Is there a solution to this geometry problem?
$begingroup$
The point $G$ is centroid in the triangle $ABC$.
$m(BGC)=90^o$
$|AB|=8 ;cm$
$|CG|=? , cm$
I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?
geometry euclidean-geometry
edited Dec 4 '18 at 18:50
greedoid
40.7k1149100
asked Dec 4 '18 at 11:51
Eldar RahimliEldar Rahimli
1269
$endgroup$
add a comment |
$begingroup$
The point $G$ is centroid in the triangle $ABC$.
$m(BGC)=90^o$
$|AB|=8 ;cm$
$|CG|=? , cm$
I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?
geometry euclidean-geometry
edited Dec 4 '18 at 18:50
greedoid
40.7k1149100
asked Dec 4 '18 at 11:51
Eldar RahimliEldar Rahimli
1269
$endgroup$
$begingroup$
I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
$endgroup$
– user10354138
Dec 4 '18 at 12:20
$begingroup$
BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
$endgroup$
– Andrei
Dec 4 '18 at 15:35
add a comment |
$begingroup$
The point $G$ is centroid in the triangle $ABC$.
$m(BGC)=90^o$
$|AB|=8 ;cm$
$|CG|=? , cm$
I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?
geometry euclidean-geometry
edited Dec 4 '18 at 18:50
greedoid
40.7k1149100
asked Dec 4 '18 at 11:51
Eldar RahimliEldar Rahimli
1269
$endgroup$
The point $G$ is centroid in the triangle $ABC$.
$m(BGC)=90^o$
$|AB|=8 ;cm$
$|CG|=? , cm$
I know that $|GD|=|BD|=|DC|$ and $|AG|=|BC|=2|GD|$. I'm stuck after this point. I've tried to find the answer by solving system of equations, but it didn't work out. How do I solve this question?
geometry euclidean-geometry
geometry euclidean-geometry
edited Dec 4 '18 at 18:50
greedoid
40.7k1149100
asked Dec 4 '18 at 11:51
Eldar RahimliEldar Rahimli
1269
edited Dec 4 '18 at 18:50
greedoid
40.7k1149100
asked Dec 4 '18 at 11:51
Eldar RahimliEldar Rahimli
1269
edited Dec 4 '18 at 18:50
greedoid
40.7k1149100
edited Dec 4 '18 at 18:50
greedoid
40.7k1149100
edited Dec 4 '18 at 18:50
greedoid
40.7k1149100
40.7k1149100
asked Dec 4 '18 at 11:51
Eldar RahimliEldar Rahimli
1269
asked Dec 4 '18 at 11:51
Eldar RahimliEldar Rahimli
1269
asked Dec 4 '18 at 11:51
Eldar RahimliEldar Rahimli
1269
1269
$begingroup$
I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
$endgroup$
– user10354138
Dec 4 '18 at 12:20
$begingroup$
BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
$endgroup$
– Andrei
Dec 4 '18 at 15:35
add a comment |
$begingroup$
I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
$endgroup$
– user10354138
Dec 4 '18 at 12:20
$begingroup$
BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
$endgroup$
– Andrei
Dec 4 '18 at 15:35
$begingroup$
I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
$endgroup$
– user10354138
Dec 4 '18 at 12:20
$begingroup$
I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
$endgroup$
– user10354138
Dec 4 '18 at 12:20
$begingroup$
BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
$endgroup$
– Andrei
Dec 4 '18 at 15:35
$begingroup$
BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
$endgroup$
– Andrei
Dec 4 '18 at 15:35
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The point $D$ is not uniqely determined. Everything is on a picture:
We have $$4b^2+c^2=16$$
but $$ CG = 2c $$
answered Dec 4 '18 at 18:07
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)
Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.
Thus, the perpendicular medians tell us this about $D$:
$$|AD| = 3,|BD|$$
Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.
Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.
However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$
answered Dec 4 '18 at 17:40
BlueBlue
48k870153
$endgroup$
add a comment |
$begingroup$
Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$
Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)
Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.
Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.
answered Dec 4 '18 at 16:32
liaombroliaombro
18017
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The point $D$ is not uniqely determined. Everything is on a picture:
We have $$4b^2+c^2=16$$
but $$ CG = 2c $$
answered Dec 4 '18 at 18:07
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
The point $D$ is not uniqely determined. Everything is on a picture:
We have $$4b^2+c^2=16$$
but $$ CG = 2c $$
answered Dec 4 '18 at 18:07
greedoidgreedoid
40.7k1149100
$endgroup$
add a comment |
$begingroup$
The point $D$ is not uniqely determined. Everything is on a picture:
We have $$4b^2+c^2=16$$
but $$ CG = 2c $$
answered Dec 4 '18 at 18:07
greedoidgreedoid
40.7k1149100
$endgroup$
The point $D$ is not uniqely determined. Everything is on a picture:
We have $$4b^2+c^2=16$$
but $$ CG = 2c $$
answered Dec 4 '18 at 18:07
greedoidgreedoid
40.7k1149100
edited Dec 4 '18 at 18:20
answered Dec 4 '18 at 18:07
greedoidgreedoid
40.7k1149100
answered Dec 4 '18 at 18:07
greedoidgreedoid
40.7k1149100
answered Dec 4 '18 at 18:07
greedoidgreedoid
40.7k1149100
40.7k1149100
add a comment |
add a comment |
$begingroup$
I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)
Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.
