Inverse Pythagorean theorem

<h2 id="proof">Proof</h2>
<p>The area of triangle △<i>ABC</i> can be expressed in terms of either AC and BC, or AB and CD:
</p>

1
                      2
                    
                  
                
                A
                C
                ⋅
                B
                C
              
              
                
                =
                
                  
                    
                      1
                      2
                    
                  
                
                A
                B
                ⋅
                C
                D
              
            
            
              
                (
                A
                C
                ⋅
                B
                C
                
                  )
                  
                    2
                  
                
              
              
                
                =
                (
                A
                B
                ⋅
                C
                D
                
                  )
                  
                    2
                  
                
              
            
            
              
                
                  
                    1
                    
                      C
                      
                        D
                        
                          2
                        
                      
                    
                  
                
              
              
                
                =
                
                  
                    
                      A
                      
                        B
                        
                          2
                        
                      
                    
                    
                      A
                      
                        C
                        
                          2
                        
                      
                      ⋅
                      B
                      
                        C
                        
                          2
                        
                      
                    
                  
                
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}{\tfrac {1}{2}}AC\cdot BC&={\tfrac {1}{2}}AB\cdot CD\\[4pt](AC\cdot BC)^{2}&=(AB\cdot CD)^{2}\\[4pt]{\frac {1}{CD^{2}}}&={\frac {AB^{2}}{AC^{2}\cdot BC^{2}}}\end{aligned}}}

<p>given <i>CD</i> > 0, <i>AC</i> > 0 and <i>BC</i> > 0.
</p><p>Using the <a href="/facts/Pythagorean_theorem/Lh6x2Kr5">Pythagorean theorem</a>,
</p>

1
                    
                      C
                      
                        D
                        
                          2
                        
                      
                    
                  
                
              
              
                
                =
                
                  
                    
                      B
                      
                        C
                        
                          2
                        
                      
                      +
                      A
                      
                        C
                        
                          2
                        
                      
                    
                    
                      A
                      
                        C
                        
                          2
                        
                      
                      ⋅
                      B
                      
                        C
                        
                          2
                        
                      
                    
                  
                
              
            
            
              
              
                
                =
                
                  
                    
                      B
                      
                        C
                        
                          2
                        
                      
                    
                    
                      A
                      
                        C
                        
                          2
                        
                      
                      ⋅
                      B
                      
                        C
                        
                          2
                        
                      
                    
                  
                
                +
                
                  
                    
                      A
                      
                        C
                        
                          2
                        
                      
                    
                    
                      A
                      
                        C
                        
                          2
                        
                      
                      ⋅
                      B
                      
                        C
                        
                          2
                        
                      
                    
                  
                
              
            
            
              
                
                ∴
                
                
                
                  
                    1
                    
                      C
                      
                        D
                        
                          2
                        
                      
                    
                  
                
              
              
                
                =
                
                  
                    1
                    
                      A
                      
                        C
                        
                          2
                        
                      
                    
                  
                
                +
                
                  
                    1
                    
                      B
                      
                        C
                        
                          2
                        
                      
                    
                  
                
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}{\frac {1}{CD^{2}}}&={\frac {BC^{2}+AC^{2}}{AC^{2}\cdot BC^{2}}}\\[4pt]&={\frac {BC^{2}}{AC^{2}\cdot BC^{2}}}+{\frac {AC^{2}}{AC^{2}\cdot BC^{2}}}\\[4pt]\quad \therefore \;\;{\frac {1}{CD^{2}}}&={\frac {1}{AC^{2}}}+{\frac {1}{BC^{2}}}\end{aligned}}}

<p>as above.
</p><p>Note in particular:
</p>

1
                      2
                    
                  
                
                A
                C
                ⋅
                B
                C
              
              
                
                =
                
                  
                    
                      1
                      2
                    
                  
                
                A
                B
                ⋅
                C
                D
              
            
            
              
                C
                D
              
              
                
                =
                
                  
                    
                      
                        A
                        C
                        ⋅
                        B
                        C
                      
                      
                        A
                        B
                      
                    
                  
                
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}{\tfrac {1}{2}}AC\cdot BC&={\tfrac {1}{2}}AB\cdot CD\\[4pt]CD&={\tfrac {AC\cdot BC}{AB}}\\[4pt]\end{aligned}}}

<h2 id="special-case-of-the-cruciform-curve">Special case of the cruciform curve</h2>
<p>The <a href="/facts/Cruciform_curve/5j35LcSS">cruciform curve</a> or cross curve is a <a href="/facts/Quartic_plane_curve/5j35LcSS">quartic plane curve</a> given by the equation
</p>

x
          
            2
          
        
        
          y
          
            2
          
        
        −
        
          b
          
            2
          
        
        
          x
          
            2
          
        
        −
        
          a
          
            2
          
        
        
          y
          
            2
          
        
        =
        0
      
    
    {\displaystyle x^{2}y^{2}-b^{2}x^{2}-a^{2}y^{2}=0}

<p>where the two <a href="/facts/Parameter/zFvVFMv9">parameters</a> determining the shape of the curve, a and b are each CD.
</p><p>Substituting x with AC and y with BC gives
</p>

