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], ['\\\\(', '\\\\)']]},\n svg: {fontCache: 'global'}\n};\n\u0026lt;\/script\u0026gt;\n\n\u0026lt;script src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-svg.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;\/head\u0026gt;\n\n\u0026lt;body\u0026gt;\n\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\nIn the Cartesian coordinate plane $xOy$, consider the parabola\n$\nC: y = x^2\n$\nand the line\n$\n\\ell: y = 13x + 44.\n$\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nThe line $\\ell$ intersects the parabola $C$ at two points $P$ and $Q$, \nwith $P$ lying to the left of $Q$.\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nLet $A, B, C$ be arbitrary points on the arc of the parabola between $P$ and $Q$.\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nRotate the region of the second and third quadrants about the $y$-axis \nby $120^\\circ$ into three-dimensional space.\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nLet $O$ denote the origin. \n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\nIf the maximum volume of $OABC$ can be expressed in the form\n$\n\\frac{a\\sqrt{b} + c\\sqrt{d}}{e},\n$\nwhere $a,b,c,d,e$ are positive integers, \n$\\gcd(a,c,e)=1$, and $b,d$ are square-free integers, \nfind\n$\na+b+c+d+67e\n$\n\u0026lt;\/p\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\u0026lt;hr\u0026gt;\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\n\n\n\n\u0026lt;p\u0026gt;The intersection of the parabola $y = x^2$ and the line $y = 13x + 44$ is determined by solving:\u0026lt;\/p\u0026gt;\n$x^2 - 13x - 44 = 0$\n\u0026lt;p\u0026gt;Using the quadratic formula:\u0026lt;\/p\u0026gt;\n$x = \\frac{13 \\pm \\sqrt{169 + 176}}{2} = \\frac{13 \\pm \\sqrt{345}}{2}$\n\u0026lt;p\u0026gt;Let $x_1 = \\frac{13 - \\sqrt{345}}{2}$ and $x_2 = \\frac{13 + \\sqrt{345}}{2}$.\u0026lt;\/p\u0026gt;\n\n\u0026lt;h3\u0026gt;Casework\u0026lt;\/h3\u0026gt;\n\u0026lt;p\u0026gt;Points $A, B,$ and $C$ lie on the arc of the parabola. We analyze the volume based on their position relative to the origin $O$ after a $120^\\circ$ rotation about the $y$-axis.\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;Case 1:\u0026lt;\/strong\u0026gt; If $A, B, C$ are all to the left of the $y$-axis ($x \u0026lt; 0$), the points remain coplanar with the origin. The volume of $OABC$ is $0$.\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;Case 2:\u0026lt;\/strong\u0026gt; If $A, B, C$ are all to the right of the $y$-axis ($x \u0026gt; 0$), the points remain coplanar with the origin. The volume of $OABC$ is $0$.\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;Case 3:\u0026lt;\/strong\u0026gt; Point $A$ is to the left of $O$ \n\u0026lt;p\u0026gt;In this configuration, we place point $A$ at the left intersection $x_1$ and points $B$ and $C$ on the right side of the arc. To find the area of the base $\\triangle OBC$, we define a point $Q$ directly below $B$ on the line $\\ell$ connecting $O$ and $C$.\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;1. Area Decomposition:\u0026lt;\/strong\u0026gt;The area of $\\triangle OBC$ can be split into two triangles, $\\triangle OBQ$ and $\\triangle CBQ$, by the vertical segment $BQ$.\u0026lt;\/p\u0026gt;\u0026lt;ul\u0026gt;\u0026lt;li\u0026gt;For $\\triangle OBQ$, the base is $BQ$ and the height is the horizontal distance $x_B$. Thus, $\\text{Area}_{OBQ} = \\frac{x_B \\cdot BQ}{2}$.\u0026lt;\/li\u0026gt;\u0026lt;li\u0026gt;For $\\triangle CBQ$, the base is $BQ$ and the height is the horizontal distance $(x_C - x_B)$. Thus, $\\text{Area}_{CBQ} = \\frac{(x_2 - x_B) \\cdot BQ}{2}$.