https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Justin.ziyu.sun&feedformat=atomAoPS Wiki - User contributions [en]2024-03-30T03:30:07ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_7&diff=1945652007 AMC 12B Problems/Problem 72023-06-17T22:31:17Z<p>Justin.ziyu.sun: /* Solution */</p>
<hr />
<div>==Problem==<br />
All sides of the [[convex polygon|convex]] [[pentagon]] <math>ABCDE</math> are of equal length, and <math>\angle A = \angle B = 90^{\circ}</math>. What is the degree measure of <math>\angle E</math>? <br />
<br />
<math>\mathrm {(A)}\ 90 \qquad \mathrm {(B)}\ 108 \qquad \mathrm {(C)}\ 120 \qquad \mathrm {(D)}\ 144 \qquad \mathrm {(E)}\ 150</math><br />
<br />
==Solution==<br />
[[Image:2007_12B_AMC-7.png]]<br />
<br />
Since <math>A</math> and <math>B</math> are [[right angle]]s, and <math>AE</math> equals <math>BC</math>, and <math>AECB</math> is a [[square]]. Since <math>ED</math> and <math>CD</math> are also 5, triangle <math>CDE</math> is [[equilateral triangle|equilateral]]. Angle <math>E</math> is therefore <math>90+60=150 \Rightarrow \mathrm {(E)}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2007|ab=B|num-b=6|num-a=8}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Justin.ziyu.sunhttps://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_24&diff=1388282020 AMC 8 Problems/Problem 242020-11-30T16:46:16Z<p>Justin.ziyu.sun: /* Video Solutions */</p>
<hr />
<div>==Problem==<br />
A large square region is paved with <math>n^2</math> gray square tiles, each measuring <math>s</math> inches on a side. A border <math>d</math> inches wide surrounds each tile. The figure below shows the case for <math>n=3</math>. When <math>n=24</math>, the <math>576</math> gray tiles cover <math>64\%</math> of the area of the large square region. What is the ratio <math>\frac{d}{s}</math> for this larger value of <math>n?</math><br />
<br />
<asy><br />
draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br />
filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br />
filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br />
filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br />
filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br />
filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br />
filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br />
filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br />
filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br />
filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br />
</asy> <br />
<br />
<math>\textbf{(A) }\frac{6}{25} \qquad \textbf{(B) }\frac{1}{4} \qquad \textbf{(C) }\frac{9}{25} \qquad \textbf{(D) }\frac{7}{16} \qquad \textbf{(E) }\frac{9}{16}</math><br />
<br />
==Solution 1==<br />
The area of the shaded region is <math>(24s)^2</math>. To find the area of the large square, we note that there is a <math>d</math>-inch border between each of the <math>23</math> pairs of consecutive squares, as well as from between first/last squares and the large square, for a total of <math>23+2 = 25</math> times the length of the border, i.e. <math>25d</math>. Adding this to the total length of the consecutive squares, which is <math>24s</math>, the side length of the large square is <math>(24s+25d)</math>, yielding the equation <math>\frac{(24s)^2}{(24s+25d)^2}=\frac{64}{100}</math>. Taking the square root of both sides (and using the fact that lengths are non-negative) gives <math>\frac{24s}{24s+25d}=\frac{8}{10} = \frac{4}{5}</math>, and cross-multiplying now gives <math>120s = 96s + 100d \Rightarrow 24s = 100d \Rightarrow \frac{d}{s} = \frac{24}{100} = \boxed{\textbf{(A) }\frac{6}{25}}</math>.<br />
<br />
==Solution 2==<br />
