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Prove that the area of the triangle formed by the medians is equal to $3/4$ the area of the original triangle
How to find area of triangle from its mediansInequality in triangle involving side lenghs, medians and areaRatio of area of triangle to that formed by its mediansTriangle inscribed inside a circle: prove that $abc = 4 times area times R$Area of a circumcenter triangle equals area of medial triangleA plane contains a set of marked points, such that any three can be covered by a unit disk. Prove that the entire set can be covered by a unit disk.Inequality from IMO 2000 problem 4 question $prodlimits_cycleft(a-1+frac1bright)leq 1$ $abc=1$For all triangle prove that $sumlimits_cycm_acosfracalpha2geqfrac34(a+b+c)$Parallelogram Contest ProblemFinding Value of Expression in a TriangleFinding the area of inner triangle constructed by three cevian lines of a large triangle
$begingroup$
This is Question $2$ from this document on Olympiad Geometry.
Let $ABC$ be a triangle and $M_A,M_B,M_C$ the midpoints of the sides $BC, CA, AB$, respectively. Show that the triangle with side lengths $AM_A, BM_B, CM_C$ has area $3/4$ that of the triangle $ABC$.
This is part of a chapter that stresses that by "slicing and dicing", we can solve a lot of complicated problems. Hence, the stress is on diagrammatic proofs.
To form a triangle with the medians, I extended $AM_A$ beyond $BC$, and formed another copy of the triangle $ABC$. My diagram looks like this:
Obviously $BP=CM_C$. Hence, if $PM_B=AM_A$, we'll have created a triangle with the medians as sides.
So my question is, is $AM_A=PM_B$? A followup question would be is it easy to see that the area of the triangle $BPM_B$ is equal to $3/8$ that of the parallelogram given?
contest-math
asked 2 days ago
Anju GeorgeAnju George
564
$endgroup$
|
show 1 more comment
$begingroup$
This is Question $2$ from this document on Olympiad Geometry.
Let $ABC$ be a triangle and $M_A,M_B,M_C$ the midpoints of the sides $BC, CA, AB$, respectively. Show that the triangle with side lengths $AM_A, BM_B, CM_C$ has area $3/4$ that of the triangle $ABC$.
This is part of a chapter that stresses that by "slicing and dicing", we can solve a lot of complicated problems. Hence, the stress is on diagrammatic proofs.
To form a triangle with the medians, I extended $AM_A$ beyond $BC$, and formed another copy of the triangle $ABC$. My diagram looks like this:
Obviously $BP=CM_C$. Hence, if $PM_B=AM_A$, we'll have created a triangle with the medians as sides.
So my question is, is $AM_A=PM_B$? A followup question would be is it easy to see that the area of the triangle $BPM_B$ is equal to $3/8$ that of the parallelogram given?
contest-math
asked 2 days ago
Anju GeorgeAnju George
564
$endgroup$
$begingroup$
Hint: Yes, $AM_A = PM_B$. Look at triangles $triangle CPM_B$ and $triangle CA'A$. They are similar.
$endgroup$
– Sameer Kailasa
2 days ago
$begingroup$
@SameerKailasa- Haha yes should've seen that. How about proving that the area of $BPM_B$ is equal to $3/8$ the area of $ABCA'$? Is it easy to see from this diagram?
$endgroup$
– Anju George
2 days ago
$begingroup$
You will find very inspiring the answers of math.stackexchange.com/q/396085
$endgroup$
– Jean Marie
2 days ago
1
$begingroup$
Besides, use for example geogebra to draw your figures instead of a photograph of an approximate figure drawn on a blackboard :)
$endgroup$
– Jean Marie
2 days ago
$begingroup$
See as well jwilson.coe.uga.edu/emt725/Medians.Triangle/…
$endgroup$
– Jean Marie
yesterday
|
show 1 more comment
$begingroup$
This is Question $2$ from this document on Olympiad Geometry.