A
                
                  C
                  
                    2
                  
                
                B
                
                  C
                  
                    2
                  
                
                −
                C
                
                  D
                  
                    2
                  
                
                A
                
                  C
                  
                    2
                  
                
                −
                C
                
                  D
                  
                    2
                  
                
                B
                
                  C
                  
                    2
                  
                
              
              
                
                =
                0
              
            
            
              
                A
                
                  C
                  
                    2
                  
                
                B
                
                  C
                  
                    2
                  
                
              
              
                
                =
                C
                
                  D
                  
                    2
                  
                
                B
                
                  C
                  
                    2
                  
                
                +
                C
                
                  D
                  
                    2
                  
                
                A
                
                  C
                  
                    2
                  
                
              
            
            
              
                
                  
                    1
                    
                      C
                      
                        D
                        
                          2
                        
                      
                    
                  
                
              
              
                
                =
                
                  
                    
                      B
                      
                        C
                        
                          2
                        
                      
                    
                    
                      A
                      
                        C
                        
                          2
                        
                      
                      ⋅
                      B
                      
                        C
                        
                          2
                        
                      
                    
                  
                
                +
                
                  
                    
                      A
                      
                        C
                        
                          2
                        
                      
                    
                    
                      A
                      
                        C
                        
                          2
                        
                      
                      ⋅
                      B
                      
                        C
                        
                          2
                        
                      
                    
                  
                
              
            
            
              
                ∴
                
                
                
                  
                    1
                    
                      C
                      
                        D
                        
                          2
                        
                      
                    
                  
                
              
              
                
                =
                
                  
                    1
                    
                      A
                      
                        C
                        
                          2
                        
                      
                    
                  
                
                +
                
                  
                    1
                    
                      B
                      
                        C
                        
                          2
                        
                      
                    
                  
                
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}AC^{2}BC^{2}-CD^{2}AC^{2}-CD^{2}BC^{2}&=0\\[4pt]AC^{2}BC^{2}&=CD^{2}BC^{2}+CD^{2}AC^{2}\\[4pt]{\frac {1}{CD^{2}}}&={\frac {BC^{2}}{AC^{2}\cdot BC^{2}}}+{\frac {AC^{2}}{AC^{2}\cdot BC^{2}}}\\[4pt]\therefore \;\;{\frac {1}{CD^{2}}}&={\frac {1}{AC^{2}}}+{\frac {1}{BC^{2}}}\end{aligned}}}

<p>Inverse-Pythagorean triples can be generated using <a href="/facts/Integer/PUwwrawx">integer</a> parameters t and u as follows.<a class="footnote-ref" id="fnref:4" href="#fn:4"><sup>4</sup></a>
</p>

A
                C
              
              
                
                =
                (
                
                  t
                  
                    2
                  
                
                +
                
                  u
                  
                    2
                  
                
                )
                (
                
                  t
                  
                    2
                  
                
                −
                
                  u
                  
                    2
                  
                
                )
              
            
            
              
                B
                C
              
              
                
                =
                2
                t
                u
                (
                
                  t
                  
                    2
                  
                
                +
                
                  u
                  
                    2
                  
                
                )
              
            
            
              
                C
                D
              
              
                
                =
                2
                t
                u
                (
                
                  t
                  
                    2
                  
                
                −
                
                  u
                  
                    2
                  
                
                )
              
            
          
        
      
    
    {\displaystyle {\begin{aligned}AC&=(t^{2}+u^{2})(t^{2}-u^{2})\\BC&=2tu(t^{2}+u^{2})\\CD&=2tu(t^{2}-u^{2})\end{aligned}}}

<h2 id="application">Application</h2>
<p>If two identical lamps are placed at A and B, the theorem and the <a href="/facts/Inverse-square_law/DxG4T15l">inverse-square law</a> imply that the light intensity at C is the same as when a single lamp is placed at D.
</p>
<h2 id="see-also">See also</h2>
<ul><li><a href="/facts/Geometric_mean_theorem/BQKf7ya4">Geometric mean theorem</a> – Theorem about right triangles</li>
<li><a href="/facts/Pythagorean_theorem/Lh6x2Kr5">Pythagorean theorem</a> – Relation between sides of a right triangle</li></ul>

<h2 id="references">References</h2>

<ol>
<li id="fn:1"><p>R. B. Nelsen, Proof Without Words: A Reciprocal Pythagorean Theorem, Mathematics Magazine, 82, December 2009, p. 370 <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li>
<li id="fn:2"><p>The upside-down Pythagorean theorem, Jennifer Richinick, The Mathematical Gazette, Vol. 92, No. 524 (July 2008), pp. 313-316 <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li>
<li id="fn:3"><p>Johan Wästlund, "Summing inverse squares by euclidean
geometry", http://www.math.chalmers.se/~wastlund/Cosmic.pdf, pp. 4–5. <a href="http://www.math.chalmers.se/~wastlund/Cosmic.pdf" target="_blank">http://www.math.chalmers.se/~wastlund/Cosmic.pdf</a> <a href="#fnref:3" class="footnote-back-ref">↩</a></p></li>
<li id="fn:4"><p>"Diophantine equation of three variables". <a href="http://math.stackexchange.com/a/2688836" target="_blank">http://math.stackexchange.com/a/2688836</a> <a href="#fnref:4" class="footnote-back-ref">↩</a></p></li>
</ol>

Inverse Pythagorean theorem open-in-new

Inverse Pythagorean theorem