Without loss of generality, we may let <math>s=1</math> (since <math>d</math> will be determined by the scale of <math>s</math>, and we are only interested in the ratio <math>\frac{d}{s}</math>). Then, as the total area of the <math>576</math> gray tiles is simply <math>576</math>, the large square has area <math>\frac{576}{0.64} = 900</math>, making the side of the large square <math>\sqrt{900}=30</math>. As in Solution 1, the the side length of the large square consists of the total length of the gray tiles and <math>25</math> lots of the border, so the length of the border is <math>d = \frac{30-24}{25} = \frac{6}{25}</math>. Since <math>\frac{d}{s}=d</math> if <math>s=1</math>, the answer is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>.<br />
<br />
==Solution 3 (using answer choices)==<br />
As in Solution 2, we let <math>s = 1</math> without loss of generality. For sufficiently large <math>n</math>, we can approximate the percentage of the area covered by the gray tiles by subdividing most of the region into congruent squares, as shown:<br />
<asy><br />
draw((0,0)--(13,0)--(13,13)--(0,13)--cycle);<br />
filldraw((1,1)--(4,1)--(4,4)--(1,4)--cycle, mediumgray);<br />
filldraw((1,5)--(4,5)--(4,8)--(1,8)--cycle, mediumgray);<br />
filldraw((1,9)--(4,9)--(4,12)--(1,12)--cycle, mediumgray);<br />
filldraw((5,1)--(8,1)--(8,4)--(5,4)--cycle, mediumgray);<br />
filldraw((5,5)--(8,5)--(8,8)--(5,8)--cycle, mediumgray);<br />
filldraw((5,9)--(8,9)--(8,12)--(5,12)--cycle, mediumgray);<br />
filldraw((9,1)--(12,1)--(12,4)--(9,4)--cycle, mediumgray);<br />
filldraw((9,5)--(12,5)--(12,8)--(9,8)--cycle, mediumgray);<br />
filldraw((9,9)--(12,9)--(12,12)--(9,12)--cycle, mediumgray);<br />
<br />
for(int i = 1; i <= 13; i += 4){<br />
draw((1,i)--(13,i), red);<br />
draw((i,1)--(i,13), red);<br />
}<br />
</asy> <br />
Each red square has side length <math>(1+d)</math>, so by solving <math>\frac{1^2}{(1+d)^2} = \frac{64}{100} \iff \frac{1}{1+d} = \frac{4}{5}</math>, we obtain <math>d = \frac{1}{4}</math>. The actual fraction of the total area covered by the gray tiles will be slightly less than <math>\frac{1}{(1+d)^2}</math>, which implies <math>\frac{1}{(1+d)^2} > \frac{64}{100} \iff \frac{1}{1+d} > \frac{4}{5} \iff d < \frac{1}{4}</math>. Hence <math>d</math> (and thus <math>\frac{d}{s}</math>, since we are assuming <math>s=1</math>) is less than <math>\frac{1}{4}</math>, and the only choice that satisfies this is <math>\boxed{\textbf{(A) }\frac{6}{25}}</math>.<br />
<br />
==Video Solutions==<br />
https://youtu.be/9tPTBxadV90 <br /><br />
https://youtu.be/KG_LI4jPAU4<br />
<br />
==See also==<br />
{{AMC8 box|year=2020|num-b=23|num-a=25}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Justin.ziyu.sunhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_22&diff=1130302003 AMC 10A Problems/Problem 222019-12-18T05:01:10Z<p>Justin.ziyu.sun: /* Solution 5 */</p>
<hr />
<div>== Problem ==<br />
In rectangle <math>ABCD</math>, we have <math>AB=8</math>, <math>BC=9</math>, <math>H</math> is on <math>BC</math> with <math>BH=6</math>, <math>E</math> is on <math>AD</math> with <math>DE=4</math>, line <math>EC</math> intersects line <math>AH</math> at <math>G</math>, and <math>F</math> is on line <math>AD</math> with <math>GF \perp AF</math>. Find the length of <math>GF</math>. <br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0);<br />
draw(D--A--B--C--D--F--G--Ep);<br />
draw(A--G);<br />
label("$F$",F,W);<br />
label("$G$",G,W);<br />
label("$C$",C,WSW);<br />
label("$H$",H,NNE);<br />
label("$6$",(6,8),N);<br />
label("$B$",B,NE);<br />
label("$A$",A,SW);<br />
label("$E$",Ep,S);<br />
label("$4$",(2,0),S);<br />
label("$D$",D,S);</asy><br />
<br />
<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 </math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
<math>\angle GHC = \angle AHB</math> (Vertical angles are equal).<br />
<br />
<math>\angle F = \angle B</math> (Both are 90 degrees).<br />
<br />
<math>\angle BHA = \angle HAD</math> (Alt. Interior Angles are congruent).<br />
<br />
Therefore <math>\triangle GFA</math> and <math>\triangle ABH</math> are similar.<br />
<math>\triangle GCH</math> and <math>\triangle GEA</math> are also similar.<br />
<br />
<math>DA</math> is 9, therefore <math>EA</math> must equal 5. Similarly, <math>CH</math> must equal 3. <br />
<br />
