https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Awesomeming327.&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T16:55:24ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_12&diff=1886802018 AIME II Problems/Problem 122023-02-05T23:58:57Z<p>Awesomeming327.: /* Solution 9 (Mindless Law of Cosines Bash) */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>.<br />
<br />
==Diagram==<br />
<br />
Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>.<br />
<asy><br />
defaultpen(fontsize(14)+0.6); unitsize(12);<br />
<br />
real x=11.25;<br />
pair B1=origin, D1=(x,0), C1=IP(CR(B1,14),CR(D1,10)), A1=OP(CR(B1,10),CR(D1,2*sqrt(65))), P1=extension(A1,C1,B1,D1);<br />
<br />
pair A=origin, C=(length(C1-A1),0), B=IP(CR(A,10),CR(C,14)), D=OP(CR(A,2*sqrt(65)),CR(C,10)), P=extension(A,C,B,D);<br />
<br />
draw(A--B--C--D--A);<br />
draw(A--C^^B--D,gray+0.4);<br />
dot("$A$",A,dir(A-P)); dot("$B$",B,dir(B-P)); dot("$C$",C,dir(C-P)); dot("$D$",D,dir(D-P)); dot("$P$",P,dir(230));<br />
<br />
pen p=fontsize(10);<br />
label("$10$",A--B,up,p); label("$10$", C--D, 2*right,p); label("$14$", B--C, N,p); label("$2\sqrt{65}$", A--D, SW,p); label("$x$", A--P,down,p); label("$\rho x$", P--C,down,p); label("$\Delta$",(A+B+P)/3, right,p); label("$\Lambda$",(A+D+P)/3, right,p);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>. We easily get <math>[PBC]=\rho \Delta</math> and <math>[PCD]=\rho\Lambda</math>. <br />
<br />
We are given that <math>[ABP] +[PCD] = [PBC]+[ADP]</math>, which we can now write as <cmath>\Delta + \rho\Lambda = \rho\Delta + \Lambda \qquad \Longrightarrow \qquad \Delta -\Lambda = \rho (\Delta -\Lambda).</cmath> Either <math>\Delta = \Lambda</math> or <math>\rho=1</math>. The former would imply that <math>ABCD</math> is a parallelogram, which it isn't; therefore we conclude <math>\rho=1</math> and <math>P</math> is the midpoint of <math>AC</math>. Let <math>\angle BAD = \theta</math> and <math>\angle BCD = \phi</math>. Then <math>[ABCD]=2\cdot [BCD]=140\sin\phi</math>. On one hand, since <math>[ABD]=[BCD]</math>, we have <cmath>\begin{align}\sqrt{65}\sin\theta = 7\sin\phi \quad \implies \quad 16+49\cos^2\phi = 65\cos^2\theta\end{align}</cmath>whereas, on the other hand, using cosine formula to get the length of <math>BD</math>, we get <cmath>10^2+4\cdot 65 - 40\sqrt{65}\cos\theta = 10^2+14^2-280\cos\phi</cmath><cmath>\begin{align}\tag{2}\implies \qquad 65\cos^2\theta = \left(7\cos\phi+ \frac{8}{5}\right)^2\end{align}</cmath>Eliminating <math>\cos\theta</math> in the above two equations and solving for <math>\cos\phi</math> we get<cmath>\cos\phi = \frac{3}{5}\qquad \implies \qquad \sin\phi = \frac{4}{5}</cmath>which finally yields <math>[ABCD]=2\cdot [BCD] = 140\sin\phi = 112</math>.<br />
<br />
==Solution 2==<br />
<br />
For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes<br />
<br />
<cmath>\frac{h_1}{2}AP + \frac{h_2}{2}CP = \frac{h_1}{2}CP + \frac{h_2}{2}AP \rightarrow \left(h_1 - h_2\right)AP = \left(h_1 - h_2\right)CP \rightarrow AP = CP</cmath><br />
<br />
