theredpea · October 25, 2023 00:37
diff --git a/equilateral-triangles-overlap-area.ipynb b/equilateral-triangles-overlap-area.ipynb
 {
  "nbformat": 4,
  "nbformat_minor": 0,
  "metadata": {
    "colab": {
      "provenance": [],
      "authorship_tag": "ABX9TyNwhdCnlMdqcw2r7dmGgWjs",
      "include_colab_link": true
    },
    "kernelspec": {
      "name": "python3",
      "display_name": "Python 3"
    },
    "language_info": {
      "name": "python"
    }
  },
  "cells": [
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "view-in-github",
        "colab_type": "text"
      },
      "source": [
        "<a href=\"https://colab.research.google.com/gist/theredpea/cfe6cf2d3ab8cbfed2ec165d4450e61f/equilateral-triangles-overlap-area.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>"
      ]
    },
    {
      "cell_type": "markdown",
      "source": [
        "These tweets propose a math/geometry/trigonometry puzzle.\n",
        " - [Original puzzle tweet from Diego Rattaggi @diegorattaggi and inspired by Michael Penn](https://twitter.com/diegorattaggi/status/1450903893899202575)\n",
        " - [Solution tweet from Per Henrik Christiansen @PerHenrikChris1](https://twitter.com/PerHenrikChris1/status/1583101619344146433)\n",
        "\n",
        "Here is the original puzzle:\n",
        "\n",
        "<img src=\"https://pbs.twimg.com/media/FCKlZAdX0AYKAOn?format=png&name=4096x4096\" alt=\"Two equilateral triangles overlap (one of them is flipped 180 degrees) so the intersections of their points and edges form the diagonal of a square\" width=\"400\"/>\n"
      ],
      "metadata": {
        "id": "YbCmZK4A3dMA"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "\n",
        " Here is a helpful diagram I use to help understand the solution from Per...\n",
        "\n",
        "Per's solution is posted on [desmos, here](https://t.co/QAwd41j8G5). In my diagram below, I use Per's conventions where:\n",
        "- $S$ is the length of the \"main\" equilateral triangle\n",
        "- $s$ is the length of the square formed in the overlap\n",
        "- $d$ is a diagonal of the square\n",
        "- $X$ is the length of the \"medium\" equilateral triangle **whose height is $s$**\n",
        "- $(S-X)$ is the length of the \"small\" equilateral triangle excluded by the square\n",
        "\n",
        "<img src=\"https://drive.google.com/uc?export=view&id=1vuJXcnQS52-aRG68QY5LQb0vzMahHTF4\" alt=\"Measuring some angles from the equilateral triangles in above example\" width=\"400\"/>\n",
        "\n",
        "\n",
        "I get stuck  when Per says this:\n",
        "$$\n",
        "s=\\frac{S\\sqrt{3}}{2\\sqrt{2}\\cos15}\n",
        "$$\n",
        "\n",
        "<img src=\"https://drive.google.com/uc?export=view&id=1ougmQolOhDIw51nmogKMHreYsAs6efXF\" alt=\"Per's solution on desmos \" width=\"600\"/>\n",
        "\n"
      ],
      "metadata": {
        "id": "xiFJiDqoo9vY"
      }
    },
    {
      "cell_type": "markdown",
      "source": [
        "Let's try it...\n",
        "\n",
        "We have two ways of calculating the diagonal length $d$. We use the [law of sines](https://en.wikipedia.org/wiki/Law_of_sines), which is an \"equation relating the lengths of the sides of any triangle to the sines of its angles\", (nicely explained [here](https://youtu.be/4Dv5IFqATrc?t=490))\n",