Because <math>GCH</math> and <math>GEA</math> are similar, the ratio of <math>CH\; =\; 3</math> and <math>EA\; =\; 5</math>, must also hold true for <math>GH</math> and <math>HA</math>. <math>\frac{GH}{GA} = \frac{3}{5}</math>, so <math>HA</math> is <math>\frac{2}{5}</math> of <math>GA</math>. By Pythagorean theorem, <math>(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10</math>.<br />
<br />
<math>HA\: =\: 10 =\: \frac{2}{5}*(GA)</math>. <br />
<br />
<math>GA\: =\: 25.</math><br />
<br />
So <math>\frac{GA}{HA}\: =\: \frac{GF}{BA}</math>.<br />
<br />
<math>\frac{25}{10}\: =\: \frac{GF}{8}</math>.<br />
<br />
Therefore <math>GF= \boxed{\mathrm{(B)}\ 20}</math>.<br />
<br />
=== Solution 2 ===<br />
Since <math>ABCD</math> is a rectangle, <math>CD=AB=8</math>. <br />
<br />
Since <math>ABCD</math> is a rectangle and <math>GF \perp AF</math>, <math>\angle GFE = \angle CDE = \angle ABC = 90^\circ </math>. <br />
<br />
Since <math>ABCD</math> is a rectangle, <math>AD || BC</math>. <br />
<br />
So, <math>AH</math> is a [[transversal]], and <math>\angle GAF = \angle AHB</math>. <br />
<br />
This is sufficient to prove that <math> GFE \approx CDE</math> and <math> GFA \approx ABH</math>.<br />
<br />
Using ratios: <br />
<br />
<math>\frac{GF}{FE}=\frac{CD}{DE}</math><br />
<br />
<math>\frac{GF}{FD+4}=\frac{8}{4}=2</math><br />
<br />
<math>GF=2 \cdot (FD+4)=2 \cdot FD+8</math><br />
<br />
<math>\frac{GF}{FA}=\frac{AB}{BH}</math><br />
<br />
<math>\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}</math><br />
<br />
<math>GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12</math><br />
<br />
Since <math>GF</math> can't have 2 different lengths, both expressions for <math>GF</math> must be equal. <br />
<br />
<math>2 \cdot FD+8=\frac{4}{3} \cdot FD+12</math><br />
<br />
<math>\frac{2}{3} \cdot FD=4</math><br />
<br />
<math>FD=6</math><br />
<br />
<math>GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 3 ===<br />
Since <math>ABCD</math> is a rectangle, <math>CH=3</math>, <math>EA=5</math>, and <math>CD=8</math>. From the [[Pythagorean Theorem]], <math>CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}</math>.<br />
==== Solution 4 ====<br />
Statement: <math>GCH \approx GEA</math><br />
<br />
Proof: <math>\angle CGH=\angle EGA</math>, obviously.<br />
<br />
<cmath>\begin{eqnarray*}<br />
\angle HCE&=&180^{\circ}-\angle CHG\\<br />
\angle DCE&=&\angle CHG-90^{\circ}\\<br />
\angle CEED&=&180-\angle CHG\\<br />
\angle GEA&=&\angle GCH<br />
\end{eqnarray*}</cmath><br />
<br />
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.<br />
<br />
<br />
<br />
Let <math>GC=x</math>.<br />
<br />
<cmath>\begin{eqnarray*}<br />
\dfrac{x}{3}&=&\dfrac{x+4\sqrt{5}}{5}\\<br />
5x&=&3x+12\sqrt{5}\\<br />
2x&=&12\sqrt{5}\\<br />
x&=&6\sqrt{5}<br />
\end{eqnarray*}</cmath><br />
<br />
Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore<br />
<br />
<cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath><br />
<br />
We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF</math> is twice of 10, or <math>\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 5 ===<br />
We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 6 ===<br />
<br />
Since <math>GF\perp AF</math> and <math>AF\perp CD, </math> we have <math>GF\parallel CD\parallel AB.</math> Thus, <math>\triangle CDE\sim GFE.</math> Suppose <math>GF=x</math> and <math>FD=y.</math> Thus, we have <math>\dfrac{x}{8}=\dfrac{y+4}{4}.</math> Additionally, now note that <math>\triangle GAF\sim AHB,</math> which is pretty obvious from insight, but can be proven by AA with extending <math>BH</math> to meet <math>GF.</math> From this new pair of similar triangles, we have <math>\dfrac{x}{8}=\dfrac{y+9}{6}.</math> Therefore, we have by combining those two equations, <cmath>\dfrac{y+9}{6}=\dfrac{y+4}{4}.</cmath> Solving, we have <math>y=6,</math> and therefore <math>x=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
== See Also ==<br />