\u0026lt;\/li\u0026gt;\u0026lt;\/ul\u0026gt;\u0026lt;p\u0026gt;Summing these areas, we get the total base area:$\\text{Area}_{OBC} = \\frac{x_B \\cdot BQ}{2} + \\frac{(x_2 - x_B) \\cdot BQ}{2} = \\frac{x_2 \\cdot BQ}{2}$\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;2. Maximizing $BQ$:\u0026lt;\/strong\u0026gt;The length $BQ$ is the vertical distance between the parabola $y=x^2$ and the segment $OC$. The line $OC$ has the equation $y = \\frac{x_2^2}{x_2}x = x_2 x$. Therefore:$BQ = x_2 x_B - x_B^2$This quadratic is maximized when $x_B = \\frac{x_2}{2}$. At this point, $BQ = x_2(\\frac{x_2}{2}) - (\\frac{x_2}{2})^2 = \\frac{x_2^2}{4}$.Substituting this back into the area formula:$\\text{Area}_{OBC} = \\frac{x_2}{2} \\cdot \\frac{x_2^2}{4} = \\frac{x_2^3}{8}$\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;3. Volume and Vieta's Theorem:\u0026lt;\/strong\u0026gt;The height of the tetrahedron in the rotated 3D space is $h = -x_1 \\frac{\\sqrt{3}}{2}$ (accounting for the $120^\\circ$ rotation). The volume is:$V = \\frac{1}{3} \\cdot \\left( \\frac{x_2^3}{8} \\right) \\cdot \\left( \\frac{-x_1\\sqrt{3}}{2} \\right) = \\frac{-x_1 x_2^3 \\sqrt{3}}{48}$By Vieta's Theorem for the equation $x^2 - 13x - 44 = 0$, we know $x_1 x_2 = -44$. We can rewrite the volume as:$V = \\frac{-(x_1 x_2) x_2^2 \\sqrt{3}}{48} = \\frac{-(-44) x_2^2 \\sqrt{3}}{48} = \\frac{44 \\sqrt{3} x_2^2}{48} = \\frac{11 \\sqrt{3} x_2^2}{12}$\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;Using $x_2^2 = \\frac{257 + 13\\sqrt{345}}{2}$:$V = \\frac{11\\sqrt{3}}{12} \\cdot \\frac{257 + 13\\sqrt{345}}{2} = \\frac{2827\\sqrt{3} + 143\\sqrt{1035}}{24} = \\frac{2827\\sqrt{3} + 429\\sqrt{115}}{24}$\u0026lt;\/p\u0026gt;\n\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;\u0026lt;strong\u0026gt;Case 4:\u0026lt;\/strong\u0026gt; Point $A$ is to the right of $O$, while points $B$ and $C$ are to the left. The volume is $\\frac{11\\sqrt{3}x_1^2}{12}$. The resulting volume is:\n$V = \\frac{2827\\sqrt{3} - 429\\sqrt{115}}{24}$\n\u0026lt;\/p\u0026gt;\n\n\n\u0026lt;p\u0026gt;Comparing Case 3 and Case 4, the maximum volume is given by Case 3. Expressing this in the form $\\frac{a\\sqrt{b} + c\\sqrt{d}}{e}$:\u0026lt;\/p\u0026gt;\n\u0026lt;ul\u0026gt;\n \u0026lt;li\u0026gt;$a = 2827, b = 3$\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;$c = 429, d = 115$\u0026lt;\/li\u0026gt;\n \u0026lt;li\u0026gt;$e = 24$\u0026lt;\/li\u0026gt;\n\u0026lt;\/ul\u0026gt;\n\n\u0026lt;p\u0026gt;The final value is:\n$a + b + c + d + 67e = 2827 + 3 + 429 + 115 + 1608 = \\boxed{4982}$\n\u0026lt;\/p\u0026gt;\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9227546}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_818046a6-1d5a-482a-a00f-af2972f97232","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_41ef131a-26e2-46ac-a695-9ffdc390633a","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"1b977d06-058c-489c-8c46-80d794f6ad0d","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"80a831dc-5c11-4696-a6fa-789cb8130130","items":[{"type":"BlockComponentItem","id":"7c7a9a30-a9e2-4974-b807-183fc8044f48","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"b20b5f62-f8a2-4cc9-be3e-476e2c369434","items":[{"type":"HtmlComponent","id":26575749,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page21022026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9227547}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"21 Feb 2026","uid":"58c69a8a-09d7-4b77-a88f-ba5fa2ae0ff9","path":"\/21feb2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_26fa4fa7-bd57-46bd-b54b-3959bb0e5ea2","sections":[{"type":"Slide","id":"f_0cc98249-fbce-4d51-86f2-966397a3e4b6","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_b390b2e2-9fc8-417d-9403-798ccd270c2f","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_8e4d2a35-ec8c-4b81-b7e9-887e7bc76565","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_9ee279b7-4516-45b3-bb79-d0700c871ce2","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_bb12efdb-8eb7-4dea-aa2c-59471cff8cf2","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_25dcbfd2-84b0-4796-8608-32ca4102ae40","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_3c21012a-7fe5-4d38-a21a-9c5b416f36bc","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_5cc860cc-b599-4700-a501-4275e781569b","defaultValue":null,"items":[{"type":"HtmlComponent","id":26629714,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Geometry Problem\u0026lt;\/title\u0026gt;\n\n \u0026lt;!-- MathJax --\u0026gt;\n \u0026lt;script\u0026gt;\n window.MathJax = {\n tex: {\n inlineMath: [['