        "\n",
        "\\begin{align} \\tag{1}\n",
        "d&=(S-X) \\frac{\\sin{120}}{\\sin{15}} \\\\\n",
        "\\\\\\tag{2}\n",
        "d&=X\\frac{\\sin{120}}{\\sin{45}} \\\\\n",
        "\\end{align}\n",
        "\n",
        "Then we can rearrange equation 2 to get the value of $X$:\n",
        "\\begin{align}\n",
        "\\\\\n",
        "X&=d\\frac{\\sin{45}}{\\sin{120}} \\\\\n",
        "\\end{align}\n",
        "\n",
        "Now substitute the value of $X$ into equation 1, and isolate $d$ on the left side of the equation, and $S$ on the right side...\n",
        "\\begin{align}\n",
        "\\\\\n",
        "d&=(S-d\\frac{\\sin{45}}{\\sin{120}}) \\frac{\\sin{120}}{\\sin{15}} \\\\\n",
        "\\\\\n",
        "d\\frac{\\sin{15}}{\\sin{120}}&=S-d\\frac{\\sin{45}}{\\sin{120}} \\\\\n",
        "\\\\\n",
        "d\\sin{15}&=S\\sin{120}-d\\sin{45} \\\\\n",
        "\\\\\n",
        "d\\sin{15} + d\\sin{45}&=S\\sin{120} \\\\\n",
        "\\\\\n",
        "d(\\sin{15} + \\sin{45})&=S\\sin{120} \\\\\n",
        "\\end{align}\n",
        "\n",
        "Now for the left side of the equation, use the [\"sum and product formulae\"](https://openstax.org/books/precalculus-2e/pages/7-4-sum-to-product-and-product-to-sum-formulas) to convert the **sum** $\\sin{15} + \\sin{45}$ into a **multiplication** of $\\sin$ and $\\cos$... (Note the [\"sine angle addition identity\"](https://youtu.be/R0EQg9vgbQw), aka the [\"compound angle formulae\"](https://www.liverpool.ac.uk/~maryrees/homepagemath191/trigid.pdf) did not apply here; that would apply if we were handling $\\sin{(15 + 45)}$ )\n",
        "\n",
        "For the right side of the equation, use the [\"cofunction identities\"](https://openstax.org/books/precalculus-2e/pages/7-2-sum-and-difference-identities) that $\\sin{\\theta} = \\cos{\\frac{\\pi}{2}-\\theta}$\n",
        "\n",
        "\\begin{align}\n",
        "d(2\\sin{\\frac{15+45}{2}}\\cos{\\frac{15-45}{2}})&=S\\sin{120} \\\\\n",
        "\\\\\n",
        "d(2\\sin{30}\\cos{-15})&=S\\cos{90-120} \\\\\n",
        "\\\\\n",
        "d(2\\sin{30}\\cos{15})&=S\\cos{30} \\\\\n",
        "\\\\\n",
        "d&=\\frac{S\\cos{30}}{(2\\sin{30}\\cos{15})} \\\\\n",
        "\\end{align}\n",
        "\n",
        "Now remember some basic values on [the unit circle](https://openstax.org/books/precalculus-2e/pages/7-2-sum-and-difference-identities), $\\cos{30}=\\frac{\\sqrt{3}}{2}$ and $\\sin{30}=\\frac{1}{2}$ to find the length of $d$\n",
        "\n",
        "\\begin{align}\n",
        "\\\\\n",
        "d&=\\frac{S(\\frac{\\sqrt{3}}{2})}{(2\\frac{1}{2}\\cos{15})} \\\\\n",
        "\\\\\n",
        "d&=\\frac{S\\sqrt{3}}{2\\cos{15}} \\\\\n",
        "\\\\\n",
        "\\end{align}\n",
        "\n",
        "Then remember that you can calculate [the area of a square from its diagonal $d$](https://byjus.com/maths/area-of-square-using-diagonal/), and the area of the square from its side $s$, so set the areas to equivalent values to express $s$ in terms of $S$...\n",
        "\n",
        "\\begin{align}\n",
        "\\frac{1}{2}d^2&=\\frac{1}{2}(\\frac{S\\sqrt{3}}{2\\cos{15}})^2 = s^2\\\\\n",
        "\\\\\n",
