Finding all possible proofs
I'm now working on a geometry problem I'll have to explain in front of my class this week (I'm in the $10^{th}$ grade). I've found so far some proofs, which might, nevertheless, be a bit complicated for my classmates (since they've barely worked with geometry).
I was wondering hence, whether there might be a "more elementary" proof or an animation (I don't know how to program it) which somehow makes it clearer or at least visual...
The problem is the following
Consider a square $ABCD$ with side $[AB]=a$ and a circumference $omega$ with radius $r$, such that both are "semi inscribed" in each other (the image below might clarify this).
Proof that the following relation holds $$a=frac{8r}{5}$$
My proofs so far are the following
Consider the isosceles triangle $Delta ABE$ inscribed in $omega$.
It suficies to use the Pythagorean theorem and the relation $$R=frac{abc}{4S} $$ in order to prove the statement.
$$$$
An analytic approach might be also helpful (let $A$ be the origin of coordinates). The rest is just a matter of solving equations (quite simple).
As said, any explanations, suggestions and approaches (and of course animations if possible) are welcomed. If you, however, find another approach (such as trigonometry for instance), even if it's more difficult, I would also be glad to receive it, since my teacher told me to find al possible ways to prove it.
Thanks in advance
proof-explanation euclidean-geometry circle ratio
asked Dec 2 '18 at 17:36
Dr. Mathva
919316
add a comment |
I'm now working on a geometry problem I'll have to explain in front of my class this week (I'm in the $10^{th}$ grade). I've found so far some proofs, which might, nevertheless, be a bit complicated for my classmates (since they've barely worked with geometry).
I was wondering hence, whether there might be a "more elementary" proof or an animation (I don't know how to program it) which somehow makes it clearer or at least visual...
The problem is the following
Consider a square $ABCD$ with side $[AB]=a$ and a circumference $omega$ with radius $r$, such that both are "semi inscribed" in each other (the image below might clarify this).
Proof that the following relation holds $$a=frac{8r}{5}$$
My proofs so far are the following
Consider the isosceles triangle $Delta ABE$ inscribed in $omega$.
It suficies to use the Pythagorean theorem and the relation $$R=frac{abc}{4S} $$ in order to prove the statement.
$$$$
An analytic approach might be also helpful (let $A$ be the origin of coordinates). The rest is just a matter of solving equations (quite simple).
As said, any explanations, suggestions and approaches (and of course animations if possible) are welcomed. If you, however, find another approach (such as trigonometry for instance), even if it's more difficult, I would also be glad to receive it, since my teacher told me to find al possible ways to prove it.
Thanks in advance
proof-explanation euclidean-geometry circle ratio
asked Dec 2 '18 at 17:36
Dr. Mathva
919316
1
see math.stackexchange.com/questions/1188845/… (possible duplicate?)
– Carmeister
Dec 2 '18 at 22:42
add a comment |
I'm now working on a geometry problem I'll have to explain in front of my class this week (I'm in the $10^{th}$ grade). I've found so far some proofs, which might, nevertheless, be a bit complicated for my classmates (since they've barely worked with geometry).
I was wondering hence, whether there might be a "more elementary" proof or an animation (I don't know how to program it) which somehow makes it clearer or at least visual...
The problem is the following
Consider a square $ABCD$ with side $[AB]=a$ and a circumference $omega$ with radius $r$, such that both are "semi inscribed" in each other (the image below might clarify this).
Proof that the following relation holds $$a=frac{8r}{5}$$
My proofs so far are the following
Consider the isosceles triangle $Delta ABE$ inscribed in $omega$.
It suficies to use the Pythagorean theorem and the relation $$R=frac{abc}{4S} $$ in order to prove the statement.
$$$$
An analytic approach might be also helpful (let $A$ be the origin of coordinates). The rest is just a matter of solving equations (quite simple).
