https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Ermscoach&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T02:00:12ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_8_Problems/Problem_16&diff=1885162023 AMC 8 Problems/Problem 162023-02-02T23:20:43Z<p>Ermscoach: Using symmetry</p>
<hr />
<div>==Problem==<br />
<br />
The letters <math>\text{P}, \text{Q},</math> and <math>\text{R}</math> are entered into a <math>20\times20</math> table according to the pattern shown below. How many <math>\text{P}</math>s, <math>\text{Q}</math>s, and <math>\text{R}</math>s will appear in the completed table? <br />
<asy><br />
/* Made by MRENTHUSIASM, Edited by Kante314 */<br />
usepackage("mathdots");<br />
size(5cm);<br />
draw((0,0)--(6,0),linewidth(1.5)+mediumgray);<br />
draw((0,1)--(6,1),linewidth(1.5)+mediumgray);<br />
draw((0,2)--(6,2),linewidth(1.5)+mediumgray);<br />
draw((0,3)--(6,3),linewidth(1.5)+mediumgray);<br />
draw((0,4)--(6,4),linewidth(1.5)+mediumgray);<br />
draw((0,5)--(6,5),linewidth(1.5)+mediumgray);<br />
<br />
draw((0,0)--(0,6),linewidth(1.5)+mediumgray);<br />
draw((1,0)--(1,6),linewidth(1.5)+mediumgray);<br />
draw((2,0)--(2,6),linewidth(1.5)+mediumgray);<br />
draw((3,0)--(3,6),linewidth(1.5)+mediumgray);<br />
draw((4,0)--(4,6),linewidth(1.5)+mediumgray);<br />
draw((5,0)--(5,6),linewidth(1.5)+mediumgray);<br />
<br />
label(scale(.9)*"\textsf{P}", (.5,.5));<br />
label(scale(.9)*"\textsf{Q}", (.5,1.5));<br />
label(scale(.9)*"\textsf{R}", (.5,2.5));<br />
label(scale(.9)*"\textsf{P}", (.5,3.5));<br />
label(scale(.9)*"\textsf{Q}", (.5,4.5));<br />
label("$\vdots$", (.5,5.6));<br />
<br />
label(scale(.9)*"\textsf{Q}", (1.5,.5));<br />
label(scale(.9)*"\textsf{R}", (1.5,1.5));<br />
label(scale(.9)*"\textsf{P}", (1.5,2.5));<br />
label(scale(.9)*"\textsf{Q}", (1.5,3.5));<br />
label(scale(.9)*"\textsf{R}", (1.5,4.5));<br />
label("$\vdots$", (1.5,5.6));<br />
<br />
label(scale(.9)*"\textsf{R}", (2.5,.5));<br />
label(scale(.9)*"\textsf{P}", (2.5,1.5));<br />
label(scale(.9)*"\textsf{Q}", (2.5,2.5));<br />
label(scale(.9)*"\textsf{R}", (2.5,3.5));<br />
label(scale(.9)*"\textsf{P}", (2.5,4.5));<br />
label("$\vdots$", (2.5,5.6));<br />
<br />
label(scale(.9)*"\textsf{P}", (3.5,.5));<br />
label(scale(.9)*"\textsf{Q}", (3.5,1.5));<br />
label(scale(.9)*"\textsf{R}", (3.5,2.5));<br />
label(scale(.9)*"\textsf{P}", (3.5,3.5));<br />
label(scale(.9)*"\textsf{Q}", (3.5,4.5));<br />
label("$\vdots$", (3.5,5.6));<br />
<br />
label(scale(.9)*"\textsf{Q}", (4.5,.5));<br />
label(scale(.9)*"\textsf{R}", (4.5,1.5));<br />
label(scale(.9)*"\textsf{P}", (4.5,2.5));<br />
label(scale(.9)*"\textsf{Q}", (4.5,3.5));<br />
label(scale(.9)*"\textsf{R}", (4.5,4.5));<br />
label("$\vdots$", (4.5,5.6));<br />
<br />
label(scale(.9)*"$\dots$", (5.5,.5));<br />
label(scale(.9)*"$\dots$", (5.5,1.5));<br />
label(scale(.9)*"$\dots$", (5.5,2.5));<br />
label(scale(.9)*"$\dots$", (5.5,3.5));<br />
label(scale(.9)*"$\dots$", (5.5,4.5));<br />
label(scale(.9)*"$\iddots$", (5.5,5.6));<br />
</asy><br />
<math>\textbf{(A)}~132\text{ Ps, }134\text{ Qs, }134\text{ Rs}</math> <br />
<br />
<math>\textbf{(B)}~133\text{ Ps, }133\text{ Qs, }134\text{ Rs}</math><br />
<br />
<math>\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}</math><br />
<br />
<math>\textbf{(D)}~134\text{ Ps, }132\text{ Qs, }134\text{ Rs}</math><br />
<br />
<math>\textbf{(E)}~134\text{ Ps, }133\text{ Qs, }133\text{ Rs}</math><br />
<br />
== Solution 1 ==<br />
<br />
In our <math>5 \times 5</math> grid we can see there are <math>8,9</math> and <math>8</math> of the letters <math>\text{P}, \text{Q},</math> and <math>\text{R}</math>’s respectively. We can see our pattern between each is <math>x, x+1,</math> and <math>x</math> for the <math>\text{P}, \text{Q},</math> and <math>\text{R}</math>’s respectively. This such pattern will follow in our bigger example, so we can see that the only answer choice which satisfies this condition is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math><br />
<br />
(Note: you could also "cheese" this problem by listing out all of the letters horizontally in a single line and looking at the repeating pattern.)<br />
<br />
~apex304, SohumUttamchandani, wuwang2002, TaeKim, Cxrupptedpat<br />
<br />
== Solution 2 ==<br />
<br />
We think about which letter is in the diagonal with <math>20</math> of a letter. We find that it is <math>2(2 + 5 + 8 + 11 + 14 + 17) + 20 = 134.</math> The rest of the grid with the <math>\text{P}</math>'s and <math>\text{R}</math>'s is symmetric. Therefore, the answer is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math><br />
<br />
~[[User:ILoveMath31415926535|ILoveMath31415926535]]<br />
<br />
== Solution 3 ==<br />
<br />
Notice that rows <math>x</math> and <math>x+3</math> are the same, for any <math>1 \leq x \leq 17.</math> Additionally, rows <math>1, 2,</math> and <math>3</math> collectively contain the same number of <math>\text{P}</math>s, <math>\text{Q}</math>s, and <math>\text{R}</math>s, because the letters are just substituted for one another. Therefore, the number of <math>\text{P}</math>s, <math>\text{Q}</math>s, and <math>\text{R}</math>s in the first <math>18</math> rows is <math>120</math>. The first row has <math>7</math> <math>\text{P}</math>s, <math>7</math> <math>\text{Q}</math>s, and <math>6</math> <math>\text{R}</math>s, and the second row has <math>6</math> <math>\text{P}</math>s, <math>7</math> <math>\text{Q}</math>s, and <math>7</math> <math>\text{R}</math>s. Adding these up, we obtain <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}</math>.<br />
<br />
~mathboy100<br />
<br />
== Solution 4 ==<br />
<br />
From the full diagram below, the answer is <math>\boxed{\textbf{(C)}~133\text{ Ps, }134\text{ Qs, }133\text{ Rs}}.</math><br />
<asy><br />
/* Made by MRENTHUSIASM, Edited by Kante314 */<br />
usepackage("mathdots");<br />
size(16.666cm);<br />
<br />
for (int y = 0; y<=20; ++y) {<br />
for (int x = 0; x<=20; ++x) {<br />
draw((x,0)--(x,20),linewidth(1.5)+mediumgray);<br />
draw((0,y)--(20,y),linewidth(1.5)+mediumgray);<br />
}<br />
}<br />
<br />
void drawDiagonal(string s, pair p) {<br />
while (p.x >= 0 && p.x < 20 && p.y >= 0 && p.y < 20) {<br />
label(scale(.9)*("\textsf{" + s + "}"),p);<br />
p += (1,-1);<br />
}<br />
}<br />
<br />
drawDiagonal("P", (0.5,0.5));<br />
drawDiagonal("Q", (0.5,1.5));<br />
drawDiagonal("R", (0.5,2.5));<br />
drawDiagonal("P", (0.5,3.5));<br />
drawDiagonal("Q", (0.5,4.5));<br />
drawDiagonal("R", (0.5,5.5));<br />
drawDiagonal("P", (0.5,6.5));<br />
drawDiagonal("Q", (0.5,7.5));<br />
drawDiagonal("R", (0.5,8.5));<br />
<br />
drawDiagonal("P", (0.5,9.5));<br />
drawDiagonal("Q", (0.5,10.5));<br />
drawDiagonal("R", (0.5,11.5));<br />
drawDiagonal("P", (0.5,12.5));<br />
drawDiagonal("Q", (0.5,13.5));<br />
drawDiagonal("R", (0.5,14.5));<br />
drawDiagonal("P", (0.5,15.5));<br />
drawDiagonal("Q", (0.5,16.5));<br />
drawDiagonal("R", (0.5,17.5));<br />
drawDiagonal("P", (0.5,18.5));<br />
drawDiagonal("Q", (0.5,19.5));<br />
<br />
drawDiagonal("R", (1.5,19.5));<br />
<br />
drawDiagonal("P", (2.5,19.5));<br />
drawDiagonal("Q", (3.5,19.5));<br />
drawDiagonal("R", (4.5,19.5));<br />
<br />
drawDiagonal("P", (5.5,19.5));<br />
drawDiagonal("Q", (6.5,19.5));<br />
drawDiagonal("R", (7.5,19.5));<br />
<br />
drawDiagonal("P", (8.5,19.5));<br />
drawDiagonal("Q", (9.5,19.5));<br />
drawDiagonal("R", (10.5,19.5));<br />
<br />
drawDiagonal("P", (11.5,19.5));<br />
drawDiagonal("Q", (12.5,19.5));<br />
drawDiagonal("R", (13.5,19.5));<br />
<br />
drawDiagonal("P", (14.5,19.5));<br />
drawDiagonal("Q", (15.5,19.5));<br />
drawDiagonal("R", (16.5,19.5));<br />
<br />
drawDiagonal("P", (17.5,19.5));<br />
drawDiagonal("Q", (18.5,19.5));<br />
drawDiagonal("R", (19.5,19.5));<br />
</asy><br />
<b>This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options.</b><br />
<br />
~MRENTHUSIASM<br />
<br />
== Solution 5 ==<br />
Note the Q diagonals are symmetrical. The R and P diagonals are not symmetrical, but are reflections of each other- the Q diagonals of length 2 are surrounded by a P of length 1 and an R of length 3, and a P of length 3 and an R of length 1.<br />
When looking at a pair of Q diagonals of the same length x, there is a total of 2x Rs and Ps next to these 2 diagonals.<br />
The main diagonal of 20Q has 19 P and 19R next to it. Thus the total is x Q, x+1 P, x+1 R, which is 133, 134, 133 PQR.<br />
<br />
~ERMSCoach<br />
<br />
==Animated Video Solution==<br />
https://youtu.be/1tnMR0lNEFY<br />
<br />
~Star League (https://starleague.us)<br />
<br />
==Video Solution by OmegaLearn (Using Cyclic Patterns)==<br />
https://youtu.be/83FnFhe4QgQ<br />
<br />
==Video Solution by Magic Square==<br />
https://youtu.be/-N46BeEKaCQ?t=3990<br />
<br />
==See Also== <br />
{{AMC8 box|year=2023|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Ermscoachhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_14&diff=1380972020 AMC 12A Problems/Problem 142020-11-22T00:48:40Z<p>Ermscoach: /* Solution 5 */</p>
<hr />
<div>==Problem 14==<br />
Regular octagon <math>ABCDEFGH</math> has area <math>n</math>. Let <math>m</math> be the area of quadrilateral <math>ACEG</math>. What is <math>\tfrac{m}{n}?</math><br />
<br />
<math>\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}</math><br />
<br />
==Solution 1==<br />
<math>ACEG</math> is a square. WLOG <math>AB = 1,</math> then using Law of Cosines, <math>AC^2 = [ACEG] = 1^2 + 1^2 - 2 \cos{135} = 2 + \sqrt{2}.</math> The area of the octagon is just <math>[ACEG]</math> plus the area of the four congruent (by symmetry) isosceles triangles, all an angle of <math>135</math> in between two sides of length 1. Now, <cmath>\dfrac{m}{n} = \dfrac{2 + \sqrt{2}}{2 + \sqrt{2} + 4 \cdot \tfrac{1}{2} \sin{135}} = \dfrac{2 + \sqrt{2}}{2 + 2 \sqrt{2}} = \dfrac{\sqrt{2}}{2}.</cmath> The answer is <math>\boxed{\textbf{(B)}}.</math><br />
<br />
==Solution 2==<br />