What an important finding! Note that the opposite sides <math>\overline{AB}</math> and <math>\overline{CD}</math> have equal length, and note that diagonal <math>\overline{DB}</math> bisects diagonal <math>\overline{AC}</math>. This is very similar to what happens if <math>ABCD</math> were a parallelogram with <math>AB = CD = 10</math>, so let's extend <math>\overline{DB}</math> to point <math>E</math>, such that <math>AECD</math> is a parallelogram. In other words, <cmath>AE = CD = 10</cmath> and <cmath>EC = DA = 2\sqrt{65}</cmath> Now, let's examine <math>\triangle ABE</math>. Since <math>AB = AE = 10</math>, the triangle is isosceles, and <math>\angle ABE \cong \angle AEB</math>. Note that in parallelogram <math>AECD</math>, <math>\angle AED</math> and <math>\angle CDE</math> are congruent, so <math>\angle ABE \cong \angle CDE</math> and thus <cmath>\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB</cmath> Define <math>\alpha := \text{m}\angle CDB</math>, so <math>180^\circ - \alpha = \text{m}\angle ABD</math>. <br />
<br />
We use the Law of Cosines on <math>\triangle DAB</math> and <math>\triangle CDB</math>:<br />
<br />
<cmath>\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha</cmath><br />
<br />
<cmath>14^2 = 10^2 + BD^2 - 20BD\cos\alpha</cmath><br />
<br />
Subtracting the second equation from the first yields<br />
<br />
<cmath>260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}</cmath><br />
<br />
This means that dropping an altitude from <math>B</math> to some foot <math>Q</math> on <math>\overline{CD}</math> gives <math>DQ = \frac{8}{5}</math> and therefore <math>CQ = \frac{42}{5}</math>. Seeing that <math>CQ = \frac{3}{5}\cdot BC</math>, we conclude that <math>\triangle QCB</math> is a 3-4-5 right triangle, so <math>BQ = \frac{56}{5}</math>. Then, the area of <math>\triangle BCD</math> is <math>\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56</math>. Since <math>AP = CP</math>, points <math>A</math> and <math>C</math> are equidistant from <math>\overline{BD}</math>, so <cmath>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</cmath> and hence <cmath>\left[ABCD\right] = 56 + 56 = \boxed{112}</cmath> -kgator<br />
<br />
<br />
Just to be complete -- <math>h_1</math> and <math>h_2</math> can actually be equal. In this case, <math>AP \neq CP</math>, but <math>BP</math> must be equal to <math>DP</math>. We get the same result. -Mathdummy.<br />
<br />
==Solution 3 (Another way to get the middle point)==<br />
<br />
So, let the area of <math>4</math> triangles <math>\triangle {ABP}=S_{1}</math>, <math>\triangle {BCP}=S_{2}</math>, <math>\triangle {CDP}=S_{3}</math>, <math>\triangle {DAP}=S_{4}</math>. Suppose <math>S_{1}>S_{3}</math> and <math>S_{2}>S_{4}</math>, then it is easy to show that <cmath>S_{1}\cdot S_{3}=S_{2}\cdot S_{4}.</cmath> Also, because <cmath>S_{1}+S_{3}=S_{2}+S_{4},</cmath> we will have <cmath>(S_{1}+S_{3})^2=(S_{2}+S_{4})^2.</cmath> So <cmath>(S_{1}+S_{3})^2=S_{1}^2+S_{3}^2+2\cdot S_{1}\cdot S_{3}=(S_{2}+S_{4})^2=S_{2}^2+S_{4}^2+2\cdot S_{2}\cdot S_{4}.</cmath> So <cmath>S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2.</cmath> So <cmath>S_{1}^2+S_{3}^2-2\cdot S_{1}\cdot S_{3}=S_{2}^2+S_{4}^2-2\cdot S_{2}\cdot S_{4}.</cmath> So <cmath>(S_{1}-S_{3})^2=(S_{2}-S_{4})^2.</cmath> As a result, <cmath>S_{1}-S_{3}=S_{2}-S_{4}.</cmath> Then, we have <cmath>S_{1}+S_{4}=S_{2}+S_{3}.</cmath> Combine the condition <cmath>S_{1}+S_{3}=S_{2}+S_{4},</cmath> we can find out that <cmath>S_{3}=S_{4},</cmath> so <math>P</math> is the midpoint of <math>\overline {AC}</math><br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
==Solution 4 (With yet another way to get the middle point)==<br />
<br />
Denote <math>\angle APB</math> by <math>\alpha</math>. Then <math>\sin(\angle APB)=\sin \alpha = \sin(\angle APD)</math>.<br />