Let $ABC$ be a triangle and $M_A,M_B,M_C$ the midpoints of the sides $BC, CA, AB$, respectively. Show that the triangle with side lengths $AM_A, BM_B, CM_C$ has area $3/4$ that of the triangle $ABC$.
This is part of a chapter that stresses that by "slicing and dicing", we can solve a lot of complicated problems. Hence, the stress is on diagrammatic proofs.
To form a triangle with the medians, I extended $AM_A$ beyond $BC$, and formed another copy of the triangle $ABC$. My diagram looks like this:
Obviously $BP=CM_C$. Hence, if $PM_B=AM_A$, we'll have created a triangle with the medians as sides.
So my question is, is $AM_A=PM_B$? A followup question would be is it easy to see that the area of the triangle $BPM_B$ is equal to $3/8$ that of the parallelogram given?
contest-math
asked 2 days ago
Anju GeorgeAnju George
564
$endgroup$
This is Question $2$ from this document on Olympiad Geometry.
Let $ABC$ be a triangle and $M_A,M_B,M_C$ the midpoints of the sides $BC, CA, AB$, respectively. Show that the triangle with side lengths $AM_A, BM_B, CM_C$ has area $3/4$ that of the triangle $ABC$.
This is part of a chapter that stresses that by "slicing and dicing", we can solve a lot of complicated problems. Hence, the stress is on diagrammatic proofs.
To form a triangle with the medians, I extended $AM_A$ beyond $BC$, and formed another copy of the triangle $ABC$. My diagram looks like this:
Obviously $BP=CM_C$. Hence, if $PM_B=AM_A$, we'll have created a triangle with the medians as sides.
So my question is, is $AM_A=PM_B$? A followup question would be is it easy to see that the area of the triangle $BPM_B$ is equal to $3/8$ that of the parallelogram given?
contest-math
contest-math
asked 2 days ago
Anju GeorgeAnju George
564
asked 2 days ago
Anju GeorgeAnju George
564
asked 2 days ago
Anju GeorgeAnju George
564
asked 2 days ago
Anju GeorgeAnju George
564
asked 2 days ago
Anju GeorgeAnju George
564
564
$begingroup$
Hint: Yes, $AM_A = PM_B$. Look at triangles $triangle CPM_B$ and $triangle CA'A$. They are similar.
$endgroup$
– Sameer Kailasa
2 days ago
$begingroup$
@SameerKailasa- Haha yes should've seen that. How about proving that the area of $BPM_B$ is equal to $3/8$ the area of $ABCA'$? Is it easy to see from this diagram?
$endgroup$
– Anju George
2 days ago
$begingroup$
You will find very inspiring the answers of math.stackexchange.com/q/396085
$endgroup$
– Jean Marie
2 days ago
1
$begingroup$
Besides, use for example geogebra to draw your figures instead of a photograph of an approximate figure drawn on a blackboard :)
$endgroup$
– Jean Marie
2 days ago
$begingroup$
See as well jwilson.coe.uga.edu/emt725/Medians.Triangle/…
$endgroup$
– Jean Marie
yesterday
|
show 1 more comment
$begingroup$
Hint: Yes, $AM_A = PM_B$. Look at triangles $triangle CPM_B$ and $triangle CA'A$. They are similar.
$endgroup$
– Sameer Kailasa
2 days ago
$begingroup$
@SameerKailasa- Haha yes should've seen that. How about proving that the area of $BPM_B$ is equal to $3/8$ the area of $ABCA'$? Is it easy to see from this diagram?
$endgroup$
– Anju George
2 days ago
$begingroup$
You will find very inspiring the answers of math.stackexchange.com/q/396085
$endgroup$
– Jean Marie
2 days ago
1
$begingroup$
Besides, use for example geogebra to draw your figures instead of a photograph of an approximate figure drawn on a blackboard :)
$endgroup$
– Jean Marie
2 days ago
$begingroup$
See as well jwilson.coe.uga.edu/emt725/Medians.Triangle/…
$endgroup$
– Jean Marie
yesterday
$begingroup$
Hint: Yes, $AM_A = PM_B$. Look at triangles $triangle CPM_B$ and $triangle CA'A$. They are similar.