Refer to the diagram. Call one of the side lengths of the square <math>s</math>. Since quadrilateral ACEG is a square, the area of the square would just be <math>s^2</math>, which we can find by applying Law of Cosines on one of the four triangles. Assume each of the sides of the octagon has length <math>1</math>. Since each angle measures <math>135^{\circ}</math> in an octagon, then <math>s^2 = 1^2+1^2-2*\cos(135^{\circ}) = 2+\sqrt{2}</math> <br />
<br />
There are many ways to find the area of the octagon, but one way is to split the octagon into two trapezoids and one rectangle. We can easily compute AF to be <math>1+\sqrt{2}</math> from splitting one of the sides into two <math>45-45-90</math> triangles. So the area of the octagon is <math>2*\frac{1+\sqrt{2}}{2}+1+\sqrt{2} \Rightarrow 2+2\sqrt{2}</math>. <br />
<br />
Hence, <cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath> <br />
<cmath>\Rightarrow \frac{2+\sqrt{2}}{2+2\sqrt{2}} \cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}}</cmath><br />
<cmath>\Rightarrow \frac{-2\sqrt{2}}{-4}</cmath><br />
<cmath>\Rightarrow \boxed{\frac{\sqrt{2}}{2}}</cmath><br />
<br />
~Solution by IronicNinja<br />
<br />
== Solution 3 (Deriving Formulas) ==<br />
<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
</asy><br />
<br />
The first thing to notice is that <math>ACEG</math> is a square. This is because, as <math>\bigtriangleup ABC \cong \bigtriangleup CDE \cong \bigtriangleup EFG \cong \bigtriangleup GHE</math>, they all have the same base, meaning that <math>AC = DE = EG = GA</math>. Hence, we have that it is a square. To determine the area of this square, we can determine the length of its diagonals.<br />
<br />
In order to do this, we first determine the area of the octagon. Letting the side length be <math>a</math>, we can create a square of length <math>s</math> around it (see figure).<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
label("a",(1, 2.41421356)--(2.41421356, 1),S);<br />
draw((-2.41421356, 2.41421356)--(2.41421356, 2.41421356)--(2.41421356, -2.41421356)--(-2.41421356, -2.41421356)--cycle);<br />
label("s",(-2.41421356, 2.41421356)--(2.41421356, 2.41421356),N);<br />
</asy><br />
<br />
Creating a small square of side length <math>a</math> from the corners of this figure gives us an area of <math>a^2</math>. Thus, <math>s^2 - a^2 = n</math> where <math>n</math> is the area of the octagon. We know from the Pythagorean Theorem that <math>s = a + \frac{a}{\sqrt{2}}</math>, meaning that <math>n = (a + \frac{a}{\sqrt{2}})^2 - a^2 = 2a^2(1+\sqrt{2})</math>.<br />
<br />
Dividing this by <math>8</math> gives us the area of each triangular segment which makes up the octagon. Further dividing by <math>2</math> gives us the area of a smaller segment consisting of the right triangle with legs of the apothem and <math>\frac{a}{2}</math>. Using the area of a triangle as <math>\frac{1}{2}bh</math>, we can determine the length of apothem <math>r</math> from<br />
<br />
<cmath>\frac{2a^2(1+\sqrt{2})}{8 \times 2} = \frac{\frac{a}{2}\times r}{2}</cmath><br />
<cmath>\frac{4a^2(1+\sqrt{2})}{8 } = ar</cmath><br />
<cmath>\frac{a(1+\sqrt{2})}{2} = r</cmath>.<br />
<br />
From the apothem, we can once again use the Pythagorean Theorem, giving us the length of the circumradius <math>R</math>.<br />
<br />
<cmath>R^2 = (\frac{a(1+\sqrt{2})}{2})^2 + (\frac{a}{2})^2</cmath><br />
<cmath>R^2 = \frac{a^2(1+\sqrt{2})^2}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(3+2\sqrt{2})}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(4+2\sqrt{2})}{4}</cmath><br />
<cmath>R = \sqrt{\frac{a^2(4+2\sqrt{2})}{4}} = \frac{a\sqrt{4 + 2\sqrt2}}{2}</cmath>.<br />
<br />
Doubling this gives us the diagonal of both the square and the octagon. From here, we can use the formula <math>A = \frac{1}{2}d^2</math> for the area of the square:<br />
<br />
<cmath>A = \frac{(a\sqrt{4 + 2\sqrt2})^2}{2}</cmath><br />
<cmath>A = \frac{a^2(4+2\sqrt{2})}{2} = m</cmath>.<br />
<br />
Thus we now only need to find the ratio <math>\frac{m}{n}</math>. This can be easily done through some algebra:<br />
<br />
<cmath>\frac{m}{n} = \frac{a^2(4+2\sqrt{2})}{2(2a^2(1+\sqrt{2})}</cmath><br />
<cmath>\frac{m}{n} = \frac{4+2\sqrt{2}}{4+4\sqrt{2}}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
<br />
Rationalizing the denominator by multiplying by the conjugate, we get <math>\frac{m}{n} = \boxed{\textbf{(B) } \frac{\sqrt{2}}{2}}</math>. ~ciceronii<br />
<br />
*<math>\textbf{Note:}</math> this can more easily be done if you know any of these formulas. This was an entire derivation of the area of an octagon, it's apothem, and it's circumradius, but it can be much simpler if you have any of these memorized.<br />
<br />
==Solution 3 (Elementary Geometry)==<br />
WLOG, let <math>AE=1</math>. Let the intersection of <math>AF</math> and <math>GD</math> be point <math>I</math>. <math>GIF</math> is an isosceles right triangle (<math>m\angle HGF=135^{\circ}</math>), so <math>IG=IF=\frac{1}{\sqrt{2}}</math>. The distance between each side and the center is then <math>IF+\frac{1}{2}GH=\frac{1}{2}+\frac{1}{\sqrt{2}}</math>. <math>ABCDEFGH</math> is 8 triangles of base 1 and altitude <math>\frac{1}{2}+\frac{1}{\sqrt{2}}</math>, so its area is <math>8(\frac{1}{2})(\frac{1}{2}+\frac{1}{\sqrt{2}})</math> or <br />
<cmath>2+2\sqrt{2}</cmath><br />
Similarly, <math>ACEG</math> is clearly a square of area <math>GA^2</math>. By the Pythagorean Theorem, <math>GA^2=GI^2+IA^2=(\frac{1}{\sqrt{2}})^2+(1+\frac{1}{\sqrt{2}})^2</math> or <br />
<cmath>2+\sqrt{2}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
From some fast manipulations, see that it is <math>\boxed{\textbf{B)}\frac{\sqrt{2}}{{2}}}</math><br />
<br />
~~BJHHar<br />
<br />
==Solution 4 (Quick n' Easy)==<br />
This solution is quite efficient and unconvoluted. Though the explanation appears long, this is only because I have tried to make it very thorough. <br />
<br />
Because we are finding the ratios of the areas of the quadrilateral, and the octagon, let us assume, for simplicity's sake, that the octagon has side length 1. Call O the center of the octagon. Because it is a regular octagon, angle AOB is 360/8=45. <br />
Draw the apothem, and call where it intersects AB, I. As the apothem bisects angle AOB, angle AIB is 45/2=22.5. <br />
Then, as the apothem is perpendicular to AB, angle IBO is 90-22.5=67.5. As the apothem bisects AB, segment IB is 1/2. Thus, the apothem has length 1/2 * tan(67.5). By the formula area = apothem * perimeter / 2, the area of the octagon is then 1/2 * tan 67.5 * 8 / 2. <br />
<br />
Now, we move on to the quadrilateral. Let us take triangle AGH. By the interior angle formula, the measure of angle AGH is 180 (8-2) / 8 = 135. Because the octagon is regular, AGH is isosceles, with assumed sidelength 1. Drop an altitude from H to AG, and call the intersection point, P. As AGH is isosceles, this altitude bisects both segment AG and angle AHG. Thus, the measure of angle AHP is 135 / 2 = 67.5. Let us call the length of AG, which is also a side of the quadrilateral, s. Then, AP is s/2. Thus, sin(67.5) = (s/2)/1, so s = 2 * sin(67.5). We can do the same for each of the other sides of the quadrilateral and find that they are of the same length. One can easily see that triangle ABC is congruent to triangle AGH. Because triangle AGH is isosceles, the measure of angle HAG is (180 - 135)/2 = 22.5. The same can be said of angles BAC. Thus, angle CAG, an interior angle of the quadrilateral is 90. We can do the same for the other interior angles of CEGA and find that they are also 90. Thus, CEGA is a square, so its area is s^2 = 4 * sin^2(67.5). <br />
<br />
Now, we must find the ratios of the two areas. <br />
Area(quadrilateral)/Area(hexagon)=4*sin^2(67.5) / (2*tan(67.5)) = 2*sin^2(67.5) / (sin(67.5)/cos(67.5)) = 2*sin(67.5)*cos(67.5) = sin(135) = sqrt(2)/2, the answer.<br />
<br />
-jyatvitskiy :)<br />
<br />
==Solution 5 ==<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
draw((-1, 2.41421356)--(1,-2.41421356));<br />
draw((1, 2.41421356)--(-1,-2.41421356));<br />
draw((2.41421356, 1)--(-2.41421356,-1));<br />
draw((2.41421356, -1)--(-2.41421356,1));<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
label("O",(.1,.1),NE);<br />
label("P",(.60710678,1.60710678),SE);<br />
<br />
</asy><br />
<br />
The ratio <math>\frac{m}{n}=\frac{area OPC}{area OBC}=\frac{OP}{OB}=\frac{OP}{OC}=\frac{\sqrt {2}}{2}</math>.<br />
<br />
<<ERMS Coach>><br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Ermscoachhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12B_Problems/Problem_10&diff=1380472020 AMC 12B Problems/Problem 102020-11-21T02:12:18Z<p>Ermscoach: </p>
<hr />
<div>==Problem==<br />
<br />
In unit square <math>ABCD,</math> the inscribed circle <math>\omega</math> intersects <math>\overline{CD}</math> at <math>M,</math> and <math>\overline{AM}</math> intersects <math>\omega</math> at a point <math>P</math> different from <math>M.</math> What is <math>AP?</math><br />
<br />
<math>\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}</math><br />
<br />
==Solution 1 (Angle Chasing/Trig)==<br />
Let <math>O</math> be the center of the circle and the point of tangency between <math>\omega</math> and <math>\overline{AD}</math> be represented by <math>K</math>. We know that <math>\overline{AK} = \overline{KD} = \overline{DM} = \frac{1}{2}</math>. Consider the right triangle <math>\bigtriangleup ADM</math>. Let <math>\measuredangle AMD = \theta</math>. <br />
<br />
Since <math>\omega</math> is tangent to <math>\overline{DC}</math> at <math>M</math>, <math>\measuredangle PMO = 90 - \theta</math>. Now, consider <math>\bigtriangleup POM</math>. This triangle is iscoceles because <math>\overline{PO}</math> and <math>\overline{OM}</math> are both radii of <math>\omega</math>. Therefore, <math>\measuredangle POM = 180 - 2(90 - \theta) = 2\theta</math>. <br />
<br />