As said, any explanations, suggestions and approaches (and of course animations if possible) are welcomed. If you, however, find another approach (such as trigonometry for instance), even if it's more difficult, I would also be glad to receive it, since my teacher told me to find al possible ways to prove it.
Thanks in advance
proof-explanation euclidean-geometry circle ratio
asked Dec 2 '18 at 17:36
Dr. Mathva
919316
I'm now working on a geometry problem I'll have to explain in front of my class this week (I'm in the $10^{th}$ grade). I've found so far some proofs, which might, nevertheless, be a bit complicated for my classmates (since they've barely worked with geometry).
I was wondering hence, whether there might be a "more elementary" proof or an animation (I don't know how to program it) which somehow makes it clearer or at least visual...
The problem is the following
Consider a square $ABCD$ with side $[AB]=a$ and a circumference $omega$ with radius $r$, such that both are "semi inscribed" in each other (the image below might clarify this).
Proof that the following relation holds $$a=frac{8r}{5}$$
My proofs so far are the following
Consider the isosceles triangle $Delta ABE$ inscribed in $omega$.
It suficies to use the Pythagorean theorem and the relation $$R=frac{abc}{4S} $$ in order to prove the statement.
$$$$
An analytic approach might be also helpful (let $A$ be the origin of coordinates). The rest is just a matter of solving equations (quite simple).
As said, any explanations, suggestions and approaches (and of course animations if possible) are welcomed. If you, however, find another approach (such as trigonometry for instance), even if it's more difficult, I would also be glad to receive it, since my teacher told me to find al possible ways to prove it.
Thanks in advance
proof-explanation euclidean-geometry circle ratio
proof-explanation euclidean-geometry circle ratio
asked Dec 2 '18 at 17:36
Dr. Mathva
919316
asked Dec 2 '18 at 17:36
Dr. Mathva
919316
edited Dec 2 '18 at 17:44
asked Dec 2 '18 at 17:36
Dr. Mathva
919316
asked Dec 2 '18 at 17:36
Dr. Mathva
919316
asked Dec 2 '18 at 17:36
Dr. Mathva
919316
919316
1
see math.stackexchange.com/questions/1188845/… (possible duplicate?)
– Carmeister
Dec 2 '18 at 22:42
add a comment |
1
see math.stackexchange.com/questions/1188845/… (possible duplicate?)
– Carmeister
Dec 2 '18 at 22:42
1
1
see math.stackexchange.com/questions/1188845/… (possible duplicate?)
– Carmeister
Dec 2 '18 at 22:42
see math.stackexchange.com/questions/1188845/… (possible duplicate?)
– Carmeister
Dec 2 '18 at 22:42
add a comment |
3 Answers
3
active
oldest
votes
I find the following quite instructive:
Obviously
begin{align}
a&=r+y \
&=r+sqrt{r^2 - left(frac{a}{2}right)^2}
end{align}
and solving gives $$frac{r}{a} = frac{5}{8} , .$$
answered Dec 2 '18 at 18:09
Diger
1,5921413
add a comment |
Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
to deduce $a=8r/5$.
answered Dec 2 '18 at 18:06
BPP
2,169927
It might be nice to clarify where you got $a-r$ from.
– Servaes
Dec 2 '18 at 20:31
2
@Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
– egreg
Dec 2 '18 at 23:46
add a comment |
Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.
