hcju/proofwikiqa huggingface dataset:
From Triangle is Medial Triangle of Larger Triangle, it is possible to construct $\triangle DEF$ such that $\triangle ABC$ is the medial triangle of $\triangle DEF$. Let $AJ$, $BG$ and $CH$ be the altitudes of $\triangle ABC$. :600px By definition of medial triangle, $A$, $B$ and $C$ are all midpoints of the sides $DF$, $DE$ and $EF$ of $\triangle DEF$ respectively. As $DF$, $DE$ and $EF$ are parallel to $BC$, $AC$ and $AB$ respectively, $AJ$, $BG$ and $CH$ are perpendicular to $DF$, $DE$ and $EF$ respectively. Thus, by definition, $AJ$, $BG$ and $CH$ are the perpendicular bisectors of $DF$, $DE$ and $EF$ respectively. The result follows from Perpendicular Bisectors of Triangle Meet at Point. {{qed}}
Let the perpendicular bisectors of $AC$ and $AB$ be constructed through $D$ and $E$ respectively to meet at $F$. :500px By definition of perpendicular bisector: :$AE = EB$ and: :$\angle AEF = \angle BEF$ are right angles. From Triangle Side-Angle-Side Equality: :$\triangle AEF = \triangle BEF$ and so $AF = BF$. Similarly, by definition of perpendicular bisector: :$AD = DC$ and: :$\angle ADF = \angle CDF$ are right angles. From Triangle Side-Angle-Side Equality: :$\triangle ADF = \triangle CDF$ and so $AF = CF$. Thus: :$BF = CF$ Let $FG$ be the angle bisector of $\angle BFC$. We have: :$BF = CF$ from above and: :$\angle BFG = \angle CFG$ by construction Thus by Triangle Side-Angle-Side Equality: :$\triangle BFG = \triangle CFG$ and so $BG = CG$. Thus as $\angle BGF = \angle CGF$ it follows that they are both right angles. Thus $FG$ is the perpendicular bisector of $BC$. Thus we have all three perpendicular bisectors of the sides of $ABC$ meeting at the same point $G$. {{qed}}
:500px By definition of centroid and medial triangle, the lines $AE$, $BF$ and $CD$ intersect at $G$. It remains to be shown that $AE$, $BF$ and $CD$ bisect the sides of $DF$, $DE$ and $EF$ respectively. {{WLOG}}, let $AE$ intersect $DF$ at $H$. From the working of Triangle is Medial Triangle of Larger Triangle, we have that: :$DE \parallel AF$ :$EF \parallel AD$ and so $\Box ADEF$ is a parallelogram whose diagonals are $DF$ and $AE$. From Diameters of Parallelogram Bisect each other, $AE$ bisects $DF$ at $H$. Similarly: :$BF$ bisects $DE$ :$DC$ bisects $EF$. Hence the result. {{qed}} Category:Medial Triangles Category:Centroids of Triangles 2c11pj1gjodspx0ze0u3d6irn660nc4
:300px Let $\Box ABCD$ be embedded in the complex plane such that $B$ is identified with the origin $0 + 0 i$. Let $A$ be identified with the complex number $z_1$. Let $C$ be identified with the complex number $z_2$. By Geometrical Interpretation of Complex Subtraction: :$z_1 - z_2 = AC$ Then: :$AE = m \paren {z_1 - z_2}$ for some $m$ where $0 \le m \le 1$. Similarly, by Geometrical Interpretation of Complex Addition: :$z_1 + z_2 = BD$ Then: :$BE = n \paren {z_1 + z_2}$ for some $n$ where $0 \le n \le 1$. Then: {{begin-eqn}} {{eqn | l = BA + AE       | r = BE       | c =  }} {{eqn | ll= \leadsto       | l = z_1 + m \paren {z_1 - z_2}       | r = n \paren {z_1 + z_2}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {1 - m - n} z_1 + \paren {m - n} z_2       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = 1 - m - n       | r = 0       | c = Linear Combination of Non-Parallel Complex Numbers is Zero if Factors are Both Zero }} {{eqn | l = m - n       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = m       | r = \frac 1 2       | c =  }} {{eqn | l = n       | r = \frac 1 2       | c =  }} {{end-eqn}} and so $E$ is at the midpoint of the diameters of $\Box ABCD$. {{qed}}