        "s&=\\sqrt{\\frac{1}{2}(\\frac{S\\sqrt{3}}{2\\cos{15}})^2} \\\\\n",
        "\\\\\n",
        "s&=\\sqrt{\\frac{1}{2}}\\frac{S\\sqrt{3}}{2\\cos{15}} \\\\\n",
        "\\\\\n",
        "s&=\\frac{S\\sqrt{3}}{2\\sqrt{2}\\cos{15}} \\\\\n",
        "\\end{align}"
      ],
      "metadata": {
        "id": "k9nmLAYG3kdt"
      }
    },
    {
      "cell_type": "code",
      "source": [],
      "metadata": {
        "id": "JLnEbYaq3tcH"
      },
      "execution_count": null,
      "outputs": []
    }
  ]
 }
	{
	"nbformat": 4,
	"nbformat_minor": 0,
	"metadata": {
	"colab": {
	"provenance": [],
	"authorship_tag": "ABX9TyNwhdCnlMdqcw2r7dmGgWjs",
	"include_colab_link": true
	},
	"kernelspec": {
	"name": "python3",
	"display_name": "Python 3"
	},
	"language_info": {
	"name": "python"
	}
	},
	"cells": [
	{
	"cell_type": "markdown",
	"metadata": {
	"id": "view-in-github",
	"colab_type": "text"
	},
	"source": [
	"<a href=\"https://colab.research.google.com/gist/theredpea/cfe6cf2d3ab8cbfed2ec165d4450e61f/equilateral-triangles-overlap-area.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>"
	]
	},
	{
	"cell_type": "markdown",
	"source": [
	"These tweets propose a math/geometry/trigonometry puzzle.\n",
	" - [Original puzzle tweet from Diego Rattaggi @diegorattaggi and inspired by Michael Penn](https://twitter.com/diegorattaggi/status/1450903893899202575)\n",
	" - [Solution tweet from Per Henrik Christiansen @PerHenrikChris1](https://twitter.com/PerHenrikChris1/status/1583101619344146433)\n",
	"\n",
	"Here is the original puzzle:\n",
	"\n",
	"<img src=\"https://pbs.twimg.com/media/FCKlZAdX0AYKAOn?format=png&name=4096x4096\" alt=\"Two equilateral triangles overlap (one of them is flipped 180 degrees) so the intersections of their points and edges form the diagonal of a square\" width=\"400\"/>\n"
	],
	"metadata": {
	"id": "YbCmZK4A3dMA"
	}
	},
	{
	"cell_type": "markdown",
	"source": [
	"\n",
	" Here is a helpful diagram I use to help understand the solution from Per...\n",
	"\n",
	"Per's solution is posted on [desmos, here](https://t.co/QAwd41j8G5). In my diagram below, I use Per's conventions where:\n",
	"- $S$ is the length of the \"main\" equilateral triangle\n",
	"- $s$ is the length of the square formed in the overlap\n",
	"- $d$ is a diagonal of the square\n",
	"- $X$ is the length of the \"medium\" equilateral triangle whose height is $s$\n",
	"- $(S-X)$ is the length of the \"small\" equilateral triangle excluded by the square\n",
	"\n",
	"<img src=\"https://drive.google.com/uc?export=view&id=1vuJXcnQS52-aRG68QY5LQb0vzMahHTF4\" alt=\"Measuring some angles from the equilateral triangles in above example\" width=\"400\"/>\n",
	"\n",
	"\n",
	"I get stuck when Per says this:\n",
	"$$\n",
	"s=\\frac{S\\sqrt{3}}{2\\sqrt{2}\\cos15}\n",
	"$$\n",
	"\n",
	"<img src=\"https://drive.google.com/uc?export=view&id=1ougmQolOhDIw51nmogKMHreYsAs6efXF\" alt=\"Per's solution on desmos \" width=\"600\"/>\n",
	"\n"
	],
	"metadata": {
	"id": "xiFJiDqoo9vY"
	}
	},
	{
	"cell_type": "markdown",
	"source": [
	"Let's try it...\n",
	"\n",
	"We have two ways of calculating the diagonal length $d$. We use the [law of sines](https://en.wikipedia.org/wiki/Law_of_sines), which is an \"equation relating the lengths of the sides of any triangle to the sines of its angles\", (nicely explained [here](https://youtu.be/4Dv5IFqATrc?t=490))\n",