$endgroup$
– Sameer Kailasa
2 days ago
$begingroup$
Hint: Yes, $AM_A = PM_B$. Look at triangles $triangle CPM_B$ and $triangle CA'A$. They are similar.
$endgroup$
– Sameer Kailasa
2 days ago
$begingroup$
@SameerKailasa- Haha yes should've seen that. How about proving that the area of $BPM_B$ is equal to $3/8$ the area of $ABCA'$? Is it easy to see from this diagram?
$endgroup$
– Anju George
2 days ago
$begingroup$
@SameerKailasa- Haha yes should've seen that. How about proving that the area of $BPM_B$ is equal to $3/8$ the area of $ABCA'$? Is it easy to see from this diagram?
$endgroup$
– Anju George
2 days ago
$begingroup$
You will find very inspiring the answers of math.stackexchange.com/q/396085
$endgroup$
– Jean Marie
2 days ago
$begingroup$
You will find very inspiring the answers of math.stackexchange.com/q/396085
$endgroup$
– Jean Marie
2 days ago
1
1
$begingroup$
Besides, use for example geogebra to draw your figures instead of a photograph of an approximate figure drawn on a blackboard :)
$endgroup$
– Jean Marie
2 days ago
$begingroup$
Besides, use for example geogebra to draw your figures instead of a photograph of an approximate figure drawn on a blackboard :)
$endgroup$
– Jean Marie
2 days ago
$begingroup$
See as well jwilson.coe.uga.edu/emt725/Medians.Triangle/…
$endgroup$
– Jean Marie
yesterday
$begingroup$
See as well jwilson.coe.uga.edu/emt725/Medians.Triangle/…
$endgroup$
– Jean Marie
yesterday
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Regarding the question, whether $AM_A=PM_B$, the answer is YES.
Simply observe that $$fracCPCA'=fracCM_BCA$$ Thus, in virtue of the Intercept theorem (also known as "Thales' Theorem") $$fracPM_BAA'=fracCPCA'=frac12$$
Can you end it now?
Alternatively, here you have a proof (almost) without words.
Definition: $[...]$ denotes the area of the polygon "..."
Observe that $CE$ is a median in $Delta CDH$. Thus
$$1=fracDKKHfrac[DKE][EKH]=frac[DKC][KHC]=frac[DKC]-[DKE][KHC]-[EKH]=frac[DEC][EHC]iff [DEC]=[EHC]$$ Similarly $$[DEC]=[DHE]=[EHC]$$
Futhermore $$[DEC]=frac[ABC]4$$ Can you end it now?
answered yesterday
Dr. MathvaDr. Mathva
2,296526
$endgroup$
1
$begingroup$
Nice presentation. A little remark : "Intercept theorem" has another very common name in many countries which is "Thales theorem"
$endgroup$
– Jean Marie
yesterday
1
$begingroup$
In Spain we also know it as "Thales Theorem", but when I looked it up in the English Wikipedia, I fount "intercepting theorem"...- Edited!
$endgroup$
– Dr. Mathva
yesterday
2
$begingroup$
Thanks for the manes of this mathematician !
$endgroup$
– Jean Marie
yesterday
$begingroup$
+1 @Dr.Mathva ...
$endgroup$
– Maria Mazur
yesterday
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
Regarding the question, whether $AM_A=PM_B$, the answer is YES.
Simply observe that $$fracCPCA'=fracCM_BCA$$ Thus, in virtue of the Intercept theorem (also known as "Thales' Theorem") $$fracPM_BAA'=fracCPCA'=frac12$$
Can you end it now?
Alternatively, here you have a proof (almost) without words.
Definition: $[...]$ denotes the area of the polygon "..."
