\n
In general, if $A$ scales with $r^2$ and $P$ scales with $r$ and $P$ is the derivative of $A$, then it must be the case that:
\n\\[
\nA = \\alpha r^2\\qquad\\text{and}\\qquad P = 2\\alpha r
\n\\]for some constant of proportionality $\\alpha$. If we have a certain shape and we know formulas for its area and perimeter, then we can solve for $\\alpha$ and $r$ in terms of $A$ and $P$ and we obtain:
\n\\[
\n\\alpha = \\frac{P^2}{4A}\\qquad\\text{and}\\qquad r = \\frac{2A}{P}
\n\\]
\nWe can verify this formula for a square of sidelength $s$. Here, $P=4s$ and $A=s^2$. Substituting into the above formulas, we obtain $\\alpha = 4$ and $r = \\tfrac{s}{2}$, just as the problem stated.<\/p>\n
In general, we can solve the case for a regular polygon with $n$ sides of length $s$. The perimeter is $P=ns$ and the area is<\/a> $A = \\frac{ns^2}{4\\tan(\\pi\/n)}$. Therefore, we obtain:
\n\\[
\n\\alpha = n\\tan\\bigl(\\tfrac{\\pi}{n}\\bigr),\\qquad\\text{and}\\qquad r = \\frac{s}{2}\\cot\\bigl(\\tfrac{\\pi}{n}\\bigr)
\n\\]So, in particular, for $n=3$, we get $r=\\frac{\\sqrt{3}}{6}s$. For regular polygons, the differential radius is the apothem<\/a> (the line joining the center of the polygon to the midpoint of one of its sides. It is also the \u201cinradius\u201d (radius of the inscribed circle). Note that based on the general formula for $r$, we have:
\n\\[
\nA = \\frac{P\\cdot r}{2}
\n\\]In the case of the regular polygon, this means the area is one half of the perimeter times the differential radius (apothem). This makes sense, since we can unfold a regular polygon and compute its area using the length of the bases (the perimeter) and the height (apothem). See below for a visual demonstration:<\/p>\n
$\"\"$ <\/a><\/p>\n
And here is an animated version:<\/p>\n
$\"\"$ <\/p>\n
Extra credit<\/h3>\n
For the case of a rectangle of side lengths $a$ and $b$, the perimeter is $P=2(a+b)$ and the area is $A=ab$. Applying the formula above, we obtain:
\n\\[
\n\\alpha = \\frac{(a+b)^2}{ab}\\qquad\\text{and}\\qquad r = \\frac{ab}{a+b}
\n\\]The differential radius is half of the harmonic mean<\/a> of $a$ and $b$. It also satisfies the nice formula:
\n\\[
\n\\frac{1}{r} = \\frac{1}{a} + \\frac{1}{b}
\n\\]which is incidentally how you add resistors in parallel<\/a>. We can construct the differential radius geometrically by drawing a line connecting opposite corners, and then another line at 45 degrees to the other corner and marking the intersection. The result is illustrated in the image below:<\/p>\n
$\"\"$ <\/a><\/p>\n
The triangles on the right are what you get when you \u201cunfold\u201d the rectangle on the left. The area of the rectangle, $ab$, is simply the base of the triangles (perimeter of rectangle), times the height ($r$, or the differential radius), times one half.<\/p>\n
This works because triangles AFP and PGC are similar. It follows that:
\n\\begin{align*}
\n\\frac{AF}{FP} &= \\frac{AD-DF}{FP} = \\frac{b-r}{r},\\quad\\text{and}\\\\[2mm]
\n\\frac{PG}{GC} &= \\frac{PG}{DC-DG} = \\frac{r}{a-r}
\n\\end{align*}Equating them, we obtain:
\n\\[
\n\\frac{r}{a-r}=\\frac{b-r}{r}
\n\\]Solving this equation for $r$, we obtain $r = \\frac{ab}{a+b}$, as required.<\/p>\n
Here is an animation of the rectangle unfolding into four triangles with equal height and whose bases sum to the perimeter of the rectangle!<\/p>\n
$\"\"$ \n<\/p><\/div>\n
<\/p>\n<\/body>","protected":false},"excerpt":{"rendered":"
This week\u2019s Fiddler is about a generalized notion of \u201cradius\u201d. For a circle with radius $r$, its area is $\\pi r^2$ and its circumference is $2\\pi r$. If you take the derivative of the area formula with respect to $r$, you get the circumference formula! Let\u2019s define the term \u201cdifferential radius.\u201d The differential radius $r$ … Continue reading “When is a triangle like a circle?”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":4072,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[42],"tags":[28,43,10],"class_list":["post-4067","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-the-fiddler","tag-calculus","tag-fiddler","tag-geometry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/4067","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/comments?post=4067"}],"version-history":[{"count":10,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/4067\/revisions"}],"predecessor-version":[{"id":4083,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/posts\/4067\/revisions\/4083"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media\/4072"}],"wp:attachment":[{"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/media?parent=4067"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/categories?post=4067"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/laurentlessard.com\/bookproofs\/wp-json\/wp\/v2\/tags?post=4067"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}