{{AMC10 box|year=2003|ab=A|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Justin.ziyu.sunhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_22&diff=1130292003 AMC 10A Problems/Problem 222019-12-18T05:00:59Z<p>Justin.ziyu.sun: /* Solution 4 */</p>
<hr />
<div>== Problem ==<br />
In rectangle <math>ABCD</math>, we have <math>AB=8</math>, <math>BC=9</math>, <math>H</math> is on <math>BC</math> with <math>BH=6</math>, <math>E</math> is on <math>AD</math> with <math>DE=4</math>, line <math>EC</math> intersects line <math>AH</math> at <math>G</math>, and <math>F</math> is on line <math>AD</math> with <math>GF \perp AF</math>. Find the length of <math>GF</math>. <br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0);<br />
draw(D--A--B--C--D--F--G--Ep);<br />
draw(A--G);<br />
label("$F$",F,W);<br />
label("$G$",G,W);<br />
label("$C$",C,WSW);<br />
label("$H$",H,NNE);<br />
label("$6$",(6,8),N);<br />
label("$B$",B,NE);<br />
label("$A$",A,SW);<br />
label("$E$",Ep,S);<br />
label("$4$",(2,0),S);<br />
label("$D$",D,S);</asy><br />
<br />
<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 </math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
<math>\angle GHC = \angle AHB</math> (Vertical angles are equal).<br />
<br />
<math>\angle F = \angle B</math> (Both are 90 degrees).<br />
<br />
<math>\angle BHA = \angle HAD</math> (Alt. Interior Angles are congruent).<br />
<br />
Therefore <math>\triangle GFA</math> and <math>\triangle ABH</math> are similar.<br />
<math>\triangle GCH</math> and <math>\triangle GEA</math> are also similar.<br />
<br />
<math>DA</math> is 9, therefore <math>EA</math> must equal 5. Similarly, <math>CH</math> must equal 3. <br />
<br />
Because <math>GCH</math> and <math>GEA</math> are similar, the ratio of <math>CH\; =\; 3</math> and <math>EA\; =\; 5</math>, must also hold true for <math>GH</math> and <math>HA</math>. <math>\frac{GH}{GA} = \frac{3}{5}</math>, so <math>HA</math> is <math>\frac{2}{5}</math> of <math>GA</math>. By Pythagorean theorem, <math>(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10</math>.<br />
<br />
<math>HA\: =\: 10 =\: \frac{2}{5}*(GA)</math>. <br />
<br />
<math>GA\: =\: 25.</math><br />
<br />
So <math>\frac{GA}{HA}\: =\: \frac{GF}{BA}</math>.<br />
<br />
<math>\frac{25}{10}\: =\: \frac{GF}{8}</math>.<br />
<br />
Therefore <math>GF= \boxed{\mathrm{(B)}\ 20}</math>.<br />
<br />
=== Solution 2 ===<br />
Since <math>ABCD</math> is a rectangle, <math>CD=AB=8</math>. <br />
<br />
Since <math>ABCD</math> is a rectangle and <math>GF \perp AF</math>, <math>\angle GFE = \angle CDE = \angle ABC = 90^\circ </math>. <br />
<br />
Since <math>ABCD</math> is a rectangle, <math>AD || BC</math>. <br />
<br />
So, <math>AH</math> is a [[transversal]], and <math>\angle GAF = \angle AHB</math>. <br />
<br />
This is sufficient to prove that <math> GFE \approx CDE</math> and <math> GFA \approx ABH</math>.<br />
<br />
Using ratios: <br />
<br />
<math>\frac{GF}{FE}=\frac{CD}{DE}</math><br />
<br />
<math>\frac{GF}{FD+4}=\frac{8}{4}=2</math><br />
<br />
<math>GF=2 \cdot (FD+4)=2 \cdot FD+8</math><br />
<br />
<math>\frac{GF}{FA}=\frac{AB}{BH}</math><br />
<br />
<math>\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}</math><br />
<br />
<math>GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12</math><br />
<br />
Since <math>GF</math> can't have 2 different lengths, both expressions for <math>GF</math> must be equal. <br />
<br />
<math>2 \cdot FD+8=\frac{4}{3} \cdot FD+12</math><br />
<br />
<math>\frac{2}{3} \cdot FD=4</math><br />
<br />
<math>FD=6</math><br />
<br />
<math>GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 3 ===<br />
Since <math>ABCD</math> is a rectangle, <math>CH=3</math>, <math>EA=5</math>, and <math>CD=8</math>. From the [[Pythagorean Theorem]], <math>CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}</math>.<br />
==== Solution 4 ====<br />
Statement: <math>GCH \approx GEA</math><br />
<br />
Proof: <math>\angle CGH=\angle EGA</math>, obviously.<br />
<br />
<cmath>\begin{eqnarray*}<br />
\angle HCE&=&180^{\circ}-\angle CHG\\<br />
