My Math Forum Pythagoras problem

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October 29th, 2013, 12:50 PM   #1
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Pythagoras problem

I tried using Apollonius's theorem to relate v, w and CP.
Let the side opposite <B be b and <A be a and v+w be c. Let D be the point on AB where the altitude from <C falls and PD=y.
By Pythagoras:
b^2+a^2=v^2+w^2+2CP^2
=(v-y)^2+(v+y)^2+2CP^2
=2(v+y)^2+2y^2+2CP^2
this is where I get stuck I see that b^2+a^2=(v+w)^2 but I can't simplify it to get that v^2+w^2=2CP^2
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 October 29th, 2013, 06:19 PM #2 Senior Member   Joined: Jul 2011 Posts: 118 Thanks: 0 Re: Pythagoras problem $|CP|^2=$$\frac{s}{\sqrt{2}}$$^2+y^2 2|CP|^2=2$$\frac{s}{\sqrt{2}}$$^2+2y^2=$$\frac{s}{ \sqrt{2}}$$^2-2y$$\frac{s}{\sqrt{2}}$$+y^2+$$\frac{s}{\sqrt{2}}\ )^2+2y\(\frac{s}{\sqrt{2}}$$+y^2=$$\frac{s}{\sqrt{ 2}}-y$$^2+$$\frac{s}{\sqrt{2}}+y$$^2=v^2+w^2$
 October 29th, 2013, 06:42 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Pythagoras problem $(v\,+\,w)^2\,=\,2s^2$ $v\,+\,w\,=\,\sqrt2s$ From the law of cosines, $\bar{CP}\,=\,\sqrt{s^2\,+\,v^2\,-\,2sv\cos45}$ $\bar{CP}^2\,=\,s^2\,+\,v^2\,-\,\sqrt2sv$ Similarly, $\bar{CP}^2\,=\,s^2\,+\,w^2\,-\,\sqrt2sw$ Adding these results, $2\bar{CP}^2\,=\,v^2\,+\,w^2\,+\,2s^2\,-\,(v\,+\,w)^2\,=\,v^2\,+\,w^2$
 October 29th, 2013, 09:01 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Pythagoras problem Construct a segment from C to AB, bisecting AB at D. CP² = (AD - v)² + (w - PD)² = AD² - 2vAD + v² + w² - 2wPD + PD² = v² + w² + CP² - 2(vAD + wPD) = v² + w² + CP² - 2(v² + 2vPD + 2PD²) [as AD = PD + v and w = 2PD + v] = v² + w² + CP² - 2((v + PD)² + PD²) = v² + w² + CP² - 2CP² 2CP² = v² + w².
 October 29th, 2013, 09:15 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1039 Re: Pythagoras problem Skinning the cat #3! Code:  C A P D B AC = BC = a (hate "s"!), AP = v, PD = y, BD = w - y, CP = p, CD = d 2a^2 = v^2 + 2vw + w^2 [1] v + y = w - y : y = (w - v) / 2 [2] d^2 = p^2 - y^2 [3] d^2 = a^2 - (w - y)^2 : d^2 = a^2 - w^2 + 2wy - y^2 [4] [3][4]: p^2 = a^2 - w^2 + 2wy Substitute [2]: p^2 = a^2 - w^2 + 2w((w - v)/2) : 2p^2 = 2a^2 - 2vw Substitute [1]: 2p^2 = v^2 + 2vw + w^2 - 2vw : 2p^2 = v^2 + w^2
 October 30th, 2013, 02:13 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 Construct PD perpendicular to BC with D on BC. It is easy to show that PD = w/?2 and CD = v/?2. By Pythagoras, v²/2 + w²/2 = CP², so v² + w² = 2CP².

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