\nLet\u2019s solve a slightly more general version of the problem. Suppose the triangle has angle $\\theta$ at point $A$, and suppose Amare\u2019s next three waypoints are at $D$, $E$, and $F$, as shown below.\n

$\"\"$ <\/p>\n

Reflect the triangle about the line $AC$, so $B \\mapsto B\u2019$ and $E \\mapsto E\u2019$, as shown:<\/p>\n

$\"\"$ <\/p>\n

Now reflect the new triangle about the line $AB\u2019$, so $C\\mapsto C\u2019$ and $F \\mapsto F\u2019$:<\/p>\n

$\"\"$ <\/p>\n

The key insight is that since the reflections preserve lengths, the path $BD+DE+EF$ followed by Amare has the same length as the path $BD+DE\u2019+E\u2019F\u2019$ shown in red below:<\/p>\n

$\"\"$ <\/p>\n

If we move the points $E$ and $F$, then the points $E\u2019$ and $F\u2019$ move accordingly, and we obtain another possible path:<\/p>\n

$\"\"$ <\/p>\n

Rather than picking $D,E,F$, we can instead pick $D,E\u2019,F\u2019$. Since the goal is to minimize the total distance, it\u2019s clear that we should place $F\u2019$ such that $AF\u2019 \\perp BF\u2019$, and $D$ and $E\u2019$ should be placed so that all three points lie on a line. This produces the figure:<\/p>\n

$\"\"$ <\/p>\n

So the shortest distance Amare can travel can be found by examining the right triangle $ABF\u2019$. Since $AB$ has length $1$, we conclude that that<\/p>\n

$\\displaystyle
\n\\text{Minimum distance } = \\sin(3\\theta)
\n$<\/span><\/p>\n

In the case where $\\theta = 15^\\circ$, we get a length of $\\sin(45^\\circ) = \\frac{1}{\\sqrt{2}} \\approx 0.7071$.<\/p>\n

Note that this solution only works if $3\\theta \\lt 90^\\circ$, i.e. $0 \\lt \\theta \\lt 30^\\circ$. In the case that $\\theta \\geq 30^\\circ$, we obtain the degenerate solution $D = E = F = A$, so Amare should head directly to point $A$ and the total distance traveled is $1$.\n<\/p><\/div>\n

And the more complicated solution:
\n [Show Solution]<\/a><\/p>\n

We can also use calculus to solve the problem. Let\u2019s start with the same picture as before:<\/p>\n

$\"\"$ <\/p>\n

Suppose $|AD|=x$, $|AE|=y$, and $|AF|=z$. From the law of cosines:
\n\\begin{align}
\n|BD| &= \\sqrt{1+x^2-2x\\cos(\\theta)} \\\\
\n|DE| &= \\sqrt{x^2+y^2-2xy\\cos(\\theta)} \\\\
\n|EF| &= \\sqrt{y^2+z^2-2yz\\cos(\\theta)}
\n\\end{align}Let $f(x,y,z)$ be the sum of these three distances, which is the total distance traveled by Amare. We want to find $x,y,z$ such that $f(x,y,z)$ is minimized. A necessary condition for minimality is that the partial derivatives of $f$ with respect to $x,y,z$ should be zero. Let\u2019s start with $z$:
\n\\[
\n\\frac{\\partial}{\\partial z}f(x,y,z) = \\frac{z-y \\cos (\\theta )}{\\sqrt{y^2+z^2-2 y z \\cos (\\theta )}}
\n\\]Setting this equal to zero, we conclude that $z=y\\cos(\\theta)$. We can now substitute this value of $z$ into the definition for $f$ and we obtain a simpler expression in only two variables:
\n\\begin{align}
\ng(x,y) &= f(x,y,y\\cos(\\theta)) \\\\
\n&= \\sqrt{1+x^2-2x\\cos(\\theta)} + \\sqrt{x^2+y^2-2xy\\cos(\\theta)} + y\\sin(\\theta)
\n\\end{align}To make sure there is no funny business going on, let\u2019s plot this function to see what it looks like for $\\theta=15^\\circ$. Here is a contour plot:
\n $\"\"$ <\/p>\n

We can clearly see that there is a unique minimum that occurs in the region near $(x,y) \\approx (0.8,0.7)$. Ok, so let\u2019s proceed. Consider now the derivative with respect to $y$:
\n\\[
\n\\frac{\\partial}{\\partial y} g(x,y) = \\frac{y-x \\cos (\\theta )}{\\sqrt{x^2+y^2-2 x y \\cos (\\theta )}} + \\sin (\\theta )
\n\\]Setting this equal to zero (I\u2019ll spare you the algebra), we obtain $y = x \\cos(2\\theta)\\sec(\\theta)$. Substituting this into $g(x,y)$, we obtain a function of just $x$:
\n\\begin{align}
\nh(x) &= g(x,x \\cos(2\\theta)\\sec(\\theta)) \\\\
\n&= \\sqrt{1+x^2-2 x \\cos (\\theta )}+x \\sin (2 \\theta )
\n\\end{align}Finally, we can take the derivative with respect to $x$:
\n\\[
\n\\frac{\\partial}{\\partial x}h(x) = \\frac{x-\\cos (\\theta )}{\\sqrt{1+x^2-2 x \\cos (\\theta )}} + \\sin (2 \\theta )
\n\\]Solving for $x$, (sparing you the algebra again!) we obtain $x = \\cos (3 \\theta ) \\sec (2 \\theta )$. Putting everything together, the $(x,y,z)$ that minimize the total distance traveled by Amare is
\n\\[
\nx = \\cos (3 \\theta ) \\sec (2 \\theta ),\\quad
\ny = \\cos(3\\theta)\\sec(\\theta),\\quad
\nz = \\cos(3\\theta)
\n\\]Substituting into either $h(x)$, $g(x,y)$, or $f(x,y,z)$, we obtain the minimum distance traveled by Amare (after simplification):<\/p>\n

$\\displaystyle
\n\\text{Minimum distance } = \\sin(3\\theta)
\n$<\/span><\/p>\n

If you read the first solution, then it should come as no surprise that the total distance is also equal to $\\sqrt{1-z^2}$, since (based on the last figure of the first solution), we have $z=|AF|=|AF\u2019|$ and $|BF\u2019|^2 + |AF\u2019|^2 = 1$ by the Pythagorean theorem.\n<\/p><\/div>\n

<\/p>\n<\/body>","protected":false},"excerpt":{"rendered":"

This week\u2019s Riddler Classic is a problem involving traversing a triangle. Amare the ant is traveling within Triangle ABC, as shown below. Angle A measures 15 degrees, and sides AB and AC both have length 1. Amare must: Start at point B. Second, touch a point \u2014 any point \u2014 on side AC. Third, touch … Continue reading “Triangle Trek”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":3146,"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":[7],"tags":[28,10,2],"class_list":["post-3137","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-riddler","tag-calculus","tag-geometry","tag-riddler"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t