\angle DCE&=&\angle CHG-90^{\circ}\\<br />
\angle CEED&=&180-\angle CHG\\<br />
\angle GEA&=&\angle GCH<br />
\end{eqnarray*}</cmath><br />
<br />
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.<br />
<br />
<br />
<br />
Let <math>GC=x</math>.<br />
<br />
<cmath>\begin{eqnarray*}<br />
\dfrac{x}{3}&=&\dfrac{x+4\sqrt{5}}{5}\\<br />
5x&=&3x+12\sqrt{5}\\<br />
2x&=&12\sqrt{5}\\<br />
x&=&6\sqrt{5}<br />
\end{eqnarray*}</cmath><br />
<br />
Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore<br />
<br />
<cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath><br />
<br />
We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF</math> is twice of 10, or <math>\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 5 ===<br />
We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 5 ===<br />
<br />
Since <math>GF\perp AF</math> and <math>AF\perp CD, </math> we have <math>GF\parallel CD\parallel AB.</math> Thus, <math>\triangle CDE\sim GFE.</math> Suppose <math>GF=x</math> and <math>FD=y.</math> Thus, we have <math>\dfrac{x}{8}=\dfrac{y+4}{4}.</math> Additionally, now note that <math>\triangle GAF\sim AHB,</math> which is pretty obvious from insight, but can be proven by AA with extending <math>BH</math> to meet <math>GF.</math> From this new pair of similar triangles, we have <math>\dfrac{x}{8}=\dfrac{y+9}{6}.</math> Therefore, we have by combining those two equations, <cmath>\dfrac{y+9}{6}=\dfrac{y+4}{4}.</cmath> Solving, we have <math>y=6,</math> and therefore <math>x=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
== See Also ==<br />
{{AMC10 box|year=2003|ab=A|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Justin.ziyu.sunhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_22&diff=1130282003 AMC 10A Problems/Problem 222019-12-18T05:00:45Z<p>Justin.ziyu.sun: /* solution 4 */</p>
<hr />
<div>== Problem ==<br />
In rectangle <math>ABCD</math>, we have <math>AB=8</math>, <math>BC=9</math>, <math>H</math> is on <math>BC</math> with <math>BH=6</math>, <math>E</math> is on <math>AD</math> with <math>DE=4</math>, line <math>EC</math> intersects line <math>AH</math> at <math>G</math>, and <math>F</math> is on line <math>AD</math> with <math>GF \perp AF</math>. Find the length of <math>GF</math>. <br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0);<br />
draw(D--A--B--C--D--F--G--Ep);<br />
draw(A--G);<br />
label("$F$",F,W);<br />
label("$G$",G,W);<br />
label("$C$",C,WSW);<br />
label("$H$",H,NNE);<br />
label("$6$",(6,8),N);<br />
label("$B$",B,NE);<br />
label("$A$",A,SW);<br />
label("$E$",Ep,S);<br />
label("$4$",(2,0),S);<br />
label("$D$",D,S);</asy><br />
<br />
<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 </math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
<math>\angle GHC = \angle AHB</math> (Vertical angles are equal).<br />
<br />
<math>\angle F = \angle B</math> (Both are 90 degrees).<br />
<br />
<math>\angle BHA = \angle HAD</math> (Alt. Interior Angles are congruent).<br />
<br />
Therefore <math>\triangle GFA</math> and <math>\triangle ABH</math> are similar.<br />
<math>\triangle GCH</math> and <math>\triangle GEA</math> are also similar.<br />
<br />
<math>DA</math> is 9, therefore <math>EA</math> must equal 5. Similarly, <math>CH</math> must equal 3. <br />
<br />
Because <math>GCH</math> and <math>GEA</math> are similar, the ratio of <math>CH\; =\; 3</math> and <math>EA\; =\; 5</math>, must also hold true for <math>GH</math> and <math>HA</math>. <math>\frac{GH}{GA} = \frac{3}{5}</math>, so <math>HA</math> is <math>\frac{2}{5}</math> of <math>GA</math>. By Pythagorean theorem, <math>(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10</math>.<br />
<br />
<math>HA\: =\: 10 =\: \frac{2}{5}*(GA)</math>. <br />
<br />
<math>GA\: =\: 25.</math><br />
<br />
So <math>\frac{GA}{HA}\: =\: \frac{GF}{BA}</math>.<br />
<br />
<math>\frac{25}{10}\: =\: \frac{GF}{8}</math>.<br />
<br />
Therefore <math>GF= \boxed{\mathrm{(B)}\ 20}</math>.<br />
<br />
=== Solution 2 ===<br />
