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Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is 16. What is the sum of the two squares' areas?

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4

$begingroup$

Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is $16$. What is the sum of the two squares' areas?

This is a bonus question from my online math class that I've been trying to solve but haven't gotten very far. I named $CD$ $x$ and $HD$ $y$, then drew $OB$ and $OF$, then did the Pythagorean theorem to try to get $x^2+y^2$ but I didn't have much luck with that.

geometry algebraic-geometry circles

share|cite|improve this question

edited 8 hours ago

user546770

asked 9 hours ago

user546770user546770

4444 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think it probably has to do with this
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Take the specific case where the squares are congruent. The solution in that case is trivial.
  $endgroup$
  – Daniel Mathias
  8 hours ago
  

  
  
  
  

add a comment
 | 

4

$begingroup$

Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is $16$. What is the sum of the two squares' areas?

This is a bonus question from my online math class that I've been trying to solve but haven't gotten very far. I named $CD$ $x$ and $HD$ $y$, then drew $OB$ and $OF$, then did the Pythagorean theorem to try to get $x^2+y^2$ but I didn't have much luck with that.

geometry algebraic-geometry circles

share|cite|improve this question

edited 8 hours ago

user546770

asked 9 hours ago

user546770user546770

4444 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think it probably has to do with this
  $endgroup$
  – saulspatz
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Take the specific case where the squares are congruent. The solution in that case is trivial.
  $endgroup$
  – Daniel Mathias
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

Two side-by-side squares are inscribed in a semicircle. The diameter of the semicircle is $16$. What is the sum of the two squares' areas?

This is a bonus question from my online math class that I've been trying to solve but haven't gotten very far. I named $CD$ $x$ and $HD$ $y$, then drew $OB$ and $OF$, then did the Pythagorean theorem to try to get $x^2+y^2$ but I didn't have much luck with that.

geometry algebraic-geometry circles

share|cite|improve this question

edited 8 hours ago

user546770

asked 9 hours ago

user546770user546770

4444 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

user546770

asked 9 hours ago

user546770user546770

4444 bronze badges

share|cite|improve this question

edited 8 hours ago

user546770

asked 9 hours ago

user546770user546770

4444 bronze badges

asked 9 hours ago

user546770user546770

4444 bronze badges

asked 9 hours ago

user546770user546770

4444 bronze badges

3 Answers
3

active

oldest

votes

4

$begingroup$

Let $OD=c$ and the side lengths of the two squares $a$ and $b$. From right triangles OAB and OHF,

$$r^2=a^2+(a-c)^2;>>>>> r^2=b^2+(b+c)^2$$

Eliminate $c$ to get,

$$a-sqrt{r^2-a^2}=sqrt{r^2-b^2}-b$$

Square both sides,

$$asqrt{ r^2-a^2} =bsqrt{r^2-b^2}$$

Square again and rearrange,

$$r^2(a^2-b^2)=a^4-b^4$$

Thus, the sum of the two areas is

$$a^2+b^2=r^2=64$$

share|cite|improve this answer

answered 7 hours ago

QuantoQuanto

5,35522 silver badges1515 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

You can use the Pythagorean theorem twice:

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

SeyedSeyed

7,49144 gold badges1616 silver badges2727 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Let $V$ be the common vertex of our squares which lies on the diameter of the semicircle. We may assume that the coordinates of $V$ are $(x,0)$. The upper-right-corner $A$ is located at the intersection (the one with a positive ordinate) of $x^2+y^2=64$ and $y=x-v$. The upper-left-corner $B$ is located at the intersection (always the one with a positive ordinate) of $y=v-x$ and $x^2+y^2=64$. It follows that the ordinate of $A$ is $frac{1}{2}left(-v+sqrt{128-v^2}right)$ and the ordinate of $B$ is $frac{1}{2}left(v+sqrt{128-v^2}right)$. By summing the squares of these numbers we get that the total area of our squares is
$$ frac{1}{4}(v^2+128-v^2+v^2+128-v^2) = 64,$$
i.e. the area of a square built on a radius.