answered Dec 3 '18 at 5:37
akhmeteli
34616
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
I find the following quite instructive:
Obviously
begin{align}
a&=r+y \
&=r+sqrt{r^2 - left(frac{a}{2}right)^2}
end{align}
and solving gives $$frac{r}{a} = frac{5}{8} , .$$
answered Dec 2 '18 at 18:09
Diger
1,5921413
add a comment |
I find the following quite instructive:
Obviously
begin{align}
a&=r+y \
&=r+sqrt{r^2 - left(frac{a}{2}right)^2}
end{align}
and solving gives $$frac{r}{a} = frac{5}{8} , .$$
answered Dec 2 '18 at 18:09
Diger
1,5921413
add a comment |
I find the following quite instructive:
Obviously
begin{align}
a&=r+y \
&=r+sqrt{r^2 - left(frac{a}{2}right)^2}
end{align}
and solving gives $$frac{r}{a} = frac{5}{8} , .$$
answered Dec 2 '18 at 18:09
Diger
1,5921413
I find the following quite instructive:
Obviously
begin{align}
a&=r+y \
&=r+sqrt{r^2 - left(frac{a}{2}right)^2}
end{align}
and solving gives $$frac{r}{a} = frac{5}{8} , .$$
answered Dec 2 '18 at 18:09
Diger
1,5921413
answered Dec 2 '18 at 18:09
Diger
1,5921413
answered Dec 2 '18 at 18:09
Diger
1,5921413
answered Dec 2 '18 at 18:09
Diger
1,5921413
1,5921413
add a comment |
add a comment |
Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
to deduce $a=8r/5$.
answered Dec 2 '18 at 18:06
BPP
2,169927
It might be nice to clarify where you got $a-r$ from.
– Servaes
Dec 2 '18 at 20:31
2
@Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
– egreg
Dec 2 '18 at 23:46
add a comment |
Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
to deduce $a=8r/5$.
answered Dec 2 '18 at 18:06
BPP
2,169927
It might be nice to clarify where you got $a-r$ from.
– Servaes
Dec 2 '18 at 20:31
2
@Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
– egreg
Dec 2 '18 at 23:46
add a comment |
Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
to deduce $a=8r/5$.
answered Dec 2 '18 at 18:06
BPP
2,169927
Let $I$ be the midpoint of $[AB]$. Use the Pythagorean theorem in triangle $AIM$: $$r^2=dfrac{a^2}{4}+(a-r)^2$$
to deduce $a=8r/5$.
answered Dec 2 '18 at 18:06
BPP
2,169927
answered Dec 2 '18 at 18:06
BPP
2,169927
answered Dec 2 '18 at 18:06
BPP
2,169927
answered Dec 2 '18 at 18:06
BPP
2,169927
2,169927
It might be nice to clarify where you got $a-r$ from.
– Servaes
Dec 2 '18 at 20:31
2
@Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
– egreg
Dec 2 '18 at 23:46
add a comment |
It might be nice to clarify where you got $a-r$ from.
– Servaes
Dec 2 '18 at 20:31
2
@Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
– egreg
Dec 2 '18 at 23:46
It might be nice to clarify where you got $a-r$ from.
– Servaes
Dec 2 '18 at 20:31
It might be nice to clarify where you got $a-r$ from.
– Servaes
Dec 2 '18 at 20:31
2
2
@Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
– egreg
Dec 2 '18 at 23:46
@Servaes Obviously $IM+ME=a$, but $ME=r$. Also $I$, $M$ and $E$ are aligned.
– egreg
Dec 2 '18 at 23:46
add a comment |
Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.
answered Dec 3 '18 at 5:37
akhmeteli
34616
add a comment |
Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.
answered Dec 3 '18 at 5:37
akhmeteli
34616
add a comment |
Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.
answered Dec 3 '18 at 5:37
akhmeteli
34616
Let P be the middle of AB, then AE=$sqrt{5}/2$ from triangle AEP. Let Q be the middle of AE, then EQ=$sqrt{5}/4$, and from similarity of triangles AEP and MEQ you get the answer.
answered Dec 3 '18 at 5:37
akhmeteli
34616
answered Dec 3 '18 at 5:37
akhmeteli
34616
answered Dec 3 '18 at 5:37
akhmeteli
34616
answered Dec 3 '18 at 5:37
akhmeteli
34616
34616
add a comment |
add a comment |
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1
see math.stackexchange.com/questions/1188845/… (possible duplicate?)
– Carmeister
Dec 2 '18 at 22:42