:500px Let $FG$, $DH$ and $EJ$ be the perpendicular bisectors of the sides of $AC$, $AB$ and $BC$ respectively. From Circumscribing Circle about Triangle, the point $K$ where they intersect is the circumcenter of $\triangle ABC$. From Perpendicular Bisector of Triangle is Altitude of Medial Triangle, $FG$, $DH$ and $EJ$ are the altitudes of $\triangle DEF$. The result follows by definition of orthocenter. {{qed}} Category:Circumcenters of Triangles Category:Orthocenters of Triangles Category:Medial Triangles 71lpvdskjjluhb7te28xhi7m1cx6w8r
First it is necessary to dispose of the case where $\triangle ABC$ is equilateral. From Orthocenter, Centroid and Circumcenter Coincide iff Triangle is Equilateral, in that case $O$, $G$ and $H$ are the same point. For all other triangles, $O$, $G$ and $H$ are distinct. :500px Let $A'$ be the midpoint of $BC$. Let $B'$ be the midpoint of $AC$. By definition of centroid, $G$ is the point at which $AA'$ and $BB'$ intersect. By Circumscribing Circle about Triangle, $O$ is the point at which the perpendicular bisectors of $AC$ and $BC$ intersect. By construction, the perpendicular bisectors of $BC$ and $AC$ pass through $A'$ and $B'$ respectively. Let $OG$ be produced to $H$ such that $OG : HG = 1 : 2$. It will be demonstrated that $H$ is the orthocenter of $\triangle ABC$. Draw $AH$ and produce it to intersect $BC$ at $P$. From Medians of Triangle Meet at Point: :$A'G : AG = 1 : 2$ Consider $\triangle A'OG$ and $\triangle AHG$. From above: :$OG : HG = A'G : AG$ From Two Straight Lines make Equal Opposite Angles: :$\angle OGA' = \angle HGA$ From Triangles with One Equal Angle and Two Sides Proportional are Similar: :$\triangle A'OG$ and $\triangle AHG$ are similar. Thus: :$\angle GHA = \angle GOA'$ From Equal Alternate Angles implies Parallel Lines: :$AH \parallel A'O$ From Parallelism implies Equal Corresponding Angles: :$\angle APC = \angle OA'C$ As $OA'$ is perpendicular to $BC$, it follows that $AP$ is also perpendicular to $BC$. Thus $AP$ is a straight line through $A$ perpendicular to $BC$. So by definition $AP$ is an altitude of $\triangle ABC$ which passes through $H$. Similarly, draw $BH$ and produce it to intersect $AC$ at $Q$. By a similar analysis of the triangles $\triangle B'OG$ and $\triangle BHG$: :$BQ$ is an altitude of $\triangle ABC$ which passes through $H$. From Altitudes of Triangle Meet at Point, the altitude of $\triangle ABC$ from $C$ to $AB$ likewise will pass through $H$. Thus $H$ is by definition the orthocenter of $\triangle ABC$. By construction, $OGH$ is a straight line such that: :$OG : GH = 1 : 2$ Hence the result. {{qed}}
:500px Consider the triangles $\triangle ABC$ and $\triangle DBE$. By construction: :$BA : BD = 2 : 1$ :$BC : BE = 2 : 1$ {{improve|Use Midline Theorem}} By Parallel Line in Triangle Cuts Sides Proportionally: :$AC \parallel DE$ From Parallelism implies Equal Alternate Angles: :$\angle AFP = \angle FPE$ By construction, $\angle AFP$ is a right angle. Thus $\angle FPE$ is also a right angle. That is, $FP$ is perpendicular to $DE$. By construction, $FP$ passes through the vertex $F$ of $\triangle DEF$. The result follows by definition of altitude of triangle. {{qed}} Category:Medial Triangles sruor55yvu0w304bjnryfmu6v9ald5e