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], ['\\\\(', '\\\\)']],\n displayMath: [['$', '$'], ['\\\\[', '\\\\]']]\n }\n };\n \u0026lt;\/script\u0026gt;\n \u0026lt;script src=\"https:\/\/cdn.jsdelivr.net\/npm\/mathjax@3\/es5\/tex-chtml.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;\/head\u0026gt;\n\n\u0026lt;body\u0026gt;\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\nFind the minimum $p \\in \\mathbb{Z}^+$ such that:$20151121^{20151121} \\mid p!$\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\u0026lt;div class=\"step\"\u0026gt;\n \u0026lt;strong\u0026gt;1. Base Factorization:\u0026lt;\/strong\u0026gt;\n The number $20151121$ is a perfect power:\n $20151121 = 4489^2 = (67^2)^2 = 67^4$\n Substituting this back into the expression:\n $(67^4)^{20151121} = 67^{80604484}$\n We must find the smallest $p$ such that $p!$ contains at least $80604484$ factors of $67$.\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"step\"\u0026gt;\n \u0026lt;strong\u0026gt;2. Legendre's Formula:\u0026lt;\/strong\u0026gt;\n The exponent of a prime $q$ in $p!$ is given by:\n $E_q(p!) = \\sum_{k=1}^{\\infty} \\left\\lfloor \\frac{p}{q^k} \\right\\rfloor$\n Using the approximation $E_q(p!) \\le \\frac{p}{q-1}$:\n $\\frac{p}{66} \\ge 80604484 \\implies p \\ge 5319895944$\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"step\"\u0026gt;\n \u0026lt;strong\u0026gt;3. Conclusion:\u0026lt;\/strong\u0026gt;\n Testing the multiple $p = 5319895944$ yields an exponent of $80604483$. Adding the next multiple of $67$ gives:\n $p = \\boxed{5319896011}$\n \u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9233160}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_1ea681e2-9a55-4853-af88-e410e9da2ab0","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_213f30d9-73b7-4db4-b744-31a283a563d3","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"c1790b81-d45d-40ad-a72e-5ab1ee5d1c54","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"057ed604-fb62-4b46-87c8-0c8f9dbf36d2","items":[{"type":"BlockComponentItem","id":"0743316f-81d6-4b65-bbc1-01e25bb95d3a","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"914c5561-29b6-41bd-988f-1bef731b1429","items":[{"type":"HtmlComponent","id":26629720,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page28022026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9233161}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"28 Feb 2026","uid":"6eac4717-e233-4df6-9016-2d3570e0cb51","path":"\/28feb2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_222a9ce6-a4b8-4160-b3a7-597b9854b917","sections":[{"type":"Slide","id":"f_b21f4940-d576-42f1-a034-274a8eaa3c3a","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_bcf267cb-93f9-45de-ad3e-29a6d8323673","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_bb747b57-cb34-480f-a8bd-45d038862fcb","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_0936d558-a236-4b33-86db-6e7a7e6f067a","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_f8a601e4-0555-4ab6-a72f-7d79fa4964d5","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_1f9c2f49-2991-461d-a906-13122404f003","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_406bb2dc-d82b-4bef-a299-00883323bdb6","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_de862e98-5119-480f-a31c-624af29d5978","defaultValue":null,"items":[{"type":"HtmlComponent","id":26676340,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n\u0026lt;meta charset=\"UTF-8\"\u0026gt;\n\u0026lt;title\u0026gt;Math Problem: Tangents through a Point\u0026lt;\/title\u0026gt;\n\u0026lt;!