	"\n",
	"\\begin{align} \\tag{1}\n",
	"d&=(S-X) \\frac{\\sin{120}}{\\sin{15}} \\\\\n",
	"\\\\\\tag{2}\n",
	"d&=X\\frac{\\sin{120}}{\\sin{45}} \\\\\n",
	"\\end{align}\n",
	"\n",
	"Then we can rearrange equation 2 to get the value of $X$:\n",
	"\\begin{align}\n",
	"\\\\\n",
	"X&=d\\frac{\\sin{45}}{\\sin{120}} \\\\\n",
	"\\end{align}\n",
	"\n",
	"Now substitute the value of $X$ into equation 1, and isolate $d$ on the left side of the equation, and $S$ on the right side...\n",
	"\\begin{align}\n",
	"\\\\\n",
	"d&=(S-d\\frac{\\sin{45}}{\\sin{120}}) \\frac{\\sin{120}}{\\sin{15}} \\\\\n",
	"\\\\\n",
	"d\\frac{\\sin{15}}{\\sin{120}}&=S-d\\frac{\\sin{45}}{\\sin{120}} \\\\\n",
	"\\\\\n",
	"d\\sin{15}&=S\\sin{120}-d\\sin{45} \\\\\n",
	"\\\\\n",
	"d\\sin{15} + d\\sin{45}&=S\\sin{120} \\\\\n",
	"\\\\\n",
	"d(\\sin{15} + \\sin{45})&=S\\sin{120} \\\\\n",
	"\\end{align}\n",
	"\n",
	"Now for the left side of the equation, use the [\"sum and product formulae\"](https://openstax.org/books/precalculus-2e/pages/7-4-sum-to-product-and-product-to-sum-formulas) to convert the sum $\\sin{15} + \\sin{45}$ into a multiplication of $\\sin$ and $\\cos$... (Note the [\"sine angle addition identity\"](https://youtu.be/R0EQg9vgbQw), aka the [\"compound angle formulae\"](https://www.liverpool.ac.uk/~maryrees/homepagemath191/trigid.pdf) did not apply here; that would apply if we were handling $\\sin{(15 + 45)}$ )\n",
	"\n",
	"For the right side of the equation, use the [\"cofunction identities\"](https://openstax.org/books/precalculus-2e/pages/7-2-sum-and-difference-identities) that $\\sin{\\theta} = \\cos{\\frac{\\pi}{2}-\\theta}$\n",
	"\n",
	"\\begin{align}\n",
	"d(2\\sin{\\frac{15+45}{2}}\\cos{\\frac{15-45}{2}})&=S\\sin{120} \\\\\n",
	"\\\\\n",
	"d(2\\sin{30}\\cos{-15})&=S\\cos{90-120} \\\\\n",
	"\\\\\n",
	"d(2\\sin{30}\\cos{15})&=S\\cos{30} \\\\\n",
	"\\\\\n",
	"d&=\\frac{S\\cos{30}}{(2\\sin{30}\\cos{15})} \\\\\n",
	"\\end{align}\n",
	"\n",
	"Now remember some basic values on [the unit circle](https://openstax.org/books/precalculus-2e/pages/7-2-sum-and-difference-identities), $\\cos{30}=\\frac{\\sqrt{3}}{2}$ and $\\sin{30}=\\frac{1}{2}$ to find the length of $d$\n",
	"\n",
	"\\begin{align}\n",
	"\\\\\n",
	"d&=\\frac{S(\\frac{\\sqrt{3}}{2})}{(2\\frac{1}{2}\\cos{15})} \\\\\n",
	"\\\\\n",
	"d&=\\frac{S\\sqrt{3}}{2\\cos{15}} \\\\\n",
	"\\\\\n",
	"\\end{align}\n",
	"\n",
	"Then remember that you can calculate [the area of a square from its diagonal $d$](https://byjus.com/maths/area-of-square-using-diagonal/), and the area of the square from its side $s$, so set the areas to equivalent values to express $s$ in terms of $S$...\n",
	"\n",
	"\\begin{align}\n",
	"\\frac{1}{2}d^2&=\\frac{1}{2}(\\frac{S\\sqrt{3}}{2\\cos{15}})^2 = s^2\\\\\n",
	"\\\\\n",
	"s&=\\sqrt{\\frac{1}{2}(\\frac{S\\sqrt{3}}{2\\cos{15}})^2} \\\\\n",
	"\\\\\n",
	"s&=\\sqrt{\\frac{1}{2}}\\frac{S\\sqrt{3}}{2\\cos{15}} \\\\\n",
	"\\\\\n",
	"s&=\\frac{S\\sqrt{3}}{2\\sqrt{2}\\cos{15}} \\\\\n",
	"\\end{align}"
	],
	"metadata": {
	"id": "k9nmLAYG3kdt"
	}
	},
	{
	"cell_type": "code",
	"source": [],
	"metadata": {
	"id": "JLnEbYaq3tcH"
	},
	"execution_count": null,
	"outputs": []
	}
	]
	}