=== Necessary Condition === Let $\triangle ABC$ be an equilateral triangle. By definition, each side of $\triangle ABC$ is the base of an isosceles triangle. :400px Let $AE$, $BF$ and $CD$ be the altitudes of $\triangle ABC$ through $A$, $B$ and $C$ respectively. From Altitudes of Triangle Meet at Point, let them intersect at $G$. From Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles: :$AE$, $BF$ and $CD$ are the medians of $\triangle ABC$ and: :$AE$, $BF$ and $CD$ are the perpendicular bisectors of $\triangle ABC$. They all meet at a single point, that is $G$. Hence by definition, the circumcenter, centroid and orthocenter of $\triangle ABC$ are the same point. {{qed|lemma}} === Converse Condition === Let $\triangle ABC$ not be an equilateral triangle. Thus at least one pair of sides of $\triangle ABC$ is not equal. {{WLOG}}, suppose $AC \ne BC$. Then $AB$ is not the base of an isosceles triangle. :400px Thus from Altitude, Median and Perpendicular Bisector Coincide iff Triangle is Isosceles, the altitude from $AB$, the midpoint of $AB$ and the perpendicular bisector of $AB$ are all different. So the circumcenter, centroid and orthocenter of $\triangle ABC$ likewise cannot be the same point. {{qed}} {{improve|A clumsy proof for a trivial result.}} Category:Equilateral Triangles c0y457k3m8c19ntyeehorieam9ec8kc
=== Necessary Condition === Let $\triangle ABC$ be an isosceles triangle whose base is $AB$. Let $D$ be the midpoint of $AB$. :300px By definition of isosceles triangle, $AC = BC$. We have $AD = DB$ by construction, and $CD$ is common. So by Triangle Side-Side-Side Equality: :$\triangle ACD = \triangle BCD$ From Two Angles on Straight Line make Two Right Angles: :$\angle ADC + \angle BDC$ equals two right angles and as $\angle ADC = \angle BDC$ they are each both right angles. Thus $CD$ is the perpendicular bisector of $AB$. By definition: :As $D$ is the midpoint of $AB$, $CD$ is the median from $AB$ to $C$. :As $CD$ is perpendicular to $AB$ and passes through $C$, which is a vertex of $\triangle ABC$, $CD$ is the altitude from $AB$ to $C$. Thus all three lines coincide. {{qed|lemma}} === Converse Statement === Let $AC \ne BC$ in $\triangle ABC$. Let $CD$ be the altitude from $AB$ to $C$. From Triangle with Two Equal Angles is Isosceles, it follows that $\angle CAB \ne \angle CBA$. Thus $\triangle ACD \ne \triangle BCD$. Let $E$ be the midpoint of $AB$. Let $EF$ be the perpendicular bisector of $AB$. By definition, $CE$ is the median from $AB$ to $C$. :400px {{AimForCont}} $D = E$. Then: :$AD = BD$ :$AC = BC$ :$CD$ is common. So by Triangle Side-Side-Side Equality: :$\triangle ACD = \triangle BCD$ But we have that $\angle CAB \ne \angle CBA$. So by Proof by Contradiction it follows that $D \ne E$. It follows that $CD$, $CE$ and $EF$ are all different lines. {{qed}} {{improve|This is a horribly clumsy proof of a trivial result.}} Category:Isosceles Triangles Category:Medians of Triangles j9oizksx9gx6qiv9m4g855prczowi61
As $A$, $B$ and $C$ are not collinear, the triangle $ABC$ can be constructed by forming the lines $AB$, $BC$ and $CA$. The result follows from Circumscribing Circle about Triangle. {{qed}}