-- KaTeX CSS --\u0026gt;\n\u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\n onload=\"renderMathInElement(document.body);\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;style\u0026gt;\n body { font-family: Arial, sans-serif; padding: 20px; line-height: 1.6; }\n h1 { color: #2c3e50; }\n .formula { font-size: 1.2em; margin: 10px 0; }\n\u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;Consider the cubic function:\u0026lt;\/p\u0026gt;\n\n \\( y = x^3 - 11x^2 + 26x - 51 \\)\n\n\n\u0026lt;p\u0026gt;Let \\( Q(4, y_0) \\) be a point. There exist exactly three tangents to the curve that pass through \\( Q \\).\u0026lt;\/p\u0026gt;\n\n\u0026lt;p\u0026gt;The range of \\( y_0 \\) is \\((a,b)\\). \\( b-a \\) can be written as a reduced fraction \\(\\frac{c}{d}\\) with coprime integers \\(c\\) and \\(d\\), and then compute:\u0026lt;\/p\u0026gt;\n\n \\( 80c - 3d \\)\n\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\n \u0026lt;div class=\"solution-step\"\u0026gt;\n \n \u0026lt;p\u0026gt;Let $(t, f(t))$ be the point of tangency on the curve $y = x^3 - 11x^2 + 26x - 51$The derivative is:\u0026lt;\/p\u0026gt;\n $ f'(t) = 3t^2 - 22t + 26 $\n \u0026lt;p\u0026gt;The equation of the tangent line passing through $Q(4, y_0)$ and $(t, f(t))$ is given by the slope formula:\u0026lt;\/p\u0026gt;\n $ \\frac{f(t) - y_0}{t - 4} = f'(t) $\n \u0026lt;p\u0026gt;Substituting the functions and simplifying:\u0026lt;\/p\u0026gt;\n $ (t^3 - 11t^2 + 26t - 51) - y_0 = (3t^2 - 22t + 26)(t - 4) $\n $ t^3 - 11t^2 + 26t - 51 - y_0 = 3t^3 - 34t^2 + 114t - 104 $\n $ 2t^3 - 23t^2 + 88t - 53 - y_0 = 0 $\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"solution-step\"\u0026gt;\n \n \u0026lt;p\u0026gt;For exactly three tangents, the cubic $g(t) = 2t^3 - 23t^2 + 88t - 53 - y_0$ must have three distinct real roots. This occurs when the local extrema of g(t) lie on opposite sides of the t-axis.\u0026lt;\/p\u0026gt;\n \u0026lt;p\u0026gt;Find critical points via $g'(t) = 0$\u0026lt;\/p\u0026gt;\n $ 6t^2 - 46t + 88 = 0 \\implies 3t^2 - 23t + 44 = 0 $\n $ (3t - 11)(t - 4) = 0 $\n \u0026lt;p\u0026gt;The critical points are at $t = \\frac{11}{3}, 4$\u0026lt;\/p\u0026gt;\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"solution-step\"\u0026gt;\n \n \u0026lt;p\u0026gt;We require $g(4) \\cdot g\\left(\\frac{11}{3}\\right) \u0026lt; 0$ Evaluating the function at the critical points:\u0026lt;\/p\u0026gt;\n \u0026lt;ul\u0026gt;\n $g(4) = 59 - y_0$\n $g\\left(\\frac{11}{3}\\right) = \\frac{1594}{27} - y_0$\n \u0026lt;\/ul\u0026gt;\n \u0026lt;p\u0026gt;This gives the range $ y_0 \\in \\left( 59, \\frac{1594}{27} \\right) $ so $a = 59$ $b = \\frac{1594}{27}$\u0026lt;\/p\u0026gt;\n \u0026lt;\/div\u0026gt;\n\n \u0026lt;div class=\"solution-step\"\u0026gt;\n \n \u0026lt;p\u0026gt;The difference is:\u0026lt;\/p\u0026gt;\n $ b - a = \\frac{1594}{27} - \\frac{1593}{27} = \\frac{1}{27} $\n \u0026lt;p\u0026gt;Where $c = 1$ $d = 27$ The final value is:\u0026lt;\/p\u0026gt;\n \u0026lt;div class=\"result-box\"\u0026gt;\n $ 80c - 3d = 80(1) - 3(27) = 80 - 81 = \\boxed{-1} $\n \u0026lt;\/div\u0026gt;\n \u0026lt;\/div\u0026gt;\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9238442}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_cf362c1b-7e51-4860-8b64-e3b988ba5bf7","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_0798bcd4-9136-4f21-b05c-9ad25d1dd370","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"abc35b75-0cc0-4d7f-8936-391355245793","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"5e3e5b10-a19e-4f67-aa26-65e1ffb7d6f5","items":[{"type":"BlockComponentItem","id":"2986dc0b-a28f-458e-9474-e1e67b00e232","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"1bca910c-18ec-4d22-92c7-0960cfb61ddd","items":[{"type":"HtmlComponent","id":26676348,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page07032026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9238443}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"7 Mar 