Observe that $CE$ is a median in $Delta CDH$. Thus
$$1=fracDKKHfrac[DKE][EKH]=frac[DKC][KHC]=frac[DKC]-[DKE][KHC]-[EKH]=frac[DEC][EHC]iff [DEC]=[EHC]$$ Similarly $$[DEC]=[DHE]=[EHC]$$
Futhermore $$[DEC]=frac[ABC]4$$ Can you end it now?
answered yesterday
Dr. MathvaDr. Mathva
2,296526
$endgroup$
1
$begingroup$
Nice presentation. A little remark : "Intercept theorem" has another very common name in many countries which is "Thales theorem"
$endgroup$
– Jean Marie
yesterday
1
$begingroup$
In Spain we also know it as "Thales Theorem", but when I looked it up in the English Wikipedia, I fount "intercepting theorem"...- Edited!
$endgroup$
– Dr. Mathva
yesterday
2
$begingroup$
Thanks for the manes of this mathematician !
$endgroup$
– Jean Marie
yesterday
$begingroup$
+1 @Dr.Mathva ...
$endgroup$
– Maria Mazur
yesterday
add a comment |
$begingroup$
Regarding the question, whether $AM_A=PM_B$, the answer is YES.
Simply observe that $$fracCPCA'=fracCM_BCA$$ Thus, in virtue of the Intercept theorem (also known as "Thales' Theorem") $$fracPM_BAA'=fracCPCA'=frac12$$
Can you end it now?
Alternatively, here you have a proof (almost) without words.
Definition: $[...]$ denotes the area of the polygon "..."
Observe that $CE$ is a median in $Delta CDH$. Thus
$$1=fracDKKHfrac[DKE][EKH]=frac[DKC][KHC]=frac[DKC]-[DKE][KHC]-[EKH]=frac[DEC][EHC]iff [DEC]=[EHC]$$ Similarly $$[DEC]=[DHE]=[EHC]$$
Futhermore $$[DEC]=frac[ABC]4$$ Can you end it now?
answered yesterday
Dr. MathvaDr. Mathva
2,296526
$endgroup$
1
$begingroup$
Nice presentation. A little remark : "Intercept theorem" has another very common name in many countries which is "Thales theorem"
$endgroup$
– Jean Marie
yesterday
1
$begingroup$
In Spain we also know it as "Thales Theorem", but when I looked it up in the English Wikipedia, I fount "intercepting theorem"...- Edited!
$endgroup$
– Dr. Mathva
yesterday
2
$begingroup$
Thanks for the manes of this mathematician !
$endgroup$
– Jean Marie
yesterday
$begingroup$
+1 @Dr.Mathva ...
$endgroup$
– Maria Mazur
yesterday
add a comment |
$begingroup$
Regarding the question, whether $AM_A=PM_B$, the answer is YES.
Simply observe that $$fracCPCA'=fracCM_BCA$$ Thus, in virtue of the Intercept theorem (also known as "Thales' Theorem") $$fracPM_BAA'=fracCPCA'=frac12$$
Can you end it now?
Alternatively, here you have a proof (almost) without words.
Definition: $[...]$ denotes the area of the polygon "..."
Observe that $CE$ is a median in $Delta CDH$. Thus
$$1=fracDKKHfrac[DKE][EKH]=frac[DKC][KHC]=frac[DKC]-[DKE][KHC]-[EKH]=frac[DEC][EHC]iff [DEC]=[EHC]$$ Similarly $$[DEC]=[DHE]=[EHC]$$
Futhermore $$[DEC]=frac[ABC]4$$ Can you end it now?
answered yesterday
Dr. MathvaDr. Mathva
2,296526
$endgroup$
Regarding the question, whether $AM_A=PM_B$, the answer is YES.
Simply observe that $$fracCPCA'=fracCM_BCA$$ Thus, in virtue of the Intercept theorem (also known as "Thales' Theorem") $$fracPM_BAA'=fracCPCA'=frac12$$
Can you end it now?
Alternatively, here you have a proof (almost) without words.
Definition: $[...]$ denotes the area of the polygon "..."