Let $m$ be the number of sides of each of the regular polygons that form the regular tessellation. Let $n$ be the number of those regular polygons which meet at each vertex. From Internal Angles of Regular Polygon, the internal angles of each polygon measure $\dfrac {\paren {m - 2} 180^\circ} m$. The sum of the internal angles at a point is equal to $360^\circ$ by Sum of Angles between Straight Lines at Point form Four Right Angles So: {{begin-eqn}} {{eqn | l = n \paren {\dfrac {\paren {m - 2} 180^\circ} m}       | r = 360^\circ       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac {m - 2} m       | r = \dfrac 2 n       | c =  }} {{eqn | ll= \leadsto       | l = 1 - \dfrac 2 m       | r = \dfrac 2 n       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 m + \dfrac 1 n       | r = \dfrac 1 2       | c =  }} {{end-eqn}} But $m$ and $n$ are both greater than $2$. So: :if $m = 3$, $n = 6$. :if $m = 4$, $n = 4$. :if $m = 5$, $n = \dfrac {10} 3$, which is not an integer. :if $m = 6$, $n = 3$. Now suppose $m > 6$. We have: {{begin-eqn}} {{eqn | l = \dfrac 1 m + \dfrac 1 n       | r = \dfrac 1 2       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 6 + \dfrac 1 n       | o = >       | r = \dfrac 1 2       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac 1 n       | o = >       | r = \dfrac 1 3       | c =  }} {{eqn | ll= \leadsto       | l = 3       | o = >       | r = n       | c =  }} {{end-eqn}} But there are no integers between $2$ and $3$, so $m \not > 6$. There are $3$ possibilities in all. Therefore all regular tessellations have been accounted for. {{qed}}
From Integer as Sum of Three Odd Squares, $8 n + 3$ is the sum of $3$ odd squares. So: {{begin-eqn}} {{eqn | lo= \forall n \in \Z_{\ge 0}:       | l = 8 n + 3       | r = \paren {2 x + 1}^2 + \paren {2 y + 1}^2 + \paren {2 z + 1}^2       | c = for some $x, y, z \in \Z_{\ge 0}$ }} {{eqn | r = 4 x^2 + 4 x + 4 y^2 + 4 y + 4 z^2 + 4 z + 3       | c =  }} {{eqn | r = 4 \paren {x \paren {x + 1} + y \paren {y + 1} + z \paren {z + 1} } + 3       | c =  }} {{eqn | ll= \leadsto       | l = n       | r = \frac {x \paren {x + 1} } 2 + \frac {y \paren {y + 1} } 2 + \frac {z \paren {z + 1} } 2       | c = subtracting $3$ and dividing both sides by $8$ }} {{end-eqn}} By Closed Form for Triangular Numbers, each of $\dfrac {x \paren {x + 1} } 2$, $\dfrac {y \paren {y + 1} } 2$ and $\dfrac {z \paren {z + 1} } 2$ are triangular numbers. {{qed}}
Suppose there were an order $2$ magic square. The row and column total is $\dfrac {1 + 2 + 3 + 4} 2 = 5$. Any row or column with a $1$ in it must therefore also have a $4$ in it. But there are: :one row :one column :one diagonal all of which have a $1$ in them. Therefore the $4$ would need to go in all $3$ cells. But $4$ appears in a magic square exactly once. Hence there can be no order $2$ magic square. {{qed}}
We have: {{begin-eqn}} {{eqn | l = 3       | r = \sqrt 9       | c =  }} {{eqn | r = \sqrt {1 + 8}       | c =  }} {{eqn | r = \sqrt {1 + 2 \sqrt {16} }       | c =  }} {{eqn | r = \sqrt {1 + 2 \sqrt {1 + 15} }       | c =  }} {{eqn | r = \sqrt {1 + 2 \sqrt {1 + 3 \sqrt {25} } }       | c =  }} {{eqn | r = \sqrt {1 + 2 \sqrt {1 + 3 \sqrt {1 + 24} } }       | c =  }} {{eqn | r = \sqrt {1 + 2 \sqrt {1 + 3 \sqrt {1 + 4 \sqrt {36} } } }       | c =  }} {{end-eqn}} In order for this sequence to continue, it needs to be shown that: :$n + 1 = \sqrt {1 + n \left({n + 2}\right)}$ Thus: {{begin-eqn}} {{eqn | l = \sqrt {1 + n \left({n + 2}\right)}       | r = \sqrt {1 + n^2 + 2 n}       | c =  }} {{eqn | r = \sqrt {\left({n + 1}\right)^2}       | c = Square of Sum }} {{eqn | r = n + 1       | c =  }} {{end-eqn}} The result follows. {{qed}} {{Namedfor|Srinivasa Ramanujan|cat = Ramanujan}}
