# orderedItems[].object.content
Let ABC be an isosceles triangle with |AB| = |AC|, and let D be point on side BC. Then
|𝐴𝐵|²=|𝐴𝐷|²+|𝐵𝐷|⋅|𝐷𝐶| .

This reduces to the Pythagorean theorem when |BD| = |DC|. But can you prove the result without assuming the Pythagorean theorem?

# orderedItems[].object.contentMap.en
Let ABC be an isosceles triangle with |AB| = |AC|, and let D be point on side BC. Then
|𝐴𝐵|²=|𝐴𝐷|²+|𝐵𝐷|⋅|𝐷𝐶| .

This reduces to the Pythagorean theorem when |BD| = |DC|. But can you prove the result without assuming the Pythagorean theorem?

# orderedItems[].object.content
@icecolbeveridge I can see that

\[12^3 + 1^3 \equiv (-2)^3 + 1^3 \equiv 0 \pmod{7},\]

but I'm not sure how satisfying that is.

# orderedItems[].object.contentMap.en
@icecolbeveridge I can see that

\[12^3 + 1^3 \equiv (-2)^3 + 1^3 \equiv 0 \pmod{7},\]

but I'm not sure how satisfying that is.

# orderedItems[].object.content
Proof:

By the Axiom of Choice, there exists a subset 𝑆 of ℝ that contains exactly one element from each coset of ℤ+√2ℤ .

For each 𝑥∈ℝ there exist unique elements 𝑠ₓ∈𝑆 and 𝑚ₓ,𝑛ₓ∈ℤ so that 𝑥=𝑠ₓ+𝑚ₓ+𝑛ₓ√2 .

Define 𝑔(𝑥)=𝑚ₓ and \(h(x)=\lfloor s_x+n_x\sqrt2\rfloor\).

Then 𝑓(𝑥)=𝑔(𝑥)+h(𝑥) , 𝑔(𝑥) has period \(\sqrt2\) , and h(𝑥) has period 1.

# orderedItems[].object.contentMap.en
Proof:

By the Axiom of Choice, there exists a subset 𝑆 of ℝ that contains exactly one element from each coset of ℤ+√2ℤ .

For each 𝑥∈ℝ there exist unique elements 𝑠ₓ∈𝑆 and 𝑚ₓ,𝑛ₓ∈ℤ so that 𝑥=𝑠ₓ+𝑚ₓ+𝑛ₓ√2 .

Define 𝑔(𝑥)=𝑚ₓ and \(h(x)=\lfloor s_x+n_x\sqrt2\rfloor\).

Then 𝑓(𝑥)=𝑔(𝑥)+h(𝑥) , 𝑔(𝑥) has period \(\sqrt2\) , and h(𝑥) has period 1.

# orderedItems[].object.content
Can you show that
\[f(x) := \lfloor x\rfloor\]
is the sum of two periodic functions?

# orderedItems[].object.contentMap.en
Can you show that
\[f(x) := \lfloor x\rfloor\]
is the sum of two periodic functions?

# orderedItems[].object.content
Testing.

\[0^{0^{0^{0^{0^{0^{0^{\dots}}}}}}} = \frac12\]

# orderedItems[].object.contentMap.en
Testing.

\[0^{0^{0^{0^{0^{0^{0^{\dots}}}}}}} = \frac12\]

# orderedItems[].object.content
@christianp A forward slash is sometimes used for this, e.g. https://en.wikipedia.org/wiki/Wheel_th eory

# orderedItems[].object.contentMap.en
@christianp A forward slash is sometimes used for this, e.g. https://en.wikipedia.org/wiki/Wheel_th eory

# orderedItems[].object.content
@christianp https://mathstodon.xyz/media/X10PRmz90 jAgwtyx1wY

# orderedItems[].object.contentMap.en
@christianp https://mathstodon.xyz/media/X10PRmz90 jAgwtyx1wY

# orderedItems[].object.content
\[\sum_{n=1}^{\infty} \frac{n(n+1)/2}{(2n+1)!!^2} = \frac18\]

# orderedItems[].object.contentMap.en
\[\sum_{n=1}^{\infty} \frac{n(n+1)/2}{(2n+1)!!^2} = \frac18\]

# orderedItems[].object.content
Current mood:

class Circle:
pass

circle = Circle()
circle.circle = circle

# orderedItems[].object.contentMap.ja
Current mood:

class Circle:
pass

circle = Circle()
circle.circle = circle

# orderedItems[].object.content
\[ \sum_{n=1}^{\infty} \sin^2(\pi n! e) < \infty \]

# orderedItems[].object.contentMap.en
\[ \sum_{n=1}^{\infty} \sin^2(\pi n! e) < \infty \]

# orderedItems[].object.content
@chase Hi Chase! Your equation looks very familiar. Did we meet on Tuesday at a Chinese restaurant?

# orderedItems[].object.contentMap.en
@chase Hi Chase! Your equation looks very familiar. Did we meet on Tuesday at a Chinese restaurant?

# orderedItems[].object.content
I wrote a Python script to look for circle packings. I can fit 20 circles of diameter 1 inside a circle of diameter 5.124. This is very close to the best known solution, which has diameter 5.122. Maybe .I can find a new record if I keep searching. https://mathstodon.xyz/media/-owErmTna xnK2FgTVI4

# orderedItems[].object.contentMap.en
I wrote a Python script to look for circle packings. I can fit 20 circles of diameter 1 inside a circle of diameter 5.124. This is very close to the best known solution, which has diameter 5.122. Maybe .I can find a new record if I keep searching. https://mathstodon.xyz/media/-owErmTna xnK2FgTVI4

# orderedItems[].object.content
Proof without words that a pentagonal number is 1/3 of a triangular number. https://mathstodon.xyz/media/_eUukzGyW YYzMNg3rh8

# orderedItems[].object.contentMap.en
Proof without words that a pentagonal number is 1/3 of a triangular number. https://mathstodon.xyz/media/_eUukzGyW YYzMNg3rh8

# orderedItems[].object.content
O hai.

# orderedItems[].object.contentMap.hiLatn
O hai.