Thus, the perpendicular medians tell us this about $D$:
$$|AD| = 3,|BD|$$
Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.
Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.
However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$
answered Dec 4 '18 at 17:40
BlueBlue
48k870153
$endgroup$
add a comment |
$begingroup$
I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)
Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.
Thus, the perpendicular medians tell us this about $D$:
$$|AD| = 3,|BD|$$
Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.
Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.
However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$
answered Dec 4 '18 at 17:40
BlueBlue
48k870153
$endgroup$
add a comment |
$begingroup$
I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)
Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.
Thus, the perpendicular medians tell us this about $D$:
$$|AD| = 3,|BD|$$
Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.
Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.
However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$
answered Dec 4 '18 at 17:40
BlueBlue
48k870153
$endgroup$
I'll start by recapping, with explanation, some things OP already knows. (I believe OP made a typo in asserting $|AG|=|BC|=frac12|GD|$. The "$frac12$" should (obviously?) be a "$2$".)
Since $overline{AD}$ is a median, the condition that $angle BGC$ is right tells us that $D$ is the midpoint of the hypotenuse of right triangle $triangle BGC$; therefore, $D$ is the circumcenter of that triangle: $|BD|=|GD|$. Moreover, we know that a centroid trisects each median: $|AD|=3,|GD|$.
Thus, the perpendicular medians tell us this about $D$:
$$|AD| = 3,|BD|$$
Now, there's a whole family of points $D$ whose distance from $A$ is three-times their distance from $B$: they live on the circle with diameter $overline{D_1D_2}$ where $D_1$ and $D_2$ are the two easily-located points on $overleftrightarrow{AB}$ (one between $A$ and $B$, the other on the "far side" of $B$) satisfying that property.
Every point $D$ on this circle (except perhaps $D_1$ and $D_2$ themselves, depending upon the reader's tolerance for degenerate cases) gives rise to a viable $triangle ABC$.
However, the target distance, $|CG|$, varies, from a minimum of $0$ (when $D=D_1$) to a maximum of $|AB|$ (when $D=D_2$). Consequently, there is no unique solution to the problem described. $square$
answered Dec 4 '18 at 17:40
BlueBlue
48k870153
answered Dec 4 '18 at 17:40
BlueBlue
48k870153
answered Dec 4 '18 at 17:40
BlueBlue
48k870153
answered Dec 4 '18 at 17:40
BlueBlue
48k870153
48k870153
add a comment |
add a comment |
$begingroup$
Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$
Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)
Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.
Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.
answered Dec 4 '18 at 16:32
liaombroliaombro
18017
$endgroup$
add a comment |
$begingroup$
Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$
Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)
Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.
Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.
answered Dec 4 '18 at 16:32
liaombroliaombro
18017
$endgroup$
add a comment |
$begingroup$
Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$
Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)
Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.
Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.
answered Dec 4 '18 at 16:32
liaombroliaombro
18017
$endgroup$
Perpendicularity of the two medians $m_b$ and $m_c$ is equivalent to $b^2+c^2=5a^2$
Knowing $c=8$ and $b^2+c^2=5a^2$ does not allow us to determine $2a^2+2b^2-c^2$ (that is, the $4m_c^2$, which we would know if we knew $|CG|$)
Consider two triangles, one having $a=sqrt{20}, b=6, c=8$ and the other $a= sqrt{45}, b=sqrt{161}, c=8$.
Both of them satisfy $b^2+c^2=5a^2$, and yet the quantity $2a^2+2b^2-c^2 = 12a^2 - 3c^2$ is different for them, since their $a$'s differ.
answered Dec 4 '18 at 16:32
liaombroliaombro
18017
answered Dec 4 '18 at 16:32
liaombroliaombro
18017
answered Dec 4 '18 at 16:32
liaombroliaombro
18017
answered Dec 4 '18 at 16:32
liaombroliaombro
18017
18017
add a comment |
add a comment |
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$begingroup$
I don't think a (unique) solution exist, e.g. $G$ very close to $B$ would mean $CGapprox 2AB$, whereas $G$ very close to $C$ would mean $CG$ must be tiny.
$endgroup$
– user10354138
Dec 4 '18 at 12:20
$begingroup$
BD=DC=GD means BC=(BD+DC)=2GD in the first equation, while in the second BC=GD/2. This can be true only if BC=0
$endgroup$
– Andrei
Dec 4 '18 at 15:35