I personally love the idea of Proofs Without Words! I like how simple, elegant and creative they can be. So here is another one for all you proof enthusiasts. Link
Expansion
In this proof, we see that \(\frac{1}{1-r}\) = \(1 + r + r^2 + r^3 ...\).
\(\triangle TPS\) is cut into multiple similar triangles with a ratio r. \({ST}\) is \(1 + r + r^2 + r^3 ...\). For the purposes of this proof, let the point where line \({QR}\) meets \(ST\) be \(A\).
\(\angle PRQ = \angle TRA\) because they're vertical angles. \(\angle PQR = \angle TAR\) since they are both right angles. Thus \(\triangle PRQ \sim \triangle TRA\) by \(AA\) Similarity. Since \(\triangle TPS \sim \triangle TRA\), \(\triangle PRQ \sim \triangle TPS\).
Tinkering with our ratios, we get \(\frac{PQ}{QR} = \frac{TS}{SP}\)
\(\frac{1}{1-r} = \frac{1 + r + r^2 + r^3 ...}{1}\)
Q.E.D
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Top NewestThis is awesome! Although perhaps it is worth noting that \(|r| < 1\) for this to work. I wonder if I could think up a similar geometric proof for the more general formula: \(1 + r + r^{2} +... = \dfrac{1-r^{n}}{1-r}\)... Hmm... – Raj Magesh · 3 years, 7 months ago
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Thank you for the feedback, I forgot to add that detail! – Sherry Sarkar · 3 years, 7 months ago
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Elegant Maths... that's what i call it – Rohitas Bansal · 3 years, 6 months ago
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It's really amazing, – Muh. Amin Widyatama · 3 years, 7 months ago
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Thanks! It's my first post, so I'm on the lines here. :D – Sherry Sarkar · 3 years, 7 months ago
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Nice one for the first post. Keep posting! – Muh. Amin Widyatama · 3 years, 7 months ago
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We will get the same answer if we extend the dotted triangle towards the right:
If we extend upto r^2, we get \[(1+r)/(1- r^{2} )\] which can be simplified to get the same answer.
If we extend upto r^3, we get \[(1+r+r^{2})/(1-r^{3})\] which again can be simplified to get the same result. And so on...... – Tamil Marx Subash · 3 years, 4 months ago
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Wow. Nice proof! – Faraz Masroor · 3 years, 6 months ago
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that's fantastic!!!!!!!!! – Shantanu Mandhane · 3 years, 6 months ago
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