Observe that $CE$ is a median in $Delta CDH$. Thus
$$1=fracDKKHfrac[DKE][EKH]=frac[DKC][KHC]=frac[DKC]-[DKE][KHC]-[EKH]=frac[DEC][EHC]iff [DEC]=[EHC]$$ Similarly $$[DEC]=[DHE]=[EHC]$$
Futhermore $$[DEC]=frac[ABC]4$$ Can you end it now?
answered yesterday
Dr. MathvaDr. Mathva
2,296526
edited yesterday
answered yesterday
Dr. MathvaDr. Mathva
2,296526
answered yesterday
Dr. MathvaDr. Mathva
2,296526
answered yesterday
Dr. MathvaDr. Mathva
2,296526
2,296526
1
$begingroup$
Nice presentation. A little remark : "Intercept theorem" has another very common name in many countries which is "Thales theorem"
$endgroup$
– Jean Marie
yesterday
1
$begingroup$
In Spain we also know it as "Thales Theorem", but when I looked it up in the English Wikipedia, I fount "intercepting theorem"...- Edited!
$endgroup$
– Dr. Mathva
yesterday
2
$begingroup$
Thanks for the manes of this mathematician !
$endgroup$
– Jean Marie
yesterday
$begingroup$
+1 @Dr.Mathva ...
$endgroup$
– Maria Mazur
yesterday
add a comment |
1
$begingroup$
Nice presentation. A little remark : "Intercept theorem" has another very common name in many countries which is "Thales theorem"
$endgroup$
– Jean Marie
yesterday
1
$begingroup$
In Spain we also know it as "Thales Theorem", but when I looked it up in the English Wikipedia, I fount "intercepting theorem"...- Edited!
$endgroup$
– Dr. Mathva
yesterday
2
$begingroup$
Thanks for the manes of this mathematician !
$endgroup$
– Jean Marie
yesterday
$begingroup$
+1 @Dr.Mathva ...
$endgroup$
– Maria Mazur
yesterday
1
1
$begingroup$
Nice presentation. A little remark : "Intercept theorem" has another very common name in many countries which is "Thales theorem"
$endgroup$
– Jean Marie
yesterday
$begingroup$
Nice presentation. A little remark : "Intercept theorem" has another very common name in many countries which is "Thales theorem"
$endgroup$
– Jean Marie
yesterday
1
1
$begingroup$
In Spain we also know it as "Thales Theorem", but when I looked it up in the English Wikipedia, I fount "intercepting theorem"...- Edited!
$endgroup$
– Dr. Mathva
yesterday
$begingroup$
In Spain we also know it as "Thales Theorem", but when I looked it up in the English Wikipedia, I fount "intercepting theorem"...- Edited!
$endgroup$
– Dr. Mathva
yesterday
2
2
$begingroup$
Thanks for the manes of this mathematician !
$endgroup$
– Jean Marie
yesterday
$begingroup$
Thanks for the manes of this mathematician !
$endgroup$
– Jean Marie
yesterday
$begingroup$
+1 @Dr.Mathva ...
$endgroup$
– Maria Mazur
yesterday
$begingroup$
+1 @Dr.Mathva ...
$endgroup$
– Maria Mazur
yesterday
add a comment |
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$begingroup$
Hint: Yes, $AM_A = PM_B$. Look at triangles $triangle CPM_B$ and $triangle CA'A$. They are similar.
$endgroup$
– Sameer Kailasa
2 days ago
$begingroup$
@SameerKailasa- Haha yes should've seen that. How about proving that the area of $BPM_B$ is equal to $3/8$ the area of $ABCA'$? Is it easy to see from this diagram?
$endgroup$
– Anju George
2 days ago
$begingroup$
You will find very inspiring the answers of math.stackexchange.com/q/396085
$endgroup$
– Jean Marie
2 days ago
1
$begingroup$
Besides, use for example geogebra to draw your figures instead of a photograph of an approximate figure drawn on a blackboard :)
$endgroup$
– Jean Marie
2 days ago
$begingroup$
See as well jwilson.coe.uga.edu/emt725/Medians.Triangle/…
$endgroup$
– Jean Marie
yesterday