In order to produce an elementary proof, we just have to show that the length of $AB=sqrt{2}sqrt{AA'^2+BB'^2}$ does not depend on the position of $V$ on the diameter. But this is trivial since the symmetric of $B$ with respect to the diameter "sees" the $AB$-chord under an angle of $45^circ$, such that $widehat{AOB}$ always is a right angle (and $AOVB$ is a cyclic quadrilateral).

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Jack D'AurizioJack D'Aurizio

300k3333 gold badges295295 silver badges692692 bronze badges

$endgroup$

add a comment
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4

$begingroup$

Let $OD=c$ and the side lengths of the two squares $a$ and $b$. From right triangles OAB and OHF,

$$r^2=a^2+(a-c)^2;>>>>> r^2=b^2+(b+c)^2$$

Eliminate $c$ to get,

$$a-sqrt{r^2-a^2}=sqrt{r^2-b^2}-b$$

Square both sides,

$$asqrt{ r^2-a^2} =bsqrt{r^2-b^2}$$

Square again and rearrange,

$$r^2(a^2-b^2)=a^4-b^4$$

Thus, the sum of the two areas is

$$a^2+b^2=r^2=64$$

share|cite|improve this answer

answered 7 hours ago

QuantoQuanto

5,35522 silver badges1515 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

Let $OD=c$ and the side lengths of the two squares $a$ and $b$. From right triangles OAB and OHF,

$$r^2=a^2+(a-c)^2;>>>>> r^2=b^2+(b+c)^2$$

Eliminate $c$ to get,

$$a-sqrt{r^2-a^2}=sqrt{r^2-b^2}-b$$

Square both sides,

$$asqrt{ r^2-a^2} =bsqrt{r^2-b^2}$$

Square again and rearrange,

$$r^2(a^2-b^2)=a^4-b^4$$

Thus, the sum of the two areas is

$$a^2+b^2=r^2=64$$

share|cite|improve this answer

answered 7 hours ago

QuantoQuanto

5,35522 silver badges1515 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Let $OD=c$ and the side lengths of the two squares $a$ and $b$. From right triangles OAB and OHF,

$$r^2=a^2+(a-c)^2;>>>>> r^2=b^2+(b+c)^2$$

Eliminate $c$ to get,

$$a-sqrt{r^2-a^2}=sqrt{r^2-b^2}-b$$

Square both sides,

$$asqrt{ r^2-a^2} =bsqrt{r^2-b^2}$$

Square again and rearrange,

$$r^2(a^2-b^2)=a^4-b^4$$

Thus, the sum of the two areas is

$$a^2+b^2=r^2=64$$

share|cite|improve this answer

answered 7 hours ago

QuantoQuanto

5,35522 silver badges1515 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

QuantoQuanto

5,35522 silver badges1515 bronze badges

answered 7 hours ago

QuantoQuanto

5,35522 silver badges1515 bronze badges

answered 7 hours ago

QuantoQuanto

5,35522 silver badges1515 bronze badges

4

$begingroup$

You can use the Pythagorean theorem twice:

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

SeyedSeyed

7,49144 gold badges1616 silver badges2727 bronze badges

$endgroup$

add a comment
 | 

4

$begingroup$

You can use the Pythagorean theorem twice:

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

SeyedSeyed

7,49144 gold badges1616 silver badges2727 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

You can use the Pythagorean theorem twice:

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

SeyedSeyed

7,49144 gold badges1616 silver badges2727 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 7 hours ago

SeyedSeyed

7,49144 gold badges1616 silver badges2727 bronze badges

answered 7 hours ago

SeyedSeyed

7,49144 gold badges1616 silver badges2727 bronze badges

answered 7 hours ago

SeyedSeyed

7,49144 gold badges1616 silver badges2727 bronze badges

3

$begingroup$

Let $V$ be the common vertex of our squares which lies on the diameter of the semicircle. We may assume that the coordinates of $V$ are $(x,0)$. The upper-right-corner $A$ is located at the intersection (the one with a positive ordinate) of $x^2+y^2=64$ and $y=x-v$. The upper-left-corner $B$ is located at the intersection (always the one with a positive ordinate) of $y=v-x$ and $x^2+y^2=64$. It follows that the ordinate of $A$ is $frac{1}{2}left(-v+sqrt{128-v^2}right)$ and the ordinate of $B$ is $frac{1}{2}left(v+sqrt{128-v^2}right)$. By summing the squares of these numbers we get that the total area of our squares is
$$ frac{1}{4}(v^2+128-v^2+v^2+128-v^2) = 64,$$
i.e. the area of a square built on a radius.