{{AimForCont}} $\paren {n - 1} n \paren {n + 1} = a^k$ for $a, k \in \Z$ where $k \ge 2$. We have that: :$\gcd \set {n − 1, n} = 1 = \gcd \set {n, n + 1}$ Thus $n$ must itself be a perfect power of the form $z^k$ for some $z \in \Z$. That means $\paren {n - 1} \paren {n + 1} = n^2 - 1$ must also be a perfect power of the same form. Let: :$n = r^k$ and $n^2 − 1 = s^k$ for $r, s \in \Z$. Then: :$\paren {r^2}^k = 1 + s^k$ But the only consecutive integers that are $k$th powers are (trivially) $0$ and $1$. Hence by Proof by Contradiction there can be no such $n$. {{qed}}
Let $\tan \alpha = \dfrac 1 5$. Then: {{begin-eqn}} {{eqn | l = \tan 2 \alpha       | r = \frac {2 \tan \alpha} {1 - \tan^2 \alpha}       | c = Double Angle Formula for Tangent }} {{eqn | r = \frac {2 / 5} {1 - 1 / 25}       | c =  }} {{eqn | r = \frac 5 {12}       | c =  }} {{eqn | ll= \leadsto       | l = \tan 4 \alpha       | r = \frac {2 \tan 2 \alpha} {1 - \tan^2 2 \alpha}       | c = Double Angle Formula for Tangent }} {{eqn | r = \frac {2 \times 5 / 12} {1 - 25 / 144}       | c =  }} {{eqn | r = \frac {120} {119}       | c =  }} {{eqn | ll= \leadsto       | l = \tan \left({4 \alpha - \frac \pi 4}\right)       | r = \frac {\tan 4 \alpha - \tan \frac \pi 4} {1 + \tan 4 \alpha \tan \frac \pi 4}       | c = Tangent of Difference }} {{eqn | r = \frac {\tan 4 \alpha - 1} {1 + \tan 4 \alpha}       | c = Tangent of $45^\circ$ }} {{eqn | r = \frac {120 / 119 - 1} {120 / 119 + 1}       | c =  }} {{eqn | r = \frac 1 {239}       | c =  }} {{eqn | ll= \leadsto       | l = 4 \alpha - \frac \pi 4       | r = \arctan \frac 1 {239}       | c =  }} {{eqn | ll= \leadsto       | l = \frac \pi 4       | r = 4 \arctan \frac 1 5 - \arctan \frac 1 {239}       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \arg \left({\left({5 + i}\right)^4 \left({239 - i}\right)}\right)       | r = \arg \left({\left({5 + i}\right)^4}\right) + \arg \left({239 - i}\right)       | c = Argument of Product equals Sum of Arguments }} {{eqn | r = 4 \arg \left({5 + i}\right) + \arg \left({239 - i}\right)       | c = De Moivre's Theorem }} {{eqn | r = 4 \arctan \frac 1 5 + \arctan \frac {-1} {239} }} {{eqn | r = 4 \arctan \frac 1 5 - \arctan \frac 1 {239}       | c = Inverse Tangent is Odd Function }} {{end-eqn}} Furthermore, because: {{begin-eqn}} {{eqn | l = \left({5 + i}\right)^4       | r = 5^4 + 4 \times 5^3 i - 6 \times 5^2 - 4 \times 5 i + 1       | c = Binomial Theorem }} {{eqn | r = 476 + 480i       | c = simplifying }} {{end-eqn}} we can write: {{begin-eqn}} {{eqn | l = \left({5 + i}\right)^4 \left({-239 + i}\right)       | r = \left({476 + 480 i}\right) \left({-239 + i}\right) }} {{eqn | r = -113764 - 480 - 114720i + 476i }} {{eqn | r = -114244 - 114244i }} {{end-eqn}} {{explain|Between the first line of this proof and the above block, the sign of the second factor has changed sign from $\left({239 - i}\right)$ to $\left({-239 + i}\right)$. This needs to be reconciled.}} From there, we substitute: {{begin-eqn}} {{eqn | l = \arg \left({\left({5 + i}\right)^4 \left({-239 + i}\right)}\right)       | r = \arg \left({-114244 - 114244 i}\right) }} {{eqn | r = \arctan \frac {-114244} {-114244} }} {{eqn | r = \arctan 1 }} {{eqn | r = \frac \pi 4 }} {{end-eqn}} By transitivity: :$\dfrac \pi 4 = 4 \arctan \dfrac 1 5 - \arctan \dfrac 1 {239}$ {{qed}}