2026","uid":"69777895-0258-4183-a297-6c832978fb48","path":"\/7mar2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_61742f60-2f5c-46a5-a074-5ba1309341b5","sections":[{"type":"Slide","id":"f_4cc583f4-8de3-4d49-a20f-541eec20631c","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_943c3c83-91c1-4c45-9f34-9883ff01ebe8","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_cb85c8c0-e3be-4d13-9ce6-fb0bbd975f4c","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_fcc879bc-7a3e-41f8-adeb-716bcaec40bf","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_aca23767-705b-40da-9dd4-a63576daeea3","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_b5ca18a4-9070-4063-bf1e-3a5ecc636fa1","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_6419c6c2-7f9d-4848-b2b1-2780c6b3adb8","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_00454226-f64b-4cdc-b208-20cae5e8fb56","defaultValue":null,"items":[{"type":"HtmlComponent","id":26721757,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n\u0026lt;meta charset=\"UTF-8\"\u0026gt;\n\u0026lt;title\u0026gt;Math Problem: Tangents through a Point\u0026lt;\/title\u0026gt;\n\u0026lt;!-- KaTeX CSS --\u0026gt;\n\u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\n onload=\"renderMathInElement(document.body);\"\u0026gt;\u0026lt;\/script\u0026gt;\n\u0026lt;style\u0026gt;\n body { font-family: Arial, sans-serif; padding: 20px; line-height: 1.6; }\n h1 { color: #2c3e50; }\n .formula { font-size: 1.2em; margin: 10px 0; }\n\u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n\n\u0026lt;p\u0026gt;\n\u0026lt;div class=\"problem-box\"\u0026gt;\n \u0026lt;p\u0026gt;\n Kevin selects three distinct integers $a, b, c$ from the set:\n \u0026lt;\/p\u0026gt;\n $ \\{1, 2, 3, \\dots, 2026\\} $\n \u0026lt;p\u0026gt;\n The probability that these numbers can form the side lengths of a triangle with non-zero area can be written as $\\frac{a}{b}$ which a and b are coprime. Find a+b.\n \u0026lt;\/p\u0026gt;\n\u0026lt;\/div\u0026gt;\n\u0026lt;div class=\"solution-box\"\u0026gt;\n \u0026lt;p\u0026gt;\u0026lt;span class=\"step\"\u0026gt;Step 1: Define the Sample Space.\u0026lt;\/span\u0026gt;\u0026lt;br\u0026gt;\n We are choosing 3 distinct integers from the set $\\{1, 2, \\dots, 2026\\}$The total number of ways is:\u0026lt;\/p\u0026gt;\n $ S = \\binom{2026}{3} = \\frac{2026 \\times 2025 \\times 2024}{6} $\n\n \u0026lt;p\u0026gt;\u0026lt;span class=\"step\"\u0026gt;Step 2: Apply the Triangle Inequality.\u0026lt;\/span\u0026gt;\u0026lt;br\u0026gt;\n For three numbers $x \u0026lt; y \u0026lt; z$ to form a triangle, we must have $x + y \u0026gt; z$. The number of such combinations $T_n$ for an even $n$ is given by:\u0026lt;\/p\u0026gt;\n $ T_n = \\frac{n(n-2)(2n-5)}{24} $\n\n \u0026lt;p\u0026gt;\u0026lt;span class=\"step\"\u0026gt;Step 3: Calculate the Probability.\u0026lt;\/span\u0026gt;\u0026lt;br\u0026gt;\n simplifies algebraically to:\u0026lt;\/p\u0026gt;\n $ P = \\frac{T_n}{S} = \\frac{2n - 5}{4(n - 1)} $\n\n \u0026lt;p\u0026gt;\u0026lt;span class=\"step\"\u0026gt;Step 4: Substitute $n = 2026$.\u0026lt;\/span\u0026gt;\u0026lt;\/p\u0026gt;\n $ P = \\frac{2(2026) - 5}{4(2026 - 1)} = \\frac{4052 - 5}{4(2025)} = \\frac{4047}{8100} = \\frac{1349}{2700} $\n\n \n \n $ a + b = 1349 + 2700 = \\boxed{4049} $\n \u0026lt;\/div\u0026gt;\n\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9242502}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_e05d8fde-c33f-4aa7-836c-c86fb143074c","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_7ca839d8-a57d-4485-ab82-e2373f37215b","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"9c41d104-c15a-4fea-96ce-c3a21cf39cf5","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"2479ead5-73e4-48f8-8a2b-5407e9128f8f","items":[{"type":"BlockComponentItem","id":"8232dee5-dcea-4031-ac82-1a6faefd0674","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"9e6c5106-06d6-418f-847a-358d42062ddb","items":[{"type":"HtmlComponent","id":26721758,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page14032026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9242503}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"14 