We can now use Law of Cosines on <math>\angle{POM}</math> to find the length of <math>{PM}</math> and subtract it from the length of <math>{AM}</math> to find <math>{AP}</math>. Since <math>\cos{\theta} = \frac{1}{\sqrt{5}}</math> and <math>\sin{\theta} = \frac{2}{\sqrt{5}}</math>, the double angle formula tells us that <math>\cos{2\theta} = -\frac{3}{5}</math>. We have<br />
<cmath><br />
PM^2 = \frac{1}{2} - \frac{1}{2}\cos{2\theta} \implies PM = \frac{2\sqrt{5}}{5}<br />
</cmath><br />
By Pythagorean theorem, we find that <math>AM = \frac{\sqrt{5}}{2} \implies \boxed{\textbf{(B) } \frac{\sqrt5}{10}}</math><br />
<br />
~awesome1st<br />
==Solution 2(Coordinate Bash)==<br />
<br />
Place circle <math>\omega</math> in the Cartesian plane such that the center lies on the origin. Then we can easily find the equation for <math>\omega</math> as <math>x^2+y^2=\frac{1}{4}</math>, because it is not translated and the radius is <math>\frac{1}{2}</math>. <br />
<br />
We have <math>A=\left(-\frac{1}{2}, \frac{1}{2}\right)</math> and <math>M=\left(0, -\frac{1}{2}\right)</math>. The slope of the line passing through these two points is <math>\frac{\frac{1}{2}+\frac{1}{2}}{-\frac{1}{2}-0}=\frac{1}{-\frac{1}{2}}=-2</math>, and the <math>y</math>-intercept is simply <math>M</math>. This gives us the line passing through both points as <math>y=-2x-\frac{1}{2}</math>. <br />
<br />
We substitute this into the equation for the circle to get <math>x^2+\left(-2x-\frac{1}{2}\right)^2=\frac{1}{4}</math>, or <math>x^2+4x^2+2x+\frac{1}{4}=\frac{1}{4}</math>. Simplifying gives <math>x(5x+2)=0</math>. The roots of this quadratic are <math>x=0</math> and <math>x=-\frac{2}{5}</math>, but if <math>x=0</math> we get point <math>M</math>, so we only want <math>x=-\frac{2}{5}</math>.<br />
<br />
We plug this back into the linear equation to find <math>y=\frac{3}{10}</math>, and so <math>P=\left(-\frac{2}{5}, \frac{3}{10}\right)</math>. Finally, we use distance formula on <math>A</math> and <math>P</math> to get <math>AP=\sqrt{\left(-\frac{5}{10}+\frac{4}{10}\right)^2+\left(\frac{5}{10}-\frac{3}{10}\right)^2}=\sqrt{\frac{1}{100}+\frac{4}{100}}=\boxed{\mathbf{(B) } \frac{\sqrt{5}}{10}}</math>.<br />
<br />
~Argonauts16<br />
<br />
==Solution 3(Power of a Point)==<br />
<br />
Let circle <math>\omega</math> intersect <math>\overline{AB}</math> at point <math>N</math>. By Power of a Point, we have <math>AN^2=AP\cdot AM</math>. We know <math>AN=\frac{1}{2}</math> because <math>N</math> is the midpoint of <math>\overline{AB}</math>, and we can easily find <math>AM</math> by the Pythagorean Theorem, which gives us <math>AM=\sqrt{1^2+\left(\frac{1}{2}\right)^2}=\frac{\sqrt{5}}{2}</math>. Our equation is now <math>\frac{1}{4}=AP\cdot \frac{\sqrt{5}}{2}</math>, or <math>AP=\frac{2}{4\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{2\cdot 5}</math>, thus our answer is <br />
<math>\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.</math><br />
<br />
~Argonauts16<br />
<br />
==Solution 4==<br />
Take <math>O</math> as the center and draw segment <math>ON</math> perpendicular to <math>AM</math>, <math>ON\cap AM=N</math>, link <math>OM</math>. Then we have <math>OM\parallel AD</math>. So <math>\angle DAM=\angle OMA</math>. Since <math>AD=2AM=2OM=1</math>, we have <math>\cos\angle DAM=\cos\angle OMP=\frac{2}{\sqrt{5}}</math>. As a result, <math>NM=OM\cos\angle OMP=\frac{1}{2}\cdot \frac{2}{\sqrt{5}}=\frac{1}{\sqrt{5}}.</math> Thus <math>PM=2NM=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}</math>. Since <math>AM=\frac{\sqrt{5}}{2}</math>, we have <math>AP=AM-PM=\frac{\sqrt{5}}{10}</math>. Put <math>\boxed{B}</math>.<br />
<br />
~FANYUCHEN20020715<br />
<br />
==Solution 5 (Similar Triangles)==<br />
Call the midpoint of <math>\overline{AB}</math> point <math>N</math>. Draw in <math>\overline{NM}</math> and <math>\overline{NP}</math>. Note that <math>\angle{NPM}=90^{\circ}</math> due to Thales's Theorem.<br />
<br />
<asy><br />
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);<br />
draw(circle((0.5,0.5),0.5));<br />
draw((0,0)--(0,0.5)--(1,0.5)--cycle);<br />
label("A",(0,0),SW);<br />
label("B",(0,1),NW);<br />
label("C",(1,1),NE);<br />
label("D",(1,0),SE);<br />
label("M",(1,0.5),E);<br />
label("P",(0.2,0.1),S);<br />
label("N",(0,0.5),W);<br />
draw((0,0.5)--(0.2,0.1));<br />
markscalefactor=0.007;<br />
draw(rightanglemark((0,0.5),(0.2,0.1),(1,0.5)));<br />
</asy><br />
Using the Pythagorean theorem, <math>AM=\frac{\sqrt{5}}{2}</math>. Now we just need to find <math>AP</math> using similar triangles.<br />
<cmath>\triangle APN\sim\triangle ANM\Rightarrow\frac{AP}{AN}=\frac{AN}{AM}\Rightarrow\frac{AP}{\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}}\Rightarrow AP=\boxed{\textbf{(B) }\frac{\sqrt{5}}{10}}</cmath><br />
~QIDb602<br />
<br />
==Video Solution==<br />
https://youtu.be/6ujfjGLzVoE<br />
<br />
~IceMatrix<br />
<br />
==Solution 6 intersecting chords==<br />
Label the midpoint of <math>AD</math> as <math>N</math>, and let <math>Q</math> be the intersection of <math>ON</math> and <math>AM</math><br />
Then <math>MQ=AQ=\frac{1}{2}AM = \frac{\sqrt{5}}{4}</math> and <math>NQ=\frac{1}{4}</math><br />
<br />
Then <math>\frac{1}{4}*\frac{3}{4}=PQ*QM</math> and <math>AP=AQ-PQ</math><br />
<br />
~ERMSCoach<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=B|num-b=9|num-a=11}}<br />
{{MAA Notice}}</div>Ermscoachhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_14&diff=1379082020 AMC 12A Problems/Problem 142020-11-20T02:03:40Z<p>Ermscoach: /* Solution 5 */</p>
<hr />
<div>==Problem 14==<br />
Regular octagon <math>ABCDEFGH</math> has area <math>n</math>. Let <math>m</math> be the area of quadrilateral <math>ACEG</math>. What is <math>\tfrac{m}{n}?</math><br />
<br />
<math>\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}</math><br />
<br />
==Solution 1==<br />
<math>ACEG</math> is a square. WLOG <math>AB = 1,</math> then using Law of Cosines, <math>AC^2 = [ACEG] = 1^2 + 1^2 - 2 \cos{135} = 2 + \sqrt{2}.</math> The area of the octagon is just <math>[ACEG]</math> plus the area of the four congruent (by symmetry) isosceles triangles, all an angle of <math>135</math> in between two sides of length 1. Now, <cmath>\dfrac{m}{n} = \dfrac{2 + \sqrt{2}}{2 + \sqrt{2} + 4 \cdot \tfrac{1}{2} \sin{135}} = \dfrac{2 + \sqrt{2}}{2 + 2 \sqrt{2}} = \dfrac{\sqrt{2}}{2}.</cmath> The answer is <math>\boxed{\textbf{(B)}}.</math><br />
<br />
==Solution 2==<br />
Refer to the diagram. Call one of the side lengths of the square <math>s</math>. Since quadrilateral ACEG is a square, the area of the square would just be <math>s^2</math>, which we can find by applying Law of Cosines on one of the four triangles. Assume each of the sides of the octagon has length <math>1</math>. Since each angle measures <math>135^{\circ}</math> in an octagon, then <math>s^2 = 1^2+1^2-2*\cos(135^{\circ}) = 2+\sqrt{2}</math> <br />
<br />
There are many ways to find the area of the octagon, but one way is to split the octagon into two trapezoids and one rectangle. We can easily compute AF to be <math>1+\sqrt{2}</math> from splitting one of the sides into two <math>45-45-90</math> triangles. So the area of the octagon is <math>2*\frac{1+\sqrt{2}}{2}+1+\sqrt{2} \Rightarrow 2+2\sqrt{2}</math>. <br />
<br />
Hence, <cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath> <br />
<cmath>\Rightarrow \frac{2+\sqrt{2}}{2+2\sqrt{2}} \cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}}</cmath><br />
<cmath>\Rightarrow \frac{-2\sqrt{2}}{-4}</cmath><br />
<cmath>\Rightarrow \boxed{\frac{\sqrt{2}}{2}}</cmath><br />
<br />
~Solution by IronicNinja<br />
<br />
== Solution 3 (Deriving Formulas) ==<br />
<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
</asy><br />
<br />
The first thing to notice is that <math>ACEG</math> is a square. This is because, as <math>\bigtriangleup ABC \cong \bigtriangleup CDE \cong \bigtriangleup EFG \cong \bigtriangleup GHE</math>, they all have the same base, meaning that <math>AC = DE = EG = GA</math>. Hence, we have that it is a square. To determine the area of this square, we can determine the length of its diagonals.<br />
<br />
In order to do this, we first determine the area of the octagon. Letting the side length be <math>a</math>, we can create a square of length <math>s</math> around it (see figure).<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
label("a",(1, 2.41421356)--(2.41421356, 1),S);<br />
draw((-2.41421356, 2.41421356)--(2.41421356, 2.41421356)--(2.41421356, -2.41421356)--(-2.41421356, -2.41421356)--cycle);<br />
label("s",(-2.41421356, 2.41421356)--(2.41421356, 2.41421356),N);<br />
</asy><br />
<br />
Creating a small square of side length <math>a</math> from the corners of this figure gives us an area of <math>a^2</math>. Thus, <math>s^2 - a^2 = n</math> where <math>n</math> is the area of the octagon. We know from the Pythagorean Theorem that <math>s = a + \frac{a}{\sqrt{2}}</math>, meaning that <math>n = (a + \frac{a}{\sqrt{2}})^2 - a^2 = 2a^2(1+\sqrt{2})</math>.<br />
<br />
Dividing this by <math>8</math> gives us the area of each triangular segment which makes up the octagon. Further dividing by <math>2</math> gives us the area of a smaller segment consisting of the right triangle with legs of the apothem and <math>\frac{a}{2}</math>. Using the area of a triangle as <math>\frac{1}{2}bh</math>, we can determine the length of apothem <math>r</math> from<br />
<br />
<cmath>\frac{2a^2(1+\sqrt{2})}{8 \times 2} = \frac{\frac{a}{2}\times r}{2}</cmath><br />
<cmath>\frac{4a^2(1+\sqrt{2})}{8 } = ar</cmath><br />
<cmath>\frac{a(1+\sqrt{2})}{2} = r</cmath>.<br />
<br />
From the apothem, we can once again use the Pythagorean Theorem, giving us the length of the circumradius <math>R</math>.<br />
<br />
<cmath>R^2 = (\frac{a(1+\sqrt{2})}{2})^2 + (\frac{a}{2})^2</cmath><br />
<cmath>R^2 = \frac{a^2(1+\sqrt{2})^2}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(3+2\sqrt{2})}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(4+2\sqrt{2})}{4}</cmath><br />
<cmath>R = \sqrt{\frac{a^2(4+2\sqrt{2})}{4}} = \frac{a\sqrt{4 + 2\sqrt2}}{2}</cmath>.<br />
<br />
