4$.
(In the back of my mind I'm worrying about the $sqrtc-1$ which is just a single term; I'm not sure at this point what will happen.)
$(a -5)^2 + (b-4) - 2sqrtb-4 + 1 +|sqrtc-1 - 2|=3-4 + 1$
$(a-5)^2 + (sqrtb-4 - 1)^2 +|sqrtc-1 - 2| = 0$.
Oh......
We have three things that can't be negative adding up to $0$. So they must each equal $0$.
So $(a-5)^2 = 0$ and $a =5$. And $(sqrtb-4-1)^2=0$ and $b=5$. And $|sqrtc-1 -2| = 0$ so $c=5$.
Well.... okay then......
answered 2 hours ago
fleablood
62.3k22678
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Noodling:
$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.
So I get
$a^2 - 10a +25+b +|sqrtc-1 - 2|=(a-5)^2 + b+ |sqrtc-1 -2|= 2sqrtb-4 + 3$
And, well this seems a bit weird but that is an even number in front of the $sqrtb-4$ and we have $sqrtb-4$ and $b$ variables to deal with so we can complete the square again with $v = sqrtb-4$ and $v^2 = b-4$.
(In the back of my mind I'm worrying about the $sqrtc-1$ which is just a single term; I'm not sure at this point what will happen.)
$(a -5)^2 + (b-4) - 2sqrtb-4 + 1 +|sqrtc-1 - 2|=3-4 + 1$
$(a-5)^2 + (sqrtb-4 - 1)^2 +|sqrtc-1 - 2| = 0$.
Oh......
We have three things that can't be negative adding up to $0$. So they must each equal $0$.
So $(a-5)^2 = 0$ and $a =5$. And $(sqrtb-4-1)^2=0$ and $b=5$. And $|sqrtc-1 -2| = 0$ so $c=5$.
Well.... okay then......
answered 2 hours ago
fleablood
62.3k22678
Noodling:
$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.
So I get
$a^2 - 10a +25+b +|sqrtc-1 - 2|=(a-5)^2 + b+ |sqrtc-1 -2|= 2sqrtb-4 + 3$
And, well this seems a bit weird but that is an even number in front of the $sqrtb-4$ and we have $sqrtb-4$ and $b$ variables to deal with so we can complete the square again with $v = sqrtb-4$ and $v^2 = b-4$.
(In the back of my mind I'm worrying about the $sqrtc-1$ which is just a single term; I'm not sure at this point what will happen.)
$(a -5)^2 + (b-4) - 2sqrtb-4 + 1 +|sqrtc-1 - 2|=3-4 + 1$
$(a-5)^2 + (sqrtb-4 - 1)^2 +|sqrtc-1 - 2| = 0$.
Oh......
We have three things that can't be negative adding up to $0$. So they must each equal $0$.
So $(a-5)^2 = 0$ and $a =5$. And $(sqrtb-4-1)^2=0$ and $b=5$. And $|sqrtc-1 -2| = 0$ so $c=5$.
Well.... okay then......
answered 2 hours ago
fleablood
62.3k22678
answered 2 hours ago
fleablood
62.3k22678
answered 2 hours ago
fleablood
62.3k22678
answered 2 hours ago
fleablood
62.3k22678
62.3k22678
add a comment |Â
add a comment |Â
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