Mar 2026","uid":"13112462-6615-4c8a-bb4d-23b3d269185f","path":"\/14mar2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_f86169c2-6be2-4cfc-90b3-6b2fe771fffa","sections":[{"type":"Slide","id":"f_7ad77ed3-5348-49ed-861d-9589ccbce485","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_cd49bb59-b76a-46b8-ad00-3069f1d0c008","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_dd55aaec-8b57-428b-97e1-38da8e90ac96","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_a2ffb588-57af-4302-8162-9ed41f1d5e14","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_eeaa4ca6-6c71-47b2-8144-30c27a291fad","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_0729e330-afad-4977-87df-9c3029ead4b4","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_14efc80a-506a-48cd-9670-44d100dbe8b7","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_5ddb0c18-d876-403c-b8fa-96fcc63a0915","defaultValue":null,"items":[{"type":"HtmlComponent","id":26815313,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '
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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \u0026lt;p\u0026gt;\n A $Q$-digit number in base $P$ is said to possess Property $(P, Q)$ if all its digits are distinct, and the absolute difference between any two of its digits does not equal any digit present in the number itself.\n \u0026lt;\/p\u0026gt;\n\n \u0026lt;p\u0026gt;\n Let $S(P, Q)$ be the set of all numbers possessing Property $(P, Q)$. There exists a positive integer $T$ such that $S(T, 2026)$ is non-empty, while for all positive integers $R \u0026lt; T$, the set $S(R, 2026)$ is empty.\n \u0026lt;\/p\u0026gt;\n\n \u0026lt;p\u0026gt;\n If $M$ is the smallest element in $S(T+2028, 2027)$, find the value of:\n \u0026lt;\/p\u0026gt;\n \n \n $ (M + T) \\pmod{2026} $\n\n\u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n \u0026lt;p\u0026gt;\n 1. Prove that $T \u0026gt; 4051$:\n We set the digits of the $2026$-digits number is $a_1, a_2, a_3 ... a_{2026}(a_1 \u0026lt; a_2 \u0026lt; a_3 ... \u0026lt; a_{2026})$, and the biggest digit is $a_{2026}$.\n\nConstruct a sequence $b_1, b_2, b_3 ... b_{2025}(b_k = a_{2026} - a_k)$ If there is two distinct positive integers $i,j$, such that $a_i = b_j$. We can find that $a_i + a_j = a_{2026}$\n\n \nThus, we have to make that $a_1, a_2, a_3 ... a_{2026}, b_1, b_2 ... b_{2025}$ $4051 $ elements are all distinct. When $T \\le 4051$, there are at least two same numbers.\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\n 2. Construct an example when $T = 4052$:\n $\na_1 = 1,\na_2 = 3,\na_3 = 5,\na_4 = 7\n...\na_{2026} = 4051\n$\nIt is easy to prove that it has Property$(4052,2026)$.\n\u0026lt;\/p\u0026gt;\n 3. Find the smallest number that has Property$(6080,2027)$ and final answer:\nSimilarly,$a_1 = 1, a_2 = 3 ... a_{2027} = 4053$.\nWe just need to find that \n$\\sum_{i=1}^{2027} 6080^{2027-i}a_i \\equiv \\sum_{i=1}^{2027} 2^{2027-i}(2i-1) \\equiv 3 \\cdot 2^{2027} - 4057 \\pmod{2026}$\n\n $(M+T) \\equiv (3 \\cdot 2^{2027} - 5) \\pmod{2026}$\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nBy Fermat's Little Theorem, \n We can find that \n$2^{2027} \\equiv 8 \\pmod{1013}$ and $2^{2027} \\equiv 8$ or $ 1021 \\pmod{2026}$\n\nBecause $2^{2027}$ is even, $2^{2027} \\equiv 8 \\pmod{2026}$\n\nThus, $(M+T) \\bmod{2026} = 3 \\cdot 8 - 5 = \\boxed{19}$\n\n\n \u0026lt;\/p\u0026gt;\n\n \n \n \n \n \n\u0026lt;\/div\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9256577}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_2bc385b4-5720-4ea8-b0d2-ccc8d539c584","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_c06c283c-c5e4-4541-b9d1-21c05ea462e7","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"27a85022-f4cc-44e7-9abe-3f7269863aae","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"5bb2e3ea-7b09-4799-96be-e30cc77b9335","items":[{"type":"BlockComponentItem","id":"f7b01d89-2444-442f-9fd5-b8a67a337e82","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"dc8cfb7d-22d5-44c7-bc3c-49167ee60a55","items":[{"type":"HtmlComponent","id":26815445,"defaultValue":false,"value":"\u0026lt;div