Using the formula for the area of a triangle, we get <cmath>\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha , </cmath> <br />
so <cmath>(AP-CP)(BP-DP)=0</cmath><br />
Hence <math>AP=CP</math> (note that <math>BP=DP</math> makes no difference here).<br />
Now, assume that <math>AP=CP=x</math>, <math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for <math>\triangle APB</math> and <math>\triangle BPC</math>, it is clear that <cmath>x^2+y^2-100=2 xy \cdot \cos{APB}=-(2 xy \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196) </cmath> or <cmath>\begin{align}x^2+y^2=148\end{align}.</cmath><br />
Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>\begin{align}\tag{2}x^2+z^2=180\end{align}.</cmath> It follows that <cmath>\begin{align}\tag{3}z^2-y^2=32\end{align}.</cmath> Since <math>\sin\alpha=\sqrt{1-\cos^2\alpha}</math>, <cmath>\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}</cmath> which simplifies to <cmath>\frac{48^2}{y^2}=\frac{80^2}{z^2} \qquad \Longrightarrow \qquad 5y=3z.</cmath> Plugging this back to equations <math>(1)</math>, <math>(2)</math>, and <math>(3)</math>, it can be solved that <math>x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}</math>. Then, the area of the quadrilateral is <cmath>x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}</cmath><br />
--Solution by MicGu<br />
<br />
==Solution 5==<br />
As in all other solutions, we can first find that either <math>AP=CP</math> or <math>BP=DP</math>, but it's an AIME problem, we can take <math>AP=CP</math>, and assume the other choice will lead to the same result (which is true). <br />
<br />
From <math>AP=CP</math>, we have <math>[DAP]=[DCP]</math>, and <math>[BAP]=[BCP] \implies [ABD] = [CBD]</math>, therefore,<br />
<cmath>\begin{align}<br />
\nonumber \frac 12 \cdot AB\cdot AD\sin A &= \frac 12 \cdot BC\cdot CD\sin C \\<br />
\Longrightarrow \hspace{1in} 7\sin C &= \sqrt{65}\sin A <br />
\end{align}</cmath><br />
By Law of Cosines,<br />
<cmath>\begin{align}<br />
\nonumber 10^2+14^2-2\cdot 10\cdot 14\cos C &= 10^2+4\cdot 65-2\cdot 10\cdot 2\sqrt{65}\cos A \\<br />
\Longrightarrow \hspace{1in} -\frac 85 - 7\cos C &= \sqrt{65}\cos A \tag{2}<br />
\end{align}</cmath><br />
Square <math>(1)</math> and <math>(2)</math>, and add them, to get<br />
<cmath>\left(\frac 85\right)^2 + 2\cdot \frac 85 \cdot 7\cos C + 7^2 = 65 </cmath><br />
Solve, <math>\cos C = 3/5 \implies \sin C = 4/5</math>,<br />
<cmath>[ABCD] = 2[BCD] = BC\cdot CD\cdot \sin C = 14\cdot 10\cdot \frac 45 = \boxed{112}</cmath><br />
-Mathdummy<br />
<br />
==Solution 6==<br />
Either <math>PA=PC</math> or <math>PD=PB</math>. Let <math>PD=PB=s</math>. Applying Stewart's Theorem on <math>\triangle ABD</math> and <math>\triangle BCD</math>, dividing by <math>2s</math> and rearranging, <cmath>\tag{1}CP^2+s^2=148</cmath> <cmath>\tag{2}AP^2+s^2=180</cmath> Applying Stewart on <math>\triangle CAB</math> and <math>\triangle CAD</math>, <cmath>\tag{3} 5CP^2=3AP^2</cmath> Substituting equations 1 and 2 into 3 and rearranging, <math>s=BP=PD\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}</math> . By Law of Cosines on <math>\triangle APB</math>, <math>\cos(\angle APB)=\frac{4\sqrt{65}}{65}</math> so <math>\sin(\angle APB)=\sin(\angle BPC)=\sin(\angle CPD)=\sin(\angle DPA)=\frac{7\sqrt{65}}{65}</math>. Using <math>[\triangle ABC]=\frac{ab\sin(\angle C)}{2}</math> to find unknown areas, <math>[ABCD]=[\triangle APB]+[\triangle BPC]+[\triangle CPD]+[\triangle DPA]=\boxed{112}</math>.<br />