Doubling this gives us the diagonal of both the square and the octagon. From here, we can use the formula <math>A = \frac{1}{2}d^2</math> for the area of the square:<br />
<br />
<cmath>A = \frac{(a\sqrt{4 + 2\sqrt2})^2}{2}</cmath><br />
<cmath>A = \frac{a^2(4+2\sqrt{2})}{2} = m</cmath>.<br />
<br />
Thus we now only need to find the ratio <math>\frac{m}{n}</math>. This can be easily done through some algebra:<br />
<br />
<cmath>\frac{m}{n} = \frac{a^2(4+2\sqrt{2})}{2(2a^2(1+\sqrt{2})}</cmath><br />
<cmath>\frac{m}{n} = \frac{4+2\sqrt{2}}{4+4\sqrt{2}}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
<br />
Rationalizing the denominator by multiplying by the conjugate, we get <math>\frac{m}{n} = \boxed{\textbf{(B) } \frac{\sqrt{2}}{2}}</math>. ~ciceronii<br />
<br />
*<math>\textbf{Note:}</math> this can more easily be done if you know any of these formulas. This was an entire derivation of the area of an octagon, it's apothem, and it's circumradius, but it can be much simpler if you have any of these memorized.<br />
<br />
==Solution 3 (Elementary Geometry)==<br />
WLOG, let <math>AE=1</math>. Let the intersection of <math>AF</math> and <math>GD</math> be point <math>I</math>. <math>GIF</math> is an isosceles right triangle (<math>m\angle HGF=135^{\circ}</math>), so <math>IG=IF=\frac{1}{\sqrt{2}}</math>. The distance between each side and the center is then <math>IF+\frac{1}{2}GH=\frac{1}{2}+\frac{1}{\sqrt{2}}</math>. <math>ABCDEFGH</math> is 8 triangles of base 1 and altitude <math>\frac{1}{2}+\frac{1}{\sqrt{2}}</math>, so its area is <math>8(\frac{1}{2})(\frac{1}{2}+\frac{1}{\sqrt{2}})</math> or <br />
<cmath>2+2\sqrt{2}</cmath><br />
Similarly, <math>ACEG</math> is clearly a square of area <math>GA^2</math>. By the Pythagorean Theorem, <math>GA^2=GI^2+IA^2=(\frac{1}{\sqrt{2}})^2+(1+\frac{1}{\sqrt{2}})^2</math> or <br />
<cmath>2+\sqrt{2}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
From some fast manipulations, see that it is <math>\boxed{\textbf{B)}\frac{\sqrt{2}}{{2}}}</math><br />
<br />
~~BJHHar<br />
<br />
==Solution 4 (Quick n' Easy)==<br />
This solution is quite efficient and unconvoluted. Though the explanation appears long, this is only because I have tried to make it very thorough. <br />
<br />
Because we are finding the ratios of the areas of the quadrilateral, and the octagon, let us assume, for simplicity's sake, that the octagon has side length 1. Call O the center of the octagon. Because it is a regular octagon, angle AOB is 360/8=45. <br />
Draw the apothem, and call where it intersects AB, I. As the apothem bisects angle AOB, angle AIB is 45/2=22.5. <br />
Then, as the apothem is perpendicular to AB, angle IBO is 90-22.5=67.5. As the apothem bisects AB, segment IB is 1/2. Thus, the apothem has length 1/2 * tan(67.5). By the formula area = apothem * perimeter / 2, the area of the octagon is then 1/2 * tan 67.5 * 8 / 2. <br />
<br />
Now, we move on to the quadrilateral. Let us take triangle AGH. By the interior angle formula, the measure of angle AGH is 180 (8-2) / 8 = 135. Because the octagon is regular, AGH is isosceles, with assumed sidelength 1. Drop an altitude from H to AG, and call the intersection point, P. As AGH is isosceles, this altitude bisects both segment AG and angle AHG. Thus, the measure of angle AHP is 135 / 2 = 67.5. Let us call the length of AG, which is also a side of the quadrilateral, s. Then, AP is s/2. Thus, sin(67.5) = (s/2)/1, so s = 2 * sin(67.5). We can do the same for each of the other sides of the quadrilateral and find that they are of the same length. One can easily see that triangle ABC is congruent to triangle AGH. Because triangle AGH is isosceles, the measure of angle HAG is (180 - 135)/2 = 22.5. The same can be said of angles BAC. Thus, angle CAG, an interior angle of the quadrilateral is 90. We can do the same for the other interior angles of CEGA and find that they are also 90. Thus, CEGA is a square, so its area is s^2 = 4 * sin^2(67.5). <br />
<br />
Now, we must find the ratios of the two areas. <br />
Area(quadrilateral)/Area(hexagon)=4*sin^2(67.5) / (2*tan(67.5)) = 2*sin^2(67.5) / (sin(67.5)/cos(67.5)) = 2*sin(67.5)*cos(67.5) = sin(135) = sqrt(2)/2, the answer.<br />
<br />
-jyatvitskiy :)<br />
<br />
==Solution 5 ==<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
draw((-1, 2.41421356)--(1,-2.41421356));<br />
draw((1, 2.41421356)--(-1,-2.41421356));<br />
draw((2.41421356, 1)--(-2.41421356,-1));<br />
draw((2.41421356, -1)--(-2.41421356,1));<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
label("O",(.1,.1),NE);<br />
label("P",(.60710678,1.60710678),SE);<br />
<br />
</asy><br />
<br />
The ratio of <math>m</math> to <math>n</math> is the ratio of <math>OPC</math> to <math>OBC</math> which is the ratio of <math>OP</math> to <math>OB</math> which is <math>\frac{\sqrt {2}}{2}</math>.<br />
<br />
<<ERMS Coach>><br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Ermscoachhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_14&diff=1379072020 AMC 12A Problems/Problem 142020-11-20T01:48:55Z<p>Ermscoach: /* Solution 5 */</p>
<hr />
<div>==Problem 14==<br />
Regular octagon <math>ABCDEFGH</math> has area <math>n</math>. Let <math>m</math> be the area of quadrilateral <math>ACEG</math>. What is <math>\tfrac{m}{n}?</math><br />
<br />
<math>\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}</math><br />
<br />
==Solution 1==<br />
<math>ACEG</math> is a square. WLOG <math>AB = 1,</math> then using Law of Cosines, <math>AC^2 = [ACEG] = 1^2 + 1^2 - 2 \cos{135} = 2 + \sqrt{2}.</math> The area of the octagon is just <math>[ACEG]</math> plus the area of the four congruent (by symmetry) isosceles triangles, all an angle of <math>135</math> in between two sides of length 1. Now, <cmath>\dfrac{m}{n} = \dfrac{2 + \sqrt{2}}{2 + \sqrt{2} + 4 \cdot \tfrac{1}{2} \sin{135}} = \dfrac{2 + \sqrt{2}}{2 + 2 \sqrt{2}} = \dfrac{\sqrt{2}}{2}.</cmath> The answer is <math>\boxed{\textbf{(B)}}.</math><br />
<br />
==Solution 2==<br />
Refer to the diagram. Call one of the side lengths of the square <math>s</math>. Since quadrilateral ACEG is a square, the area of the square would just be <math>s^2</math>, which we can find by applying Law of Cosines on one of the four triangles. Assume each of the sides of the octagon has length <math>1</math>. Since each angle measures <math>135^{\circ}</math> in an octagon, then <math>s^2 = 1^2+1^2-2*\cos(135^{\circ}) = 2+\sqrt{2}</math> <br />
<br />
There are many ways to find the area of the octagon, but one way is to split the octagon into two trapezoids and one rectangle. We can easily compute AF to be <math>1+\sqrt{2}</math> from splitting one of the sides into two <math>45-45-90</math> triangles. So the area of the octagon is <math>2*\frac{1+\sqrt{2}}{2}+1+\sqrt{2} \Rightarrow 2+2\sqrt{2}</math>. <br />
<br />
Hence, <cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath> <br />
<cmath>\Rightarrow \frac{2+\sqrt{2}}{2+2\sqrt{2}} \cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}}</cmath><br />
<cmath>\Rightarrow \frac{-2\sqrt{2}}{-4}</cmath><br />
<cmath>\Rightarrow \boxed{\frac{\sqrt{2}}{2}}</cmath><br />
<br />
~Solution by IronicNinja<br />
<br />
== Solution 3 (Deriving Formulas) ==<br />
<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
</asy><br />
<br />
The first thing to notice is that <math>ACEG</math> is a square. This is because, as <math>\bigtriangleup ABC \cong \bigtriangleup CDE \cong \bigtriangleup EFG \cong \bigtriangleup GHE</math>, they all have the same base, meaning that <math>AC = DE = EG = GA</math>. Hence, we have that it is a square. To determine the area of this square, we can determine the length of its diagonals.<br />
<br />
In order to do this, we first determine the area of the octagon. Letting the side length be <math>a</math>, we can create a square of length <math>s</math> around it (see figure).<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
label("a",(1, 2.41421356)--(2.41421356, 1),S);<br />
draw((-2.41421356, 2.41421356)--(2.41421356, 2.41421356)--(2.41421356, -2.41421356)--(-2.41421356, -2.41421356)--cycle);<br />
label("s",(-2.41421356, 2.41421356)--(2.41421356, 2.41421356),N);<br />
</asy><br />
<br />
Creating a small square of side length <math>a</math> from the corners of this figure gives us an area of <math>a^2</math>. Thus, <math>s^2 - a^2 = n</math> where <math>n</math> is the area of the octagon. We know from the Pythagorean Theorem that <math>s = a + \frac{a}{\sqrt{2}}</math>, meaning that <math>n = (a + \frac{a}{\sqrt{2}})^2 - a^2 = 2a^2(1+\sqrt{2})</math>.<br />
<br />
Dividing this by <math>8</math> gives us the area of each triangular segment which makes up the octagon. Further dividing by <math>2</math> gives us the area of a smaller segment consisting of the right triangle with legs of the apothem and <math>\frac{a}{2}</math>. Using the area of a triangle as <math>\frac{1}{2}bh</math>, we can determine the length of apothem <math>r</math> from<br />
<br />
<cmath>\frac{2a^2(1+\sqrt{2})}{8 \times 2} = \frac{\frac{a}{2}\times r}{2}</cmath><br />
<cmath>\frac{4a^2(1+\sqrt{2})}{8 } = ar</cmath><br />
<cmath>\frac{a(1+\sqrt{2})}{2} = r</cmath>.<br />
<br />
From the apothem, we can once again use the Pythagorean Theorem, giving us the length of the circumradius <math>R</math>.<br />
<br />
<cmath>R^2 = (\frac{a(1+\sqrt{2})}{2})^2 + (\frac{a}{2})^2</cmath><br />
<cmath>R^2 = \frac{a^2(1+\sqrt{2})^2}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(3+2\sqrt{2})}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(4+2\sqrt{2})}{4}</cmath><br />
<cmath>R = \sqrt{\frac{a^2(4+2\sqrt{2})}{4}} = \frac{a\sqrt{4 + 2\sqrt2}}{2}</cmath>.<br />
<br />
Doubling this gives us the diagonal of both the square and the octagon. From here, we can use the formula <math>A = \frac{1}{2}d^2</math> for the area of the square:<br />
<br />
<cmath>A = \frac{(a\sqrt{4 + 2\sqrt2})^2}{2}</cmath><br />
<cmath>A = \frac{a^2(4+2\sqrt{2})}{2} = m</cmath>.<br />
<br />