id=\"disqus_thread\"\u0026gt;\u0026lt;\/div\u0026gt;\n\u0026lt;script\u0026gt;\nvar disqus_config = function () {\nthis.page.url = window.location.href;\nthis.page.identifier = \"page21032026\";\n};\n(function() {\nvar d = document, s = d.createElement('script');\ns.src = 'https:\/\/YOURNAME.disqus.com\/embed.js';\ns.setAttribute('data-timestamp', +new Date());\n(d.head || d.body).appendChild(s);\n})();\n\n\u0026lt;\/script\u0026gt;","render_as_iframe":false,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9256585}"}]}}}],"inlineLayout":"12"}}}],"layout":[],"inlineLayout":"1"}}}],"title":"21 Mar 2026","uid":"2cefb69d-ee7f-415c-854d-4ae03989254d","path":"\/21mar2026","autoPath":false,"authorized":true},{"type":"Page","id":"f_5e7b01b0-8f45-43dd-8e25-a2845eed1454","sections":[{"type":"Slide","id":"f_43e4e8bc-b8e6-4e2a-b460-a66939bfccbb","defaultValue":null,"template_name":"s6_common_section","template_version":"s6","components":{"slideSettings":{"type":"SlideSettings","id":"f_78dc2f4b-0657-4d8d-a849-5fd6d644141a","defaultValue":null,"show_nav":true,"show_nav_multi_mode":false,"nameChanged":null,"hidden_section":false,"name":"Make Your Own","sync_key":null,"layout_variation":null,"display_settings":{},"padding":{"top":"normal","bottom":"normal"},"layout_config":{"width":"auto","height":"auto"}},"background1":{"type":"Background","id":"f_5fa03bc4-5d00-42ea-9b4d-6126eddc806e","defaultValue":true,"url":"","textColor":"light","backgroundVariation":"","sizing":"cover","userClassName":null,"linkUrl":null,"linkTarget":null,"videoUrl":"","videoHtml":"","storageKey":null,"storage":null,"format":null,"h":null,"w":null,"s":null,"useImage":null,"noCompression":null,"focus":{},"backgroundColor":{}},"block1":{"type":"BlockComponent","id":"f_ec3a2983-9cfd-4406-8b82-a32d09d8a0d9","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_8d073596-d408-4720-9407-5cd78b180180","defaultValue":null,"name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"f_efaef965-bade-41d4-8d5d-98f7fcd87cdf","defaultValue":null,"items":[{"type":"BlockComponentItem","id":"f_0875ebff-459c-4cf1-8d1b-b3bd8fc243a0","defaultValue":null,"name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"f_72fc710d-7b0e-4a42-8fc1-5128de336ce4","defaultValue":null,"items":[{"type":"HtmlComponent","id":26860037,"defaultValue":false,"value":"\u0026lt;!DOCTYPE html\u0026gt;\n\u0026lt;html lang=\"en\"\u0026gt;\n\u0026lt;head\u0026gt;\n \u0026lt;meta charset=\"UTF-8\"\u0026gt;\n \u0026lt;title\u0026gt;Math Problem: Property (P, Q)\u0026lt;\/title\u0026gt;\n \n \u0026lt;link rel=\"stylesheet\" href=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.css\"\u0026gt;\n\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/katex.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n \u0026lt;script defer src=\"https:\/\/cdn.jsdelivr.net\/npm\/katex@0.16.8\/dist\/contrib\/auto-render.min.js\"\u0026gt;\u0026lt;\/script\u0026gt;\n\n \u0026lt;script\u0026gt;\n document.addEventListener(\"DOMContentLoaded\", function() {\n renderMathInElement(document.body, {\n delimiters: [\n {left: '$', right: '$', display: true},\n {left: '
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, display: false}\n ],\n \/\/ \u95dc\u9375\u8a2d\u5b9a\uff1a\u9632\u6b62\u91cd\u8907\u986f\u793a MathML\n output: 'htmlAndMathml', \n throwOnError: false\n });\n });\n \u0026lt;\/script\u0026gt;\n\n \u0026lt;style\u0026gt;\n \/* \u984d\u5916\u4fdd\u96aa\uff1a\u5f37\u5236\u96b1\u85cf\u91cd\u8907\u7684 MathML \u5143\u7d20 *\/\n .katex-mathml {\n display: none;\n }\n \n body { font-family: 'Segoe UI', Arial, sans-serif; padding: 40px; line-height: 1.6; background-color: #fdfdfd; }\n .problem-container { max-width: 850px; margin: auto; border: 1px solid #ddd; padding: 30px; background-color: white; box-shadow: 0 4px 6px