The result follows from Ordinals under Addition form Semigroup and Subset is Compatible with Ordinal Addition. {{qed}} Category:Ordinal Arithmetic fyil7s36wzjk6cdz24akr8a1cqcj7gs
The result follows from Membership is Left Compatible with Ordinal Addition. {{qed}} Category:Ordinal Arithmetic 49mwa205l5rwaqgavvl8dr7kwd8c0x8
The result follows from Subset is Left Compatible with Ordinal Addition and Subset is Right Compatible with Ordinal Addition. {{qed}} Category:Ordinal Arithmetic iue2kz0ss2qz8vgo4upebwe7joi1kye
The result follows from Ordinal Addition is Closed and Ordinal Addition is Associative. {{qed}} Category:Ordinal Arithmetic 0ijwmgha5mzq97ungghf4v73fcn6jdm
The result follows from Ordinal Multiplication is Closed and Ordinal Multiplication is Associative. {{qed}} Category:Ordinal Arithmetic ops8niodhihijmbp8xusa38z6ay606g
The result follows from Membership is Left Compatible with Ordinal Multiplication. {{qed}} Category:Ordinal Arithmetic qo7mlzvmsidg1u8pmp0bhww3nxzeknu
The result follows from Subset is Left Compatible with Ordinal Multiplication and Subset is Right Compatible with Ordinal Multiplication. {{qed}} Category:Ordinal Arithmetic 1ncpwx7cfs7rsccsg9daxafo2yqifi4
The result follows from Ordinals under Multiplication form Semigroup and Subset is Compatible with Ordinal Multiplication. {{qed}} Category:Ordinal Arithmetic 6drdsr5rb8tagx5ccthgsibpvkxtt0o
The result follows from Ordinals under Addition form Semigroup and Ordinal Addition by Zero. {{qed}} Category:Ordinal Arithmetic 39iitp9iglinzcda4qahcsw7s50umsz
The result follows from Ordinals under Addition form Monoid and Ordinals under Addition form Ordered Semigroup. {{qed}} Category:Ordinal Arithmetic dg4jkg0i27er991ncqg8a30xowf8o80
The result follows from Ordinals under Multiplication form Semigroup and Ordinal Multiplication by One. {{qed}} Category:Ordinal Arithmetic 982gusvq55yiqfbjml4d8lfz2b0vno6
The result follows from Ordinals under Multiplication form Monoid and Ordinals under Multiplication form Ordered Semigroup. {{qed}} Category:Ordinal Arithmetic 8xmk0bsyamoab02t2essh7tyv69ii0q
For simplicity, consider the real number plane $\R^2$ divided into strips by the lines $x = k$ for each integer $k$. Then the needle would have length $1$, which is the distance between the lines. Define $\theta \in \hointr {-\dfrac \pi 2} {\dfrac \pi 2}$ as the angle between the needle and the $x$-axis. Then the horizontal component of length of the needle is $\cos \theta$ for each $\theta$. Let: :$E$ be the event where the needle falls across the vertical lines, :$\Theta_\theta$ be the event where the angle between the needle and the $x$-axis is $\theta$. Let the needle drop. {{WLOG}}, let the end with the larger $x$-coordinate have $x$-coordinate $0 \le x_n < 1$. Then for each $\theta$, the needle falls across the line $x = 0$ exactly when $0 \le x_n \le \cos \theta$. Therefore the probability that this happens is: :$\map \Pr {E \mid \Theta_\theta} = \dfrac {\cos \theta} 1 = \cos \theta$ By considering $\theta$ as a continuous random variable, {{begin-eqn}} {{eqn | l = \map \Pr E       | r = \sum_{\theta \mathop \in \hointr {-\pi / 2} {\pi / 2} } \map \Pr {E \mid \Theta_\theta} \map \Pr {\Theta_\theta}       | c = Total Probability Theorem }} {{eqn | r = \int_{-\pi / 2}^{\pi / 2} \cos \theta \frac {\d \theta} \pi }} {{eqn | r = \intlimits {\frac 1 \pi \sin\theta} {-\pi / 2} {\pi / 2}       | c = Primitive of Cosine Function }} {{eqn | r = \frac 1 \pi \paren {1 - \paren {-1} } }} {{eqn | r = \frac 2 \pi }} {{end-eqn}} {{qed}}