<br />
-Solution by Gart<br />
<br />
==Solution 7==<br />
Now we prove P is the midpoint of <math>BD</math>. Denote the height from <math>B</math> to <math>AC</math> as <math>h_1</math>, height from <math>D</math> to <math>AC</math> as <math>h_2</math>.According to the problem, <math>AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2 </math> implies <math> h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2 </math>. Then according to basic congruent triangles we get <math>BP=DP</math><br />
Firstly, denote that <math>CP=a,BP=b,CP=c,AP=d</math>. Applying Stewart theorem, getting that <math>100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c</math>, denote <math>b=5x,c=3x</math><br />
Applying Stewart Theorem, getting <math>260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)</math> solve for a, getting <math>a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}</math><br />
Now everything is clear, we can find <math>cos\angle{BPA}=\frac{4}{\sqrt{65}}</math> using LOC, <math>sin\angle{BPA}=\frac{7\sqrt{65}}{65}</math>, the whole area is <math>\sqrt{130}*8\sqrt{2}*\frac{7\sqrt{65}}{65}=\boxed{112}</math><br />
<br />
~bluesoul<br />
<br />
==Solution 8 (Simple Geometry)==<br />
[[File:AIME-II-2018-12.png|400px|right]]<br />
<math>BP = PD</math> as in another solutions.<br />
<br />
Let <math>D'</math> be the reflection of <math>D</math> across <math>C</math>.<br />
Let points <math>E, E',</math> and <math>H</math> be the foot of perpendiculars on <math>AC</math> from <math>D,D'</math>, and <math>B</math> respectively.<br />
<cmath>\begin{align*}<br />
&\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\<br />
\Rightarrow &\qquad \triangle ABH = \triangle CDE = \triangle CD'E' \\<br />
\Rightarrow &\qquad \angle BAC = \angle ACD' \\<br />
\Rightarrow &\qquad \triangle ABC = \triangle AD'C \\<br />
\Rightarrow &\qquad BC = AD'.<br />
\end{align*}</cmath><br />
The area of quadrilateral <math>ABCD</math> is equal to the area of triangle <math>ADD'</math> with sides <math>AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20</math>.<br />
The semiperimeter is <math>s = 17 + \sqrt{65},</math> the area <cmath>[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.</cmath><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
<br />
==Solution 9 (Mindless Law of Cosines Bash)==<br />
<br />
Use your favorite method to get that <math>P</math> is the midpoint of one of the two diagonals (suppose it's the midpoint of <math>\overline{AC}</math>). From here, let <math>x=AP=PC, y=BP, z=PD, a=\cos\theta</math> where <math>\theta</math> is the angle that the diagonals make. Then we have a system of four equations:<br />
<br />
\begin{align*}<br />
x^2+y^2+2xya=100 \\<br />
z^2+x^2+2xza=100 \\<br />
x^2+y^2-2xya=196 \\<br />
x^2+z^2-2xza=260 \\<br />
\end{align*}<br />
<br />
From these equations we get that <br />
\begin{align*}<br />
xya=-24 \\<br />
xza=-40 \\<br />
x^2+y^2-48=10 \\<br />
x^2+z^2-80=10<br />
\end{align*}<br />
From here we can see that <math>\frac{z}{y}={5}{3}, z^2-y^2=32,</math> so <math>z=5\sqrt{2}, y=3\sqrt{2}.</math> Furthermore, this implies <math>x=\sqrt{130}</math> and <math>xa=-4\sqrt{2},</math> which implies <math>a=\cos\theta=\frac{4}{\sqrt{65}}.</math> Then note that the area of the quadrilateral is <cmath>\frac{1}{2}\sin\theta (xy+xz+xz+xy)=\sin\theta (\sqrt{130}\cdot 3\sqrt{2}+\sqrt{130} \cdot 5\sqrt{2})=7\cdot (3\cdot 2+5\cdot 2)=7(6+10)=7\cdot 16=\boxed{112}.</cmath><br />