rgba(0,0,0,0.1); }\n h2 { color: #2c3e50; border-bottom: 2px solid #3498db; padding-bottom: 10px; }\n p { margin-bottom: 15px; text-align: justify; }\n .equation-box { text-align: center; font-size: 1.2em; margin: 20px 0; }\n \u0026lt;\/style\u0026gt;\n\u0026lt;\/head\u0026gt;\n\u0026lt;body\u0026gt;\n\n\n \n\u0026lt;body\u0026gt;\n\n\u0026lt;div class=\"problem-container\"\u0026gt;\n \u0026lt;h2\u0026gt;Problem\u0026lt;\/h2\u0026gt;\n \n \n\n \u0026lt;p\u0026gt;\n Let $A$ be a set of all right circular cones such that each cone in $A$ has an inscribed sphere with a radius of $1000$.From the set $A$, let $P$ be the cone that has the minimum volume. An inscribed sphere of radius $1000$ is placed inside cone $P$.Let $R$ be a positive integer, and let $M$ be a positive integer such that $M$ spheres, each of radius $R$, can be placed simultaneously into the gap at the bottom of the cone (the region bounded by the cone's base, its lateral surface, and the large inscribed sphere).\u0026lt;\/p\u0026gt;\u0026lt;p\u0026gt;Find the value of $RM+R+M$ when the integer $R$ is at its maximum possible value.\u0026lt;\/p\u0026gt;\n \u0026lt;h2\u0026gt;Solution\u0026lt;\/h2\u0026gt;\n\u0026lt;p\u0026gt;\n 1. find the smallest value of the volume:\n\u0026lt;p\u0026gt;$V = \\frac{\\pi r^2 h}{3}$ , which $s$ represents the base radius of the cone\u0026lt;\/p\u0026gt;\nWe can take a vertical section from the cone.\nThe section is an isosceles triangle, with the $2r$ base side and the $h$ height. \nIt is easy to prove that $2arctan \\frac{1000}{r} = arctan \\frac{h}{r}$\n\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nThus, $h = \\frac{2000r^2}{r^2-1000000}$, and plug it in to $V$ to get $V = \\frac{2000\\pi r^4}{3r^2-3000000}$ and $V' = \\frac{4000\\pi r^3(r^2 - 2000000)}{3(r^2 - 1000000)}$, when $r = 1000\\sqrt{2}$, the value of the volume is smallest.\n\n\n \u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\n2. find the biggest radius of the balls:\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nConsider a right triangle $ABC$, with $C = \\frac{\\pi}{2}, AC = 1000\\sqrt{2}, BC = 4000$, and $D$ is on the side $BC$, with $CD = 1000$. Connect $AD$. Let $x$ and point$P, M$ on $AD$ be the radius ,center and a point on segment $AD$ of the small ball, respectively, which $PA = \\sqrt{3}x, PM = x, AM = (1+\\sqrt{3})x \\le 1000(\\sqrt{3}-1)$\nThus, $x \\le 1000(2-\\sqrt{3}) \\approx 267.9492$, which means $x = 267 = R$\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\n3. find the biggest number of the balls:\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nWhen $R = 267$, let the distance between the center of the ball and the central axis of the cone be $d$. where $d \\le (1000-267)\\sqrt{2} \\approx 1036.62$.\nThus the number of balls is $M = \\lfloor \\frac{\\pi}{arcsin \\frac{R}{d}} \\rfloor = 12$\n\u0026lt;\/p\u0026gt;\n\u0026lt;p\u0026gt;\nFinally, the answer is $RM + R + M = 267 \\cdot 12 + 267 + 12 = \\boxed{3483}$\n\u0026lt;\/p\u0026gt;\n\u0026lt;\/div\u0026gt;\n\n\u0026lt;\/body\u0026gt;\n\u0026lt;\/html\u0026gt;","render_as_iframe":true,"selected_app_name":"HtmlApp","app_list":"{\"HtmlApp\":9258844}"}],"layout":[],"inlineLayout":null}}}],"layout":[{"type":"LayoutVariants","id":"f_2134036c-79b1-461c-b055-f69f32352033","defaultValue":null,"value":"two-thirds"},{"type":"LayoutVariants","id":"f_bc2b9845-108d-44f6-b386-c46792d6b415","defaultValue":null,"value":"third"}],"inlineLayout":"12"}}},{"type":"BlockComponentItem","id":"9c3f5b42-2788-42bb-bed8-9bfed3f6fab4","name":"rowBlock","components":{"block1":{"type":"BlockComponent","id":"65f65d2b-868c-46c9-b749-a1544af43526","items":[{"type":"BlockComponentItem","id":"5fa6db7e-827a-4bc4-b789-491b3dac330f","name":"columnBlock","components":{"block1":{"type":"BlockComponent","id":"2ec0fc2c-a7f9-4e9c-af35-f156fc637f68","items":[{"type":"HtmlComponent","id":26860185,"defaultValue":false,"value":"\u0026lt;div 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