In order to produce an elementary proof, we just have to show that the length of $AB=sqrt{2}sqrt{AA'^2+BB'^2}$ does not depend on the position of $V$ on the diameter. But this is trivial since the symmetric of $B$ with respect to the diameter "sees" the $AB$-chord under an angle of $45^circ$, such that $widehat{AOB}$ always is a right angle (and $AOVB$ is a cyclic quadrilateral).

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Jack D'AurizioJack D'Aurizio

300k3333 gold badges295295 silver badges692692 bronze badges

$endgroup$

add a comment
 | 

3

$begingroup$

Let $V$ be the common vertex of our squares which lies on the diameter of the semicircle. We may assume that the coordinates of $V$ are $(x,0)$. The upper-right-corner $A$ is located at the intersection (the one with a positive ordinate) of $x^2+y^2=64$ and $y=x-v$. The upper-left-corner $B$ is located at the intersection (always the one with a positive ordinate) of $y=v-x$ and $x^2+y^2=64$. It follows that the ordinate of $A$ is $frac{1}{2}left(-v+sqrt{128-v^2}right)$ and the ordinate of $B$ is $frac{1}{2}left(v+sqrt{128-v^2}right)$. By summing the squares of these numbers we get that the total area of our squares is
$$ frac{1}{4}(v^2+128-v^2+v^2+128-v^2) = 64,$$
i.e. the area of a square built on a radius.

In order to produce an elementary proof, we just have to show that the length of $AB=sqrt{2}sqrt{AA'^2+BB'^2}$ does not depend on the position of $V$ on the diameter. But this is trivial since the symmetric of $B$ with respect to the diameter "sees" the $AB$-chord under an angle of $45^circ$, such that $widehat{AOB}$ always is a right angle (and $AOVB$ is a cyclic quadrilateral).

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Jack D'AurizioJack D'Aurizio

300k3333 gold badges295295 silver badges692692 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

Let $V$ be the common vertex of our squares which lies on the diameter of the semicircle. We may assume that the coordinates of $V$ are $(x,0)$. The upper-right-corner $A$ is located at the intersection (the one with a positive ordinate) of $x^2+y^2=64$ and $y=x-v$. The upper-left-corner $B$ is located at the intersection (always the one with a positive ordinate) of $y=v-x$ and $x^2+y^2=64$. It follows that the ordinate of $A$ is $frac{1}{2}left(-v+sqrt{128-v^2}right)$ and the ordinate of $B$ is $frac{1}{2}left(v+sqrt{128-v^2}right)$. By summing the squares of these numbers we get that the total area of our squares is
$$ frac{1}{4}(v^2+128-v^2+v^2+128-v^2) = 64,$$
i.e. the area of a square built on a radius.

In order to produce an elementary proof, we just have to show that the length of $AB=sqrt{2}sqrt{AA'^2+BB'^2}$ does not depend on the position of $V$ on the diameter. But this is trivial since the symmetric of $B$ with respect to the diameter "sees" the $AB$-chord under an angle of $45^circ$, such that $widehat{AOB}$ always is a right angle (and $AOVB$ is a cyclic quadrilateral).

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Jack D'AurizioJack D'Aurizio

300k3333 gold badges295295 silver badges692692 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Jack D'AurizioJack D'Aurizio

300k3333 gold badges295295 silver badges692692 bronze badges

answered 8 hours ago

Jack D'AurizioJack D'Aurizio

300k3333 gold badges295295 silver badges692692 bronze badges

answered 8 hours ago

Jack D'AurizioJack D'Aurizio

300k3333 gold badges295295 silver badges692692 bronze badges