<br />
~Dhillonr25<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=II|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Awesomeming327.https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_II_Problems/Problem_12&diff=1886792018 AIME II Problems/Problem 122023-02-05T23:58:32Z<p>Awesomeming327.: /* Solution 9 (Mindless Law of Cosines Bash) */</p>
<hr />
<div>==Problem==<br />
<br />
Let <math>ABCD</math> be a convex quadrilateral with <math>AB = CD = 10</math>, <math>BC = 14</math>, and <math>AD = 2\sqrt{65}</math>. Assume that the diagonals of <math>ABCD</math> intersect at point <math>P</math>, and that the sum of the areas of triangles <math>APB</math> and <math>CPD</math> equals the sum of the areas of triangles <math>BPC</math> and <math>APD</math>. Find the area of quadrilateral <math>ABCD</math>.<br />
<br />
==Diagram==<br />
<br />
Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>.<br />
<asy><br />
defaultpen(fontsize(14)+0.6); unitsize(12);<br />
<br />
real x=11.25;<br />
pair B1=origin, D1=(x,0), C1=IP(CR(B1,14),CR(D1,10)), A1=OP(CR(B1,10),CR(D1,2*sqrt(65))), P1=extension(A1,C1,B1,D1);<br />
<br />
pair A=origin, C=(length(C1-A1),0), B=IP(CR(A,10),CR(C,14)), D=OP(CR(A,2*sqrt(65)),CR(C,10)), P=extension(A,C,B,D);<br />
<br />
draw(A--B--C--D--A);<br />
draw(A--C^^B--D,gray+0.4);<br />
dot("$A$",A,dir(A-P)); dot("$B$",B,dir(B-P)); dot("$C$",C,dir(C-P)); dot("$D$",D,dir(D-P)); dot("$P$",P,dir(230));<br />
<br />
pen p=fontsize(10);<br />
label("$10$",A--B,up,p); label("$10$", C--D, 2*right,p); label("$14$", B--C, N,p); label("$2\sqrt{65}$", A--D, SW,p); label("$x$", A--P,down,p); label("$\rho x$", P--C,down,p); label("$\Delta$",(A+B+P)/3, right,p); label("$\Lambda$",(A+D+P)/3, right,p);<br />
</asy><br />
<br />
==Solution 1==<br />
<br />
Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>. We easily get <math>[PBC]=\rho \Delta</math> and <math>[PCD]=\rho\Lambda</math>. <br />
<br />
We are given that <math>[ABP] +[PCD] = [PBC]+[ADP]</math>, which we can now write as <cmath>\Delta + \rho\Lambda = \rho\Delta + \Lambda \qquad \Longrightarrow \qquad \Delta -\Lambda = \rho (\Delta -\Lambda).</cmath> Either <math>\Delta = \Lambda</math> or <math>\rho=1</math>. The former would imply that <math>ABCD</math> is a parallelogram, which it isn't; therefore we conclude <math>\rho=1</math> and <math>P</math> is the midpoint of <math>AC</math>. Let <math>\angle BAD = \theta</math> and <math>\angle BCD = \phi</math>. Then <math>[ABCD]=2\cdot [BCD]=140\sin\phi</math>. On one hand, since <math>[ABD]=[BCD]</math>, we have <cmath>\begin{align}\sqrt{65}\sin\theta = 7\sin\phi \quad \implies \quad 16+49\cos^2\phi = 65\cos^2\theta\end{align}</cmath>whereas, on the other hand, using cosine formula to get the length of <math>BD</math>, we get <cmath>10^2+4\cdot 65 - 40\sqrt{65}\cos\theta = 10^2+14^2-280\cos\phi</cmath><cmath>\begin{align}\tag{2}\implies \qquad 65\cos^2\theta = \left(7\cos\phi+ \frac{8}{5}\right)^2\end{align}</cmath>Eliminating <math>\cos\theta</math> in the above two equations and solving for <math>\cos\phi</math> we get<cmath>\cos\phi = \frac{3}{5}\qquad \implies \qquad \sin\phi = \frac{4}{5}</cmath>which finally yields <math>[ABCD]=2\cdot [BCD] = 140\sin\phi = 112</math>.<br />
<br />
==Solution 2==<br />
<br />
For reference, <math>2\sqrt{65} \approx 16</math>, so <math>\overline{AD}</math> is the longest of the four sides of <math>ABCD</math>. Let <math>h_1</math> be