Since <math>ABCD</math> is a rectangle, <math>CD=AB=8</math>. <br />
<br />
Since <math>ABCD</math> is a rectangle and <math>GF \perp AF</math>, <math>\angle GFE = \angle CDE = \angle ABC = 90^\circ </math>. <br />
<br />
Since <math>ABCD</math> is a rectangle, <math>AD || BC</math>. <br />
<br />
So, <math>AH</math> is a [[transversal]], and <math>\angle GAF = \angle AHB</math>. <br />
<br />
This is sufficient to prove that <math> GFE \approx CDE</math> and <math> GFA \approx ABH</math>.<br />
<br />
Using ratios: <br />
<br />
<math>\frac{GF}{FE}=\frac{CD}{DE}</math><br />
<br />
<math>\frac{GF}{FD+4}=\frac{8}{4}=2</math><br />
<br />
<math>GF=2 \cdot (FD+4)=2 \cdot FD+8</math><br />
<br />
<math>\frac{GF}{FA}=\frac{AB}{BH}</math><br />
<br />
<math>\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}</math><br />
<br />
<math>GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12</math><br />
<br />
Since <math>GF</math> can't have 2 different lengths, both expressions for <math>GF</math> must be equal. <br />
<br />
<math>2 \cdot FD+8=\frac{4}{3} \cdot FD+12</math><br />
<br />
<math>\frac{2}{3} \cdot FD=4</math><br />
<br />
<math>FD=6</math><br />
<br />
<math>GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 3 ===<br />
Since <math>ABCD</math> is a rectangle, <math>CH=3</math>, <math>EA=5</math>, and <math>CD=8</math>. From the [[Pythagorean Theorem]], <math>CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}</math>.<br />
==== Solution 4 ====<br />
Statement: <math>GCH \approx GEA</math><br />
<br />
Proof: <math>\angle CGH=\angle EGA</math>, obviously.<br />
<br />
<cmath>\begin{eqnarray*}<br />
\angle HCE&=&180^{\circ}-\angle CHG\\<br />
\angle DCE&=&\angle CHG-90^{\circ}\\<br />
\angle CEED&=&180-\angle CHG\\<br />
\angle GEA&=&\angle GCH<br />
\end{eqnarray*}</cmath><br />
<br />
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.<br />
<br />
<br />
<br />
Let <math>GC=x</math>.<br />
<br />
<cmath>\begin{eqnarray*}<br />
\dfrac{x}{3}&=&\dfrac{x+4\sqrt{5}}{5}\\<br />
5x&=&3x+12\sqrt{5}\\<br />
2x&=&12\sqrt{5}\\<br />
x&=&6\sqrt{5}<br />
\end{eqnarray*}</cmath><br />
<br />
Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore<br />
<br />
<cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath><br />
<br />
We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF</math> is twice of 10, or <math>\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 4 ===<br />
We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 5 ===<br />
<br />
Since <math>GF\perp AF</math> and <math>AF\perp CD, </math> we have <math>GF\parallel CD\parallel AB.</math> Thus, <math>\triangle CDE\sim GFE.</math> Suppose <math>GF=x</math> and <math>FD=y.</math> Thus, we have <math>\dfrac{x}{8}=\dfrac{y+4}{4}.</math> Additionally, now note that <math>\triangle GAF\sim AHB,</math> which is pretty obvious from insight, but can be proven by AA with extending <math>BH</math> to meet <math>GF.</math> From this new pair of similar triangles, we have <math>\dfrac{x}{8}=\dfrac{y+9}{6}.</math> Therefore, we have by combining those two equations, <cmath>\dfrac{y+9}{6}=\dfrac{y+4}{4}.</cmath> Solving, we have <math>y=6,</math> and therefore <math>x=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
== See Also ==<br />
{{AMC10 box|year=2003|ab=A|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Justin.ziyu.sunhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_22&diff=1130272003 AMC 10A Problems/Problem 222019-12-18T05:00:30Z<p>Justin.ziyu.sun: /* Lemma */</p>
<hr />
<div>== Problem ==<br />
In rectangle <math>ABCD</math>, we have <math>AB=8</math>, <math>BC=9</math>, <math>H</math> is on <math>BC</math> with <math>BH=6</math>, <math>E</math> is on <math>AD</math> with <math>DE=4</math>, line <math>EC</math> intersects line <math>AH</math> at <math>G</math>, and <math>F</math> is on line <math>AD</math> with <math>GF \perp AF</math>. Find the length of <math>GF</math>. <br />
<br />
<asy><br />
unitsize(3mm);<br />
defaultpen(linewidth(.8pt)+fontsize(8pt));<br />