Let $f: \R \to \R$ be the real function defined as: :$\forall x \in \R: \map f x = \dfrac {x^1} 1 + \dfrac {x^3} 3 - \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 + \dfrac {x^{11} } {11} - \dfrac {x^{13} } {13} - \dfrac {x^{15} } {15} \cdots$ We have: {{begin-eqn}} {{eqn | l = \map f x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {\frac {x^{4 n + 1} } {4 n + 1} + \frac {x^{4 n + 3} } {4 n + 3} }       | c = }} {{eqn | ll= \leadsto       | l = \map {f'} x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {x^{4 n} + x^{4 n + 2} }       | c = Power Rule for Derivatives }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {1 + x^2} \paren {x^{4 n} }       | c = }} {{eqn | r = \paren {1 + x^2} \sum_{n \mathop = 0}^\infty \paren {-1}^n \paren {x^{4 n} }       | c = }} {{eqn | r = \dfrac {1 + x^2} {1 + x^4}       | c = Sum of Infinite Geometric Sequence }} {{eqn | ll= \leadsto       | l = \map f 1       | r = \int_0^1 \dfrac {1 + x^2} {1 + x^4} \rd x       | c = }} {{eqn | r = \intlimits {\dfrac 1 {\sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {x - \dfrac 1 x} } } 0 1       | c = Primitive of One plus x Squared over One plus Fourth Power of x }} {{eqn | r = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {1 - \dfrac 1 1} } - \dfrac 1 {\sqrt 2} \map \arctan {\dfrac 1 {\sqrt 2} \paren {0 - \dfrac 1 0} }       | c =  }} {{eqn | r = \dfrac 1 {\sqrt 2} \paren {\map \arctan 0 - \map \arctan {-\infty} }       | c = }} {{eqn | r = \dfrac 1 {\sqrt 2} \paren {0 - \paren {-\dfrac \pi 2} }       | c = {{Defof|Real Arctangent}} }} {{eqn | r = \dfrac {\pi \sqrt 2} 4       | c = }} {{end-eqn}} {{qed}} {{Help | May need further justification that using the sum of geometric progession is valid at 1}}
Let $f: \R \to \R$ be the real function defined as: :$\forall x \in \R: \map f x = x^1 - x^3 + x^5 - x^7 + x^9 - x^{11} + x^{13} - x^{15} \cdots$ We have: {{begin-eqn}} {{eqn | l = \map f x       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n + 1}        | c = }} {{eqn | r = x \sum_{n \mathop = 0}^\infty \paren {-1}^n x^{2 n}       | c = }} {{eqn | r = \dfrac x {x^2 + 1}       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} Integrating $3$ times will give us the desired series. {{begin-eqn}} {{eqn | l = \map {F''} x       | r = \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 4} } {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} }       | c = Repeated use of Integral of Power }} {{end-eqn}} === Lemma === {{:Pi as Sum of Sequence of Reciprocal of Product of Three Consecutive Integers/Lemma}} {{qed|lemma}} Substitute $x = 1$: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n} {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} }       | r = \paren 1 \map \arctan 1 + \dfrac {\paren {1^2 - 1} \map \ln {1^2 + 1} - 3 \paren 1^2} 4       | c = }} {{eqn | l = \sum_{n \mathop = 1}^\infty \dfrac {\paren {-1}^{n + 1} } {\paren {2 n} \paren {2 n + 1} \paren {2 n + 2} }       | r = \map \arctan 1 - \dfrac 3 4       | c = }} {{eqn | r = \dfrac {\pi - 3} 4       | c = }} {{end-eqn}} {{qed}} {{Help | May need further justification that using the sum of geometric progession is valid at 1}}