Thus we now only need to find the ratio <math>\frac{m}{n}</math>. This can be easily done through some algebra:<br />
<br />
<cmath>\frac{m}{n} = \frac{a^2(4+2\sqrt{2})}{2(2a^2(1+\sqrt{2})}</cmath><br />
<cmath>\frac{m}{n} = \frac{4+2\sqrt{2}}{4+4\sqrt{2}}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
<br />
Rationalizing the denominator by multiplying by the conjugate, we get <math>\frac{m}{n} = \boxed{\textbf{(B) } \frac{\sqrt{2}}{2}}</math>. ~ciceronii<br />
<br />
*<math>\textbf{Note:}</math> this can more easily be done if you know any of these formulas. This was an entire derivation of the area of an octagon, it's apothem, and it's circumradius, but it can be much simpler if you have any of these memorized.<br />
<br />
==Solution 3 (Elementary Geometry)==<br />
WLOG, let <math>AE=1</math>. Let the intersection of <math>AF</math> and <math>GD</math> be point <math>I</math>. <math>GIF</math> is an isosceles right triangle (<math>m\angle HGF=135^{\circ}</math>), so <math>IG=IF=\frac{1}{\sqrt{2}}</math>. The distance between each side and the center is then <math>IF+\frac{1}{2}GH=\frac{1}{2}+\frac{1}{\sqrt{2}}</math>. <math>ABCDEFGH</math> is 8 triangles of base 1 and altitude <math>\frac{1}{2}+\frac{1}{\sqrt{2}}</math>, so its area is <math>8(\frac{1}{2})(\frac{1}{2}+\frac{1}{\sqrt{2}})</math> or <br />
<cmath>2+2\sqrt{2}</cmath><br />
Similarly, <math>ACEG</math> is clearly a square of area <math>GA^2</math>. By the Pythagorean Theorem, <math>GA^2=GI^2+IA^2=(\frac{1}{\sqrt{2}})^2+(1+\frac{1}{\sqrt{2}})^2</math> or <br />
<cmath>2+\sqrt{2}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
From some fast manipulations, see that it is <math>\boxed{\textbf{B)}\frac{\sqrt{2}}{{2}}}</math><br />
<br />
~~BJHHar<br />
<br />
==Solution 4 (Quick n' Easy)==<br />
This solution is quite efficient and unconvoluted. Though the explanation appears long, this is only because I have tried to make it very thorough. <br />
<br />
Because we are finding the ratios of the areas of the quadrilateral, and the octagon, let us assume, for simplicity's sake, that the octagon has side length 1. Call O the center of the octagon. Because it is a regular octagon, angle AOB is 360/8=45. <br />
Draw the apothem, and call where it intersects AB, I. As the apothem bisects angle AOB, angle AIB is 45/2=22.5. <br />
Then, as the apothem is perpendicular to AB, angle IBO is 90-22.5=67.5. As the apothem bisects AB, segment IB is 1/2. Thus, the apothem has length 1/2 * tan(67.5). By the formula area = apothem * perimeter / 2, the area of the octagon is then 1/2 * tan 67.5 * 8 / 2. <br />
<br />
Now, we move on to the quadrilateral. Let us take triangle AGH. By the interior angle formula, the measure of angle AGH is 180 (8-2) / 8 = 135. Because the octagon is regular, AGH is isosceles, with assumed sidelength 1. Drop an altitude from H to AG, and call the intersection point, P. As AGH is isosceles, this altitude bisects both segment AG and angle AHG. Thus, the measure of angle AHP is 135 / 2 = 67.5. Let us call the length of AG, which is also a side of the quadrilateral, s. Then, AP is s/2. Thus, sin(67.5) = (s/2)/1, so s = 2 * sin(67.5). We can do the same for each of the other sides of the quadrilateral and find that they are of the same length. One can easily see that triangle ABC is congruent to triangle AGH. Because triangle AGH is isosceles, the measure of angle HAG is (180 - 135)/2 = 22.5. The same can be said of angles BAC. Thus, angle CAG, an interior angle of the quadrilateral is 90. We can do the same for the other interior angles of CEGA and find that they are also 90. Thus, CEGA is a square, so its area is s^2 = 4 * sin^2(67.5). <br />
<br />
Now, we must find the ratios of the two areas. <br />
Area(quadrilateral)/Area(hexagon)=4*sin^2(67.5) / (2*tan(67.5)) = 2*sin^2(67.5) / (sin(67.5)/cos(67.5)) = 2*sin(67.5)*cos(67.5) = sin(135) = sqrt(2)/2, the answer.<br />
<br />
-jyatvitskiy :)<br />
<br />
==Solution 5 ==<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
draw((-1, 2.41421356)--(1,-2.41421356));<br />
draw((1, 2.41421356)--(-1,-2.41421356));<br />
draw((2.41421356, 1)--(-2.41421356,-1));<br />
draw((2.41421356, -1)--(-2.41421356,1));<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
label("O",(0,0),NE);<br />
label("P",(.70710678,1.70710678),SE);<br />
<br />
</asy><br />
<br />
The ratio of <math>m</math> to <math>n</math> is the ratio of <math>OPC</math> to <math>OBC</math> which is the ratio of <math>OP</math> to <math>OB</math> which is <math>\frac{\sqrt {2}}{2}</math>.<br />
<br />
<<ERMS Coach>><br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Ermscoachhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_14&diff=1379052020 AMC 12A Problems/Problem 142020-11-20T01:46:14Z<p>Ermscoach: /* Solution 5 */</p>
<hr />
<div>==Problem 14==<br />
Regular octagon <math>ABCDEFGH</math> has area <math>n</math>. Let <math>m</math> be the area of quadrilateral <math>ACEG</math>. What is <math>\tfrac{m}{n}?</math><br />
<br />
<math>\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}</math><br />
<br />
==Solution 1==<br />
<math>ACEG</math> is a square. WLOG <math>AB = 1,</math> then using Law of Cosines, <math>AC^2 = [ACEG] = 1^2 + 1^2 - 2 \cos{135} = 2 + \sqrt{2}.</math> The area of the octagon is just <math>[ACEG]</math> plus the area of the four congruent (by symmetry) isosceles triangles, all an angle of <math>135</math> in between two sides of length 1. Now, <cmath>\dfrac{m}{n} = \dfrac{2 + \sqrt{2}}{2 + \sqrt{2} + 4 \cdot \tfrac{1}{2} \sin{135}} = \dfrac{2 + \sqrt{2}}{2 + 2 \sqrt{2}} = \dfrac{\sqrt{2}}{2}.</cmath> The answer is <math>\boxed{\textbf{(B)}}.</math><br />
<br />
==Solution 2==<br />
Refer to the diagram. Call one of the side lengths of the square <math>s</math>. Since quadrilateral ACEG is a square, the area of the square would just be <math>s^2</math>, which we can find by applying Law of Cosines on one of the four triangles. Assume each of the sides of the octagon has length <math>1</math>. Since each angle measures <math>135^{\circ}</math> in an octagon, then <math>s^2 = 1^2+1^2-2*\cos(135^{\circ}) = 2+\sqrt{2}</math> <br />
<br />
There are many ways to find the area of the octagon, but one way is to split the octagon into two trapezoids and one rectangle. We can easily compute AF to be <math>1+\sqrt{2}</math> from splitting one of the sides into two <math>45-45-90</math> triangles. So the area of the octagon is <math>2*\frac{1+\sqrt{2}}{2}+1+\sqrt{2} \Rightarrow 2+2\sqrt{2}</math>. <br />
<br />
Hence, <cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath> <br />
<cmath>\Rightarrow \frac{2+\sqrt{2}}{2+2\sqrt{2}} \cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}}</cmath><br />
<cmath>\Rightarrow \frac{-2\sqrt{2}}{-4}</cmath><br />
<cmath>\Rightarrow \boxed{\frac{\sqrt{2}}{2}}</cmath><br />
<br />
~Solution by IronicNinja<br />
<br />
== Solution 3 (Deriving Formulas) ==<br />
<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
</asy><br />
<br />
The first thing to notice is that <math>ACEG</math> is a square. This is because, as <math>\bigtriangleup ABC \cong \bigtriangleup CDE \cong \bigtriangleup EFG \cong \bigtriangleup GHE</math>, they all have the same base, meaning that <math>AC = DE = EG = GA</math>. Hence, we have that it is a square. To determine the area of this square, we can determine the length of its diagonals.<br />
<br />
In order to do this, we first determine the area of the octagon. Letting the side length be <math>a</math>, we can create a square of length <math>s</math> around it (see figure).<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
label("a",(1, 2.41421356)--(2.41421356, 1),S);<br />
draw((-2.41421356, 2.41421356)--(2.41421356, 2.41421356)--(2.41421356, -2.41421356)--(-2.41421356, -2.41421356)--cycle);<br />
label("s",(-2.41421356, 2.41421356)--(2.41421356, 2.41421356),N);<br />
</asy><br />
<br />
Creating a small square of side length <math>a</math> from the corners of this figure gives us an area of <math>a^2</math>. Thus, <math>s^2 - a^2 = n</math> where <math>n</math> is the area of the octagon. We know from the Pythagorean Theorem that <math>s = a + \frac{a}{\sqrt{2}}</math>, meaning that <math>n = (a + \frac{a}{\sqrt{2}})^2 - a^2 = 2a^2(1+\sqrt{2})</math>.<br />
<br />
Dividing this by <math>8</math> gives us the area of each triangular segment which makes up the octagon. Further dividing by <math>2</math> gives us the area of a smaller segment consisting of the right triangle with legs of the apothem and <math>\frac{a}{2}</math>. Using the area of a triangle as <math>\frac{1}{2}bh</math>, we can determine the length of apothem <math>r</math> from<br />
<br />
<cmath>\frac{2a^2(1+\sqrt{2})}{8 \times 2} = \frac{\frac{a}{2}\times r}{2}</cmath><br />
<cmath>\frac{4a^2(1+\sqrt{2})}{8 } = ar</cmath><br />
<cmath>\frac{a(1+\sqrt{2})}{2} = r</cmath>.<br />
<br />
From the apothem, we can once again use the Pythagorean Theorem, giving us the length of the circumradius <math>R</math>.<br />
<br />
<cmath>R^2 = (\frac{a(1+\sqrt{2})}{2})^2 + (\frac{a}{2})^2</cmath><br />
<cmath>R^2 = \frac{a^2(1+\sqrt{2})^2}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(3+2\sqrt{2})}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(4+2\sqrt{2})}{4}</cmath><br />
<cmath>R = \sqrt{\frac{a^2(4+2\sqrt{2})}{4}} = \frac{a\sqrt{4 + 2\sqrt2}}{2}</cmath>.<br />
<br />
Doubling this gives us the diagonal of both the square and the octagon. From here, we can use the formula <math>A = \frac{1}{2}d^2</math> for the area of the square:<br />
<br />
<cmath>A = \frac{(a\sqrt{4 + 2\sqrt2})^2}{2}</cmath><br />
<cmath>A = \frac{a^2(4+2\sqrt{2})}{2} = m</cmath>.<br />
<br />
Thus we now only need to find the ratio <math>\frac{m}{n}</math>. This can be easily done through some algebra:<br />
<br />
<cmath>\frac{m}{n} = \frac{a^2(4+2\sqrt{2})}{2(2a^2(1+\sqrt{2})}</cmath><br />
<cmath>\frac{m}{n} = \frac{4+2\sqrt{2}}{4+4\sqrt{2}}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
<br />