pair D=(0,0), Ep=(4,0), A=(9,0), B=(9,8), H=(3,8), C=(0,8), G=(-6,20), F=(-6,0);<br />
draw(D--A--B--C--D--F--G--Ep);<br />
draw(A--G);<br />
label("$F$",F,W);<br />
label("$G$",G,W);<br />
label("$C$",C,WSW);<br />
label("$H$",H,NNE);<br />
label("$6$",(6,8),N);<br />
label("$B$",B,NE);<br />
label("$A$",A,SW);<br />
label("$E$",Ep,S);<br />
label("$4$",(2,0),S);<br />
label("$D$",D,S);</asy><br />
<br />
<math> \mathrm{(A) \ } 16\qquad \mathrm{(B) \ } 20\qquad \mathrm{(C) \ } 24\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 30 </math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
<math>\angle GHC = \angle AHB</math> (Vertical angles are equal).<br />
<br />
<math>\angle F = \angle B</math> (Both are 90 degrees).<br />
<br />
<math>\angle BHA = \angle HAD</math> (Alt. Interior Angles are congruent).<br />
<br />
Therefore <math>\triangle GFA</math> and <math>\triangle ABH</math> are similar.<br />
<math>\triangle GCH</math> and <math>\triangle GEA</math> are also similar.<br />
<br />
<math>DA</math> is 9, therefore <math>EA</math> must equal 5. Similarly, <math>CH</math> must equal 3. <br />
<br />
Because <math>GCH</math> and <math>GEA</math> are similar, the ratio of <math>CH\; =\; 3</math> and <math>EA\; =\; 5</math>, must also hold true for <math>GH</math> and <math>HA</math>. <math>\frac{GH}{GA} = \frac{3}{5}</math>, so <math>HA</math> is <math>\frac{2}{5}</math> of <math>GA</math>. By Pythagorean theorem, <math>(HA)^2\; =\; (HB)^2\; +\; (BA)^2\;...\;HA=10</math>.<br />
<br />
<math>HA\: =\: 10 =\: \frac{2}{5}*(GA)</math>. <br />
<br />
<math>GA\: =\: 25.</math><br />
<br />
So <math>\frac{GA}{HA}\: =\: \frac{GF}{BA}</math>.<br />
<br />
<math>\frac{25}{10}\: =\: \frac{GF}{8}</math>.<br />
<br />
Therefore <math>GF= \boxed{\mathrm{(B)}\ 20}</math>.<br />
<br />
=== Solution 2 ===<br />
Since <math>ABCD</math> is a rectangle, <math>CD=AB=8</math>. <br />
<br />
Since <math>ABCD</math> is a rectangle and <math>GF \perp AF</math>, <math>\angle GFE = \angle CDE = \angle ABC = 90^\circ </math>. <br />
<br />
Since <math>ABCD</math> is a rectangle, <math>AD || BC</math>. <br />
<br />
So, <math>AH</math> is a [[transversal]], and <math>\angle GAF = \angle AHB</math>. <br />
<br />
This is sufficient to prove that <math> GFE \approx CDE</math> and <math> GFA \approx ABH</math>.<br />
<br />
Using ratios: <br />
<br />
<math>\frac{GF}{FE}=\frac{CD}{DE}</math><br />
<br />
<math>\frac{GF}{FD+4}=\frac{8}{4}=2</math><br />
<br />
<math>GF=2 \cdot (FD+4)=2 \cdot FD+8</math><br />
<br />
<math>\frac{GF}{FA}=\frac{AB}{BH}</math><br />
<br />
<math>\frac{GF}{FD+9}=\frac{8}{6}=\frac{4}{3}</math><br />
<br />
<math>GF=\frac{4}{3} \cdot (FD+9)=\frac{4}{3} \cdot FD+12</math><br />
<br />
Since <math>GF</math> can't have 2 different lengths, both expressions for <math>GF</math> must be equal. <br />
<br />
<math>2 \cdot FD+8=\frac{4}{3} \cdot FD+12</math><br />
<br />
<math>\frac{2}{3} \cdot FD=4</math><br />
<br />
<math>FD=6</math><br />
<br />
<math>GF=2 \cdot FD+8=2\cdot6+8=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 3 ===<br />
Since <math>ABCD</math> is a rectangle, <math>CH=3</math>, <math>EA=5</math>, and <math>CD=8</math>. From the [[Pythagorean Theorem]], <math>CE^2=CD^2+DE^2=80\Rightarrow CE=4\sqrt{5}</math>.<br />
==== solution 4 ====<br />
Statement: <math>GCH \approx GEA</math><br />
<br />
Proof: <math>\angle CGH=\angle EGA</math>, obviously.<br />
<br />
<cmath>\begin{eqnarray*}<br />
\angle HCE&=&180^{\circ}-\angle CHG\\<br />
\angle DCE&=&\angle CHG-90^{\circ}\\<br />
\angle CEED&=&180-\angle CHG\\<br />
\angle GEA&=&\angle GCH<br />
\end{eqnarray*}</cmath><br />
<br />
Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.<br />
<br />
<br />
<br />
Let <math>GC=x</math>.<br />
<br />
<cmath>\begin{eqnarray*}<br />
\dfrac{x}{3}&=&\dfrac{x+4\sqrt{5}}{5}\\<br />
5x&=&3x+12\sqrt{5}\\<br />
2x&=&12\sqrt{5}\\<br />
x&=&6\sqrt{5}<br />
\end{eqnarray*}</cmath><br />
<br />
Also, <math>\triangle GFE\approx \triangle CDE</math>, therefore<br />
<br />