Rationalizing the denominator by multiplying by the conjugate, we get <math>\frac{m}{n} = \boxed{\textbf{(B) } \frac{\sqrt{2}}{2}}</math>. ~ciceronii<br />
<br />
*<math>\textbf{Note:}</math> this can more easily be done if you know any of these formulas. This was an entire derivation of the area of an octagon, it's apothem, and it's circumradius, but it can be much simpler if you have any of these memorized.<br />
<br />
==Solution 3 (Elementary Geometry)==<br />
WLOG, let <math>AE=1</math>. Let the intersection of <math>AF</math> and <math>GD</math> be point <math>I</math>. <math>GIF</math> is an isosceles right triangle (<math>m\angle HGF=135^{\circ}</math>), so <math>IG=IF=\frac{1}{\sqrt{2}}</math>. The distance between each side and the center is then <math>IF+\frac{1}{2}GH=\frac{1}{2}+\frac{1}{\sqrt{2}}</math>. <math>ABCDEFGH</math> is 8 triangles of base 1 and altitude <math>\frac{1}{2}+\frac{1}{\sqrt{2}}</math>, so its area is <math>8(\frac{1}{2})(\frac{1}{2}+\frac{1}{\sqrt{2}})</math> or <br />
<cmath>2+2\sqrt{2}</cmath><br />
Similarly, <math>ACEG</math> is clearly a square of area <math>GA^2</math>. By the Pythagorean Theorem, <math>GA^2=GI^2+IA^2=(\frac{1}{\sqrt{2}})^2+(1+\frac{1}{\sqrt{2}})^2</math> or <br />
<cmath>2+\sqrt{2}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
From some fast manipulations, see that it is <math>\boxed{\textbf{B)}\frac{\sqrt{2}}{{2}}}</math><br />
<br />
~~BJHHar<br />
<br />
==Solution 4 (Quick n' Easy)==<br />
This solution is quite efficient and unconvoluted. Though the explanation appears long, this is only because I have tried to make it very thorough. <br />
<br />
Because we are finding the ratios of the areas of the quadrilateral, and the octagon, let us assume, for simplicity's sake, that the octagon has side length 1. Call O the center of the octagon. Because it is a regular octagon, angle AOB is 360/8=45. <br />
Draw the apothem, and call where it intersects AB, I. As the apothem bisects angle AOB, angle AIB is 45/2=22.5. <br />
Then, as the apothem is perpendicular to AB, angle IBO is 90-22.5=67.5. As the apothem bisects AB, segment IB is 1/2. Thus, the apothem has length 1/2 * tan(67.5). By the formula area = apothem * perimeter / 2, the area of the octagon is then 1/2 * tan 67.5 * 8 / 2. <br />
<br />
Now, we move on to the quadrilateral. Let us take triangle AGH. By the interior angle formula, the measure of angle AGH is 180 (8-2) / 8 = 135. Because the octagon is regular, AGH is isosceles, with assumed sidelength 1. Drop an altitude from H to AG, and call the intersection point, P. As AGH is isosceles, this altitude bisects both segment AG and angle AHG. Thus, the measure of angle AHP is 135 / 2 = 67.5. Let us call the length of AG, which is also a side of the quadrilateral, s. Then, AP is s/2. Thus, sin(67.5) = (s/2)/1, so s = 2 * sin(67.5). We can do the same for each of the other sides of the quadrilateral and find that they are of the same length. One can easily see that triangle ABC is congruent to triangle AGH. Because triangle AGH is isosceles, the measure of angle HAG is (180 - 135)/2 = 22.5. The same can be said of angles BAC. Thus, angle CAG, an interior angle of the quadrilateral is 90. We can do the same for the other interior angles of CEGA and find that they are also 90. Thus, CEGA is a square, so its area is s^2 = 4 * sin^2(67.5). <br />
<br />
Now, we must find the ratios of the two areas. <br />
Area(quadrilateral)/Area(hexagon)=4*sin^2(67.5) / (2*tan(67.5)) = 2*sin^2(67.5) / (sin(67.5)/cos(67.5)) = 2*sin(67.5)*cos(67.5) = sin(135) = sqrt(2)/2, the answer.<br />
<br />
-jyatvitskiy :)<br />
<br />
==Solution 5 ==<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
draw((-1, 2.41421356)--(1,-2.41421356));<br />
draw((1, 2.41421356)--(-1,-2.41421356));<br />
draw((2.41421356, 1)--(-2.41421356,-1));<br />
draw((2.41421356, -1)--(-2.41421356,1));<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
label("O",(0,0),NE);<br />
label("P",(.70710678,1.70710678),SE);<br />
<br />
</asy><br />
<br />
The ratio of <math>m</math> to <math>n</math> is the same as the ratio of <math>OP</math> to <math>OB</math> which is <math>\frac{\sqrt {2}}{2}</math>.<br />
<br />
ERMS Coach<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Ermscoachhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_14&diff=1379042020 AMC 12A Problems/Problem 142020-11-20T01:45:40Z<p>Ermscoach: /* Solution 5 */</p>
<hr />
<div>==Problem 14==<br />
Regular octagon <math>ABCDEFGH</math> has area <math>n</math>. Let <math>m</math> be the area of quadrilateral <math>ACEG</math>. What is <math>\tfrac{m}{n}?</math><br />
<br />
<math>\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}</math><br />
<br />
==Solution 1==<br />
<math>ACEG</math> is a square. WLOG <math>AB = 1,</math> then using Law of Cosines, <math>AC^2 = [ACEG] = 1^2 + 1^2 - 2 \cos{135} = 2 + \sqrt{2}.</math> The area of the octagon is just <math>[ACEG]</math> plus the area of the four congruent (by symmetry) isosceles triangles, all an angle of <math>135</math> in between two sides of length 1. Now, <cmath>\dfrac{m}{n} = \dfrac{2 + \sqrt{2}}{2 + \sqrt{2} + 4 \cdot \tfrac{1}{2} \sin{135}} = \dfrac{2 + \sqrt{2}}{2 + 2 \sqrt{2}} = \dfrac{\sqrt{2}}{2}.</cmath> The answer is <math>\boxed{\textbf{(B)}}.</math><br />
<br />
==Solution 2==<br />
Refer to the diagram. Call one of the side lengths of the square <math>s</math>. Since quadrilateral ACEG is a square, the area of the square would just be <math>s^2</math>, which we can find by applying Law of Cosines on one of the four triangles. Assume each of the sides of the octagon has length <math>1</math>. Since each angle measures <math>135^{\circ}</math> in an octagon, then <math>s^2 = 1^2+1^2-2*\cos(135^{\circ}) = 2+\sqrt{2}</math> <br />
<br />
There are many ways to find the area of the octagon, but one way is to split the octagon into two trapezoids and one rectangle. We can easily compute AF to be <math>1+\sqrt{2}</math> from splitting one of the sides into two <math>45-45-90</math> triangles. So the area of the octagon is <math>2*\frac{1+\sqrt{2}}{2}+1+\sqrt{2} \Rightarrow 2+2\sqrt{2}</math>. <br />
<br />
Hence, <cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath> <br />
<cmath>\Rightarrow \frac{2+\sqrt{2}}{2+2\sqrt{2}} \cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}}</cmath><br />
<cmath>\Rightarrow \frac{-2\sqrt{2}}{-4}</cmath><br />
<cmath>\Rightarrow \boxed{\frac{\sqrt{2}}{2}}</cmath><br />
<br />
~Solution by IronicNinja<br />
<br />
== Solution 3 (Deriving Formulas) ==<br />
<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
</asy><br />
<br />
The first thing to notice is that <math>ACEG</math> is a square. This is because, as <math>\bigtriangleup ABC \cong \bigtriangleup CDE \cong \bigtriangleup EFG \cong \bigtriangleup GHE</math>, they all have the same base, meaning that <math>AC = DE = EG = GA</math>. Hence, we have that it is a square. To determine the area of this square, we can determine the length of its diagonals.<br />
<br />
In order to do this, we first determine the area of the octagon. Letting the side length be <math>a</math>, we can create a square of length <math>s</math> around it (see figure).<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
label("a",(1, 2.41421356)--(2.41421356, 1),S);<br />
draw((-2.41421356, 2.41421356)--(2.41421356, 2.41421356)--(2.41421356, -2.41421356)--(-2.41421356, -2.41421356)--cycle);<br />
label("s",(-2.41421356, 2.41421356)--(2.41421356, 2.41421356),N);<br />
</asy><br />
<br />
Creating a small square of side length <math>a</math> from the corners of this figure gives us an area of <math>a^2</math>. Thus, <math>s^2 - a^2 = n</math> where <math>n</math> is the area of the octagon. We know from the Pythagorean Theorem that <math>s = a + \frac{a}{\sqrt{2}}</math>, meaning that <math>n = (a + \frac{a}{\sqrt{2}})^2 - a^2 = 2a^2(1+\sqrt{2})</math>.<br />
<br />
Dividing this by <math>8</math> gives us the area of each triangular segment which makes up the octagon. Further dividing by <math>2</math> gives us the area of a smaller segment consisting of the right triangle with legs of the apothem and <math>\frac{a}{2}</math>. Using the area of a triangle as <math>\frac{1}{2}bh</math>, we can determine the length of apothem <math>r</math> from<br />
<br />
<cmath>\frac{2a^2(1+\sqrt{2})}{8 \times 2} = \frac{\frac{a}{2}\times r}{2}</cmath><br />
<cmath>\frac{4a^2(1+\sqrt{2})}{8 } = ar</cmath><br />
<cmath>\frac{a(1+\sqrt{2})}{2} = r</cmath>.<br />
<br />
From the apothem, we can once again use the Pythagorean Theorem, giving us the length of the circumradius <math>R</math>.<br />
<br />
<cmath>R^2 = (\frac{a(1+\sqrt{2})}{2})^2 + (\frac{a}{2})^2</cmath><br />
<cmath>R^2 = \frac{a^2(1+\sqrt{2})^2}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(3+2\sqrt{2})}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(4+2\sqrt{2})}{4}</cmath><br />
<cmath>R = \sqrt{\frac{a^2(4+2\sqrt{2})}{4}} = \frac{a\sqrt{4 + 2\sqrt2}}{2}</cmath>.<br />
<br />
Doubling this gives us the diagonal of both the square and the octagon. From here, we can use the formula <math>A = \frac{1}{2}d^2</math> for the area of the square:<br />
<br />
<cmath>A = \frac{(a\sqrt{4 + 2\sqrt2})^2}{2}</cmath><br />
<cmath>A = \frac{a^2(4+2\sqrt{2})}{2} = m</cmath>.<br />
<br />
Thus we now only need to find the ratio <math>\frac{m}{n}</math>. This can be easily done through some algebra:<br />
<br />
<cmath>\frac{m}{n} = \frac{a^2(4+2\sqrt{2})}{2(2a^2(1+\sqrt{2})}</cmath><br />
<cmath>\frac{m}{n} = \frac{4+2\sqrt{2}}{4+4\sqrt{2}}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
<br />
Rationalizing the denominator by multiplying by the conjugate, we get <math>\frac{m}{n} = \boxed{\textbf{(B) } \frac{\sqrt{2}}{2}}</math>. ~ciceronii<br />
<br />
*<math>\textbf{Note:}</math> this can more easily be done if you know any of these formulas. This was an entire derivation of the area of an octagon, it's apothem, and it's circumradius, but it can be much simpler if you have any of these memorized.<br />
<br />