<cmath>\dfrac{8}{4\sqrt{5}}=\dfrac{GF}{10\sqrt{5}}</cmath><br />
<br />
We can multiply both sides by <math>\sqrt{5}</math> to get that <math>GF</math> is twice of 10, or <math>\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 4 ===<br />
We extend <math>BC</math> such that it intersects <math>GF</math> at <math>X</math>. Since <math>ABCD</math> is a rectangle, it follows that <math>CD=8</math>, therefore, <math>XF=8</math>. Let <math>GX=y</math>. From the similarity of triangles <math>GCH</math> and <math>GEA</math>, we have the ratio <math>3:5</math> (as <math>CH=9-6=3</math>, and <math>EA=9-4=5</math>). <math>GX</math> and <math>GF</math> are the altitudes of <math>GCH</math> and <math>GEA</math>, respectively. Thus, <math>y:y+8 = 3:5</math>, from which we have <math>y=12</math>, thus <math>GF=y+8=12+8=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
=== Solution 5 ===<br />
<br />
Since <math>GF\perp AF</math> and <math>AF\perp CD, </math> we have <math>GF\parallel CD\parallel AB.</math> Thus, <math>\triangle CDE\sim GFE.</math> Suppose <math>GF=x</math> and <math>FD=y.</math> Thus, we have <math>\dfrac{x}{8}=\dfrac{y+4}{4}.</math> Additionally, now note that <math>\triangle GAF\sim AHB,</math> which is pretty obvious from insight, but can be proven by AA with extending <math>BH</math> to meet <math>GF.</math> From this new pair of similar triangles, we have <math>\dfrac{x}{8}=\dfrac{y+9}{6}.</math> Therefore, we have by combining those two equations, <cmath>\dfrac{y+9}{6}=\dfrac{y+4}{4}.</cmath> Solving, we have <math>y=6,</math> and therefore <math>x=\boxed{\mathrm{(B)}\ 20}</math><br />
<br />
== See Also ==<br />
{{AMC10 box|year=2003|ab=A|num-b=21|num-a=23}}<br />
<br />
[[Category:Introductory Geometry Problems]]<br />
{{MAA Notice}}</div>Justin.ziyu.sunhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_20&diff=1127962003 AMC 8 Problems/Problem 202019-12-13T00:45:22Z<p>Justin.ziyu.sun: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12</math><br />
<br />
==Solution 1==<br />
<br />
Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between <math>4</math> and <math>5</math> is <math>30</math> degrees (since it is 1/12 of a full circle, 360). By <math>4:20</math>, the hour hand would have moved <math>\frac{1}{3}</math> way from 4 to 5 since <math>\frac{20}{60}</math> is reducible to <math>\frac{1}{3}</math>. One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is <math>\boxed{10, D}</math>, and we are done.<br />
<br />
==Solution 2==<br />
<br />
You may also use the formula <math>30h-5.5m</math> to yield a result in only ~10 seconds<br />
<br />
-sub_math<br />
<br />
==See Also==<br />
{{AMC8 box|year=2003|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Justin.ziyu.sunhttps://artofproblemsolving.com/wiki/index.php?title=2003_AMC_8_Problems/Problem_20&diff=1127952003 AMC 8 Problems/Problem 202019-12-13T00:44:49Z<p>Justin.ziyu.sun: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
What is the measure of the acute angle formed by the hands of the clock at 4:20 PM?<br />
<br />
<math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 12</math><br />
<br />
==Solution==<br />
<br />
Imagine the clock as a circle. The minute hand will be at the 4 at 20 minutes past the hour. The central angle formed between <math>4</math> and <math>5</math> is <math>30</math> degrees (since it is 1/12 of a full circle, 360). By <math>4:20</math>, the hour hand would have moved <math>\frac{1}{3}</math> way from 4 to 5 since <math>\frac{20}{60}</math> is reducible to <math>\frac{1}{3}</math>. One third of the way from 4 to 5 is one third of 30 degrees, which is 10 degrees past the 4. Recall that the minute hand is at the 4, so the angle between them is <math>\boxed{10, D}</math>, and we are done.<br />
<br />
==Solution 2==<br />
<br />
You may also use the formula <math>30h-5.5m</math> to yield a result in only ~10 seconds<br />
<br />
-sub_math<br />
<br />
==See Also==<br />
{{AMC8 box|year=2003|num-b=19|num-a=21}}<br />
{{MAA Notice}}</div>Justin.ziyu.sun