==Solution 3 (Elementary Geometry)==<br />
WLOG, let <math>AE=1</math>. Let the intersection of <math>AF</math> and <math>GD</math> be point <math>I</math>. <math>GIF</math> is an isosceles right triangle (<math>m\angle HGF=135^{\circ}</math>), so <math>IG=IF=\frac{1}{\sqrt{2}}</math>. The distance between each side and the center is then <math>IF+\frac{1}{2}GH=\frac{1}{2}+\frac{1}{\sqrt{2}}</math>. <math>ABCDEFGH</math> is 8 triangles of base 1 and altitude <math>\frac{1}{2}+\frac{1}{\sqrt{2}}</math>, so its area is <math>8(\frac{1}{2})(\frac{1}{2}+\frac{1}{\sqrt{2}})</math> or <br />
<cmath>2+2\sqrt{2}</cmath><br />
Similarly, <math>ACEG</math> is clearly a square of area <math>GA^2</math>. By the Pythagorean Theorem, <math>GA^2=GI^2+IA^2=(\frac{1}{\sqrt{2}})^2+(1+\frac{1}{\sqrt{2}})^2</math> or <br />
<cmath>2+\sqrt{2}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
From some fast manipulations, see that it is <math>\boxed{\textbf{B)}\frac{\sqrt{2}}{{2}}}</math><br />
<br />
~~BJHHar<br />
<br />
==Solution 4 (Quick n' Easy)==<br />
This solution is quite efficient and unconvoluted. Though the explanation appears long, this is only because I have tried to make it very thorough. <br />
<br />
Because we are finding the ratios of the areas of the quadrilateral, and the octagon, let us assume, for simplicity's sake, that the octagon has side length 1. Call O the center of the octagon. Because it is a regular octagon, angle AOB is 360/8=45. <br />
Draw the apothem, and call where it intersects AB, I. As the apothem bisects angle AOB, angle AIB is 45/2=22.5. <br />
Then, as the apothem is perpendicular to AB, angle IBO is 90-22.5=67.5. As the apothem bisects AB, segment IB is 1/2. Thus, the apothem has length 1/2 * tan(67.5). By the formula area = apothem * perimeter / 2, the area of the octagon is then 1/2 * tan 67.5 * 8 / 2. <br />
<br />
Now, we move on to the quadrilateral. Let us take triangle AGH. By the interior angle formula, the measure of angle AGH is 180 (8-2) / 8 = 135. Because the octagon is regular, AGH is isosceles, with assumed sidelength 1. Drop an altitude from H to AG, and call the intersection point, P. As AGH is isosceles, this altitude bisects both segment AG and angle AHG. Thus, the measure of angle AHP is 135 / 2 = 67.5. Let us call the length of AG, which is also a side of the quadrilateral, s. Then, AP is s/2. Thus, sin(67.5) = (s/2)/1, so s = 2 * sin(67.5). We can do the same for each of the other sides of the quadrilateral and find that they are of the same length. One can easily see that triangle ABC is congruent to triangle AGH. Because triangle AGH is isosceles, the measure of angle HAG is (180 - 135)/2 = 22.5. The same can be said of angles BAC. Thus, angle CAG, an interior angle of the quadrilateral is 90. We can do the same for the other interior angles of CEGA and find that they are also 90. Thus, CEGA is a square, so its area is s^2 = 4 * sin^2(67.5). <br />
<br />
Now, we must find the ratios of the two areas. <br />
Area(quadrilateral)/Area(hexagon)=4*sin^2(67.5) / (2*tan(67.5)) = 2*sin^2(67.5) / (sin(67.5)/cos(67.5)) = 2*sin(67.5)*cos(67.5) = sin(135) = sqrt(2)/2, the answer.<br />
<br />
-jyatvitskiy :)<br />
<br />
==Solution 5 ==<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
draw((-1, 2.41421356)--(1,-2.41421356));<br />
draw((1, 2.41421356)--(-1,-2.41421356));<br />
draw((2.41421356, 1)--(-2.41421356,-1));<br />
draw((2.41421356, -1)--(-2.41421356,1));<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
label("O",(0,0),NE);<br />
label("P",(.70710678,1.70710678),NE);<br />
<br />
</asy><br />
<br />
The ratio of <math>m</math> to <math>n</math> is the same as the ratio of <math>OP</math> to <math>OB</math> which is <math>\frac{\sqrt {2}}{2}</math>.<br />
<br />
ERMS Coach<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Ermscoachhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_14&diff=1379032020 AMC 12A Problems/Problem 142020-11-20T01:44:14Z<p>Ermscoach: </p>
<hr />
<div>==Problem 14==<br />
Regular octagon <math>ABCDEFGH</math> has area <math>n</math>. Let <math>m</math> be the area of quadrilateral <math>ACEG</math>. What is <math>\tfrac{m}{n}?</math><br />
<br />
<math>\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}</math><br />
<br />
==Solution 1==<br />
<math>ACEG</math> is a square. WLOG <math>AB = 1,</math> then using Law of Cosines, <math>AC^2 = [ACEG] = 1^2 + 1^2 - 2 \cos{135} = 2 + \sqrt{2}.</math> The area of the octagon is just <math>[ACEG]</math> plus the area of the four congruent (by symmetry) isosceles triangles, all an angle of <math>135</math> in between two sides of length 1. Now, <cmath>\dfrac{m}{n} = \dfrac{2 + \sqrt{2}}{2 + \sqrt{2} + 4 \cdot \tfrac{1}{2} \sin{135}} = \dfrac{2 + \sqrt{2}}{2 + 2 \sqrt{2}} = \dfrac{\sqrt{2}}{2}.</cmath> The answer is <math>\boxed{\textbf{(B)}}.</math><br />
<br />
==Solution 2==<br />
Refer to the diagram. Call one of the side lengths of the square <math>s</math>. Since quadrilateral ACEG is a square, the area of the square would just be <math>s^2</math>, which we can find by applying Law of Cosines on one of the four triangles. Assume each of the sides of the octagon has length <math>1</math>. Since each angle measures <math>135^{\circ}</math> in an octagon, then <math>s^2 = 1^2+1^2-2*\cos(135^{\circ}) = 2+\sqrt{2}</math> <br />
<br />
There are many ways to find the area of the octagon, but one way is to split the octagon into two trapezoids and one rectangle. We can easily compute AF to be <math>1+\sqrt{2}</math> from splitting one of the sides into two <math>45-45-90</math> triangles. So the area of the octagon is <math>2*\frac{1+\sqrt{2}}{2}+1+\sqrt{2} \Rightarrow 2+2\sqrt{2}</math>. <br />
<br />
Hence, <cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath> <br />
<cmath>\Rightarrow \frac{2+\sqrt{2}}{2+2\sqrt{2}} \cdot \frac{2-2\sqrt{2}}{2-2\sqrt{2}}</cmath><br />
<cmath>\Rightarrow \frac{-2\sqrt{2}}{-4}</cmath><br />
<cmath>\Rightarrow \boxed{\frac{\sqrt{2}}{2}}</cmath><br />
<br />
~Solution by IronicNinja<br />
<br />
== Solution 3 (Deriving Formulas) ==<br />
<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
</asy><br />
<br />
The first thing to notice is that <math>ACEG</math> is a square. This is because, as <math>\bigtriangleup ABC \cong \bigtriangleup CDE \cong \bigtriangleup EFG \cong \bigtriangleup GHE</math>, they all have the same base, meaning that <math>AC = DE = EG = GA</math>. Hence, we have that it is a square. To determine the area of this square, we can determine the length of its diagonals.<br />
<br />
In order to do this, we first determine the area of the octagon. Letting the side length be <math>a</math>, we can create a square of length <math>s</math> around it (see figure).<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
label("a",(1, 2.41421356)--(2.41421356, 1),S);<br />
draw((-2.41421356, 2.41421356)--(2.41421356, 2.41421356)--(2.41421356, -2.41421356)--(-2.41421356, -2.41421356)--cycle);<br />
label("s",(-2.41421356, 2.41421356)--(2.41421356, 2.41421356),N);<br />
</asy><br />
<br />
Creating a small square of side length <math>a</math> from the corners of this figure gives us an area of <math>a^2</math>. Thus, <math>s^2 - a^2 = n</math> where <math>n</math> is the area of the octagon. We know from the Pythagorean Theorem that <math>s = a + \frac{a}{\sqrt{2}}</math>, meaning that <math>n = (a + \frac{a}{\sqrt{2}})^2 - a^2 = 2a^2(1+\sqrt{2})</math>.<br />
<br />
Dividing this by <math>8</math> gives us the area of each triangular segment which makes up the octagon. Further dividing by <math>2</math> gives us the area of a smaller segment consisting of the right triangle with legs of the apothem and <math>\frac{a}{2}</math>. Using the area of a triangle as <math>\frac{1}{2}bh</math>, we can determine the length of apothem <math>r</math> from<br />
<br />
<cmath>\frac{2a^2(1+\sqrt{2})}{8 \times 2} = \frac{\frac{a}{2}\times r}{2}</cmath><br />
<cmath>\frac{4a^2(1+\sqrt{2})}{8 } = ar</cmath><br />
<cmath>\frac{a(1+\sqrt{2})}{2} = r</cmath>.<br />
<br />
From the apothem, we can once again use the Pythagorean Theorem, giving us the length of the circumradius <math>R</math>.<br />
<br />
<cmath>R^2 = (\frac{a(1+\sqrt{2})}{2})^2 + (\frac{a}{2})^2</cmath><br />
<cmath>R^2 = \frac{a^2(1+\sqrt{2})^2}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(3+2\sqrt{2})}{4} + \frac{a^2}{4}</cmath><br />
<cmath>R^2 = \frac{a^2(4+2\sqrt{2})}{4}</cmath><br />
<cmath>R = \sqrt{\frac{a^2(4+2\sqrt{2})}{4}} = \frac{a\sqrt{4 + 2\sqrt2}}{2}</cmath>.<br />
<br />
Doubling this gives us the diagonal of both the square and the octagon. From here, we can use the formula <math>A = \frac{1}{2}d^2</math> for the area of the square:<br />
<br />
<cmath>A = \frac{(a\sqrt{4 + 2\sqrt2})^2}{2}</cmath><br />
<cmath>A = \frac{a^2(4+2\sqrt{2})}{2} = m</cmath>.<br />
<br />
Thus we now only need to find the ratio <math>\frac{m}{n}</math>. This can be easily done through some algebra:<br />
<br />
<cmath>\frac{m}{n} = \frac{a^2(4+2\sqrt{2})}{2(2a^2(1+\sqrt{2})}</cmath><br />
<cmath>\frac{m}{n} = \frac{4+2\sqrt{2}}{4+4\sqrt{2}}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
<br />
Rationalizing the denominator by multiplying by the conjugate, we get <math>\frac{m}{n} = \boxed{\textbf{(B) } \frac{\sqrt{2}}{2}}</math>. ~ciceronii<br />
<br />
*<math>\textbf{Note:}</math> this can more easily be done if you know any of these formulas. This was an entire derivation of the area of an octagon, it's apothem, and it's circumradius, but it can be much simpler if you have any of these memorized.<br />
<br />
==Solution 3 (Elementary Geometry)==<br />
WLOG, let <math>AE=1</math>. Let the intersection of <math>AF</math> and <math>GD</math> be point <math>I</math>. <math>GIF</math> is an isosceles right triangle (<math>m\angle HGF=135^{\circ}</math>), so <math>IG=IF=\frac{1}{\sqrt{2}}</math>. The distance between each side and the center is then <math>IF+\frac{1}{2}GH=\frac{1}{2}+\frac{1}{\sqrt{2}}</math>. <math>ABCDEFGH</math> is 8 triangles of base 1 and altitude <math>\frac{1}{2}+\frac{1}{\sqrt{2}}</math>, so its area is <math>8(\frac{1}{2})(\frac{1}{2}+\frac{1}{\sqrt{2}})</math> or <br />
<cmath>2+2\sqrt{2}</cmath><br />
Similarly, <math>ACEG</math> is clearly a square of area <math>GA^2</math>. By the Pythagorean Theorem, <math>GA^2=GI^2+IA^2=(\frac{1}{\sqrt{2}})^2+(1+\frac{1}{\sqrt{2}})^2</math> or <br />
<cmath>2+\sqrt{2}</cmath><br />
<cmath>\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}</cmath><br />
From some fast manipulations, see that it is <math>\boxed{\textbf{B)}\frac{\sqrt{2}}{{2}}}</math><br />
<br />
~~BJHHar<br />
<br />
==Solution 4 (Quick n' Easy)==<br />
This solution is quite efficient and unconvoluted. Though the explanation appears long, this is only because I have tried to make it very thorough. <br />
<br />
Because we are finding the ratios of the areas of the quadrilateral, and the octagon, let us assume, for simplicity's sake, that the octagon has side length 1. Call O the center of the octagon. Because it is a regular octagon, angle AOB is 360/8=45. <br />
Draw the apothem, and call where it intersects AB, I. As the apothem bisects angle AOB, angle AIB is 45/2=22.5. <br />
Then, as the apothem is perpendicular to AB, angle IBO is 90-22.5=67.5. As the apothem bisects AB, segment IB is 1/2. Thus, the apothem has length 1/2 * tan(67.5). By the formula area = apothem * perimeter / 2, the area of the octagon is then 1/2 * tan 67.5 * 8 / 2. <br />
<br />
Now, we move on to the quadrilateral. Let us take triangle AGH. By the interior angle formula, the measure of angle AGH is 180 (8-2) / 8 = 135. Because the octagon is regular, AGH is isosceles, with assumed sidelength 1. Drop an altitude from H to AG, and call the intersection point, P. As AGH is isosceles, this altitude bisects both segment AG and angle AHG. Thus, the measure of angle AHP is 135 / 2 = 67.5. Let us call the length of AG, which is also a side of the quadrilateral, s. Then, AP is s/2. Thus, sin(67.5) = (s/2)/1, so s = 2 * sin(67.5). We can do the same for each of the other sides of the quadrilateral and find that they are of the same length. One can easily see that triangle ABC is congruent to triangle AGH. Because triangle AGH is isosceles, the measure of angle HAG is (180 - 135)/2 = 22.5. The same can be said of angles BAC. Thus, angle CAG, an interior angle of the quadrilateral is 90. We can do the same for the other interior angles of CEGA and find that they are also 90. Thus, CEGA is a square, so its area is s^2 = 4 * sin^2(67.5). <br />
<br />
Now, we must find the ratios of the two areas. <br />
Area(quadrilateral)/Area(hexagon)=4*sin^2(67.5) / (2*tan(67.5)) = 2*sin^2(67.5) / (sin(67.5)/cos(67.5)) = 2*sin(67.5)*cos(67.5) = sin(135) = sqrt(2)/2, the answer.<br />
<br />
-jyatvitskiy :)<br />
<br />
==Solution 5 ==<br />
<asy><br />
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);<br />
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);<br />
draw((-1, 2.41421356)--(1,-2.41421356));<br />
draw((1, 2.41421356)--(-1,-2.41421356));<br />
draw((2.41421356, 1)--(-2.41421356,-1));<br />
draw((2.41421356, -1)--(-2.41421356,1));<br />
label("A",(-1, 2.41421356),NW);<br />
label("B",(1, 2.41421356),NE);<br />
label("C",(2.41421356, 1),NE);<br />
label("D",(2.41421356, -1),SE);<br />
label("E",(1,-2.41421356),SE);<br />
label("F",(-1,-2.41421356),SW);<br />
label("G",(-2.41421356,-1),SW);<br />
label("H",(-2.41421356,1),NW);<br />
label("O",(0,0),NE);<br />
label("P",(.70710678,1.70710678),NE);<br />
<br />
</asy><br />
<br />
The ratio of <math>m</math> to <math>n</math> is the same as the ratio of <math>OP</math> to <math>OB</math> which is <math>\frac{\sqrt(2)}{2}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=13|num-a=15}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Ermscoachhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_12A_Problems/Problem_12&diff=1379002020 AMC 12A Problems/Problem 122020-11-20T01:07:06Z<p>Ermscoach: </p>
<hr />
<div>== Problem ==<br />
<br />
Line <math>l</math> in the coordinate plane has equation <math>3x-5y+40=0</math>. This line is rotated <math>45^{\circ}</math> counterclockwise about the point <math>(20,20)</math> to obtain line <math>k</math>. What is the <math>x</math>-coordinate of the <math>x</math>-intercept of line <math>k?</math><br />
<br />
<math>\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30</math><br />
<br />
== Solution ==<br />
<br />
The slope of the line is <math>\frac{3}{5}</math>. We must transform it by <math>45^{\circ}</math>. <br />
<br />
<math>45^{\circ}</math> creates an isosceles right triangle since the sum of the angles of the triangle must be <math>180^{\circ}</math> and one angle is <math>90^{\circ}</math> which means the last leg angle must also be <math>45^{\circ}</math>. <br />
<br />
In the isosceles right triangle, the two legs are congruent. We can, therefore, construct an isosceles right triangle with a line of <math>\frac{3}{5}</math> slope on graph paper. That line with <math>\frac{3}{5}</math> slope starts at <math>(0,0)</math> and will go to <math>(5,3)</math>, the vector <math><5,3></math>. <br />
<br />
Construct another line from <math>(0,0)</math> to <math>(3,-5)</math>, the vector <math><3,-5></math>. This is <math>\perp</math> and equal to the original line segment. The difference between the two vectors is <math><2,8></math>, which is the slope <math>4</math>, and that is the slope of line <math>k</math>. <br />
<br />
Furthermore, the equation <math>3x-5y+40=0</math> passes straight through <math>(20,20)</math> since <math>3(20)-5(20)+40=60-100+40=0</math>, which means that any rotations about <math>(20,20)</math> would contain <math>(20,20)</math>. We can create a line of slope <math>4</math> through <math>(20,20)</math>. The <math>x</math>-intercept is therefore <math>20-\frac{20}{4} = \boxed{\textbf{(B) } 15.}</math>~lopkiloinm<br />
<br />
==Solution 2==<br />
Since the slope of the line is <math>\frac{3}{5}</math>, and the angle we are rotating around is x, then <math>\tan x = \frac{3}{5}</math><br />
<math>\tan(x+45^{\circ}) = \frac{\tan x + \tan(45^{\circ})}{1-\tan x*\tan(45^{\circ})} = \frac{0.6+1}{1-0.6} = \frac{1.6}{0.4} = 4</math><br />
<br />
Hence, the slope of the rotated line is <math>4</math>. Since we know the line intersects the point <math>(20,20)</math>, then we know the line is <math>y=4x-60</math>. Set <math>y=0</math> to find the x-intercept, and so <math>x=\boxed{15}</math><br />
<br />
~Solution by IronicNinja<br />
<br />
==Solution 3==<br />
<br />
<asy><br />
draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle);<br />
draw((20, 20)--(0, 8));<br />
draw((15, 0)--(20, 20));<br />
<br />
dot("$P$", (20, 20));<br />
dot("$A$", (0, 8), dir(75));<br />
dot("$B$", (15, 0), dir(45));<br />
dot("$X$", (20, 0));<br />
dot("$Y$", (0, 20), dir(50));<br />
</asy><br />
<br />
Let <math>P</math> be <math>(20, 20)</math> and <math>X, Y</math> be <math>(20, 0)</math> and <math>(0, 20)</math> respectively. Since the slope of the line is <math>3/5</math> we know that <math>\tan{\angle{YPA}} = 3/5.</math> Segments <math>\overline{PA}</math> and <math>\overline{PB}</math> represent the before and after of rotating <math>l</math> by 45 counterclockwise. Thus, <math>\angle{XPB} = 45 - \angle{YPA}</math> and <cmath>BX = 20 \tan{\angle{XPB}} = 20 \cdot \frac{1 - 3/5}{1 + 3/5} = 5</cmath> by tangent addition formula. Since <math>BX</math> is 5 and the sidelength of the square is 20 the answer is <math>20 - 5 \implies \boxed{\textbf{B}}.</math><br />
<br />
==Solution 4 (Cheap)==<br />
<br />
Using the protractor you brought, carefully graph the equation and rotate the given line <math>45^{\circ}</math> counter-clockwise about the point <math>(20,20)</math>. Scaling everything down by a factor of 5 makes this process easier.<br />
<br />
It should then become fairly obvious that the x intercept is <math>x=\boxed{15}</math> (only use this as a last resort). <br />
<br />
~Silverdragon<br />
<br />
==Solution 5 (Rotation Matrix)==<br />
First note that the given line goes through <math>(20,20)</math> with a slope of <math>\frac{3}{5}</math>. This means that <math>(25,23)</math> is on the line. Now consider translating the graph so that <math>(20,20)</math> goes to the origin, then <math>(25,23)</math> becomes <math>(5,3)</math>. We now rotate the line <math>45^\circ</math> about the origin using a rotation matrix. This maps <math>(5,3)</math> to <br />
<cmath>\begin{bmatrix} \frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\ \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\end{bmatrix}\begin{bmatrix} 5 \\ 3\end{bmatrix}=\begin{bmatrix}\sqrt{2} \\ 4\sqrt{2}\end{bmatrix}</cmath><br />
The line through the origin and <math>(\sqrt{2}, 4\sqrt{2})</math> has slope <math>4</math>. Translating this line so that the origin is mapped to <math>(20,20)</math>, we find that the equation for the new line is <math>4x-60</math>, meaning that the <math>x</math>-intercept is <math>x=\frac{60}{4}=\boxed{\textbf{(B) }15}</math>.<br />
<br />
==Solution 6 (Angle Bisector)==<br />
Note <math>(20,20)</math> is on the line. Construct the perpendicular line <math>5x+3y-160=0</math>. This creates a right triangle that intersects the x-axis at <math>(\frac{-40}{3},0)</math> and <math>(32,0)</math> a distance of <math>136</math> apart. The <math>45^\circ</math> transformation will bisect the right angle.<br />
The angle bisector theorem tells us the <math>136</math> will split in ratio to the lengths of the sides.<br />
These are <math>\sqrt(12^2+20^2)</math> and <math>\sqrt(\frac{100}{3}^2 + 20^2) = 4\sqrt(34)</math> and <math>\frac{20}{3}\sqrt(34)</math>. Thus the x intercept will split the line from <math>\frac{-40}{3}</math> to <math>32</math> into a ratio of <math>5:3</math> making the x-intercept <math>15</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC12 box|year=2020|ab=A|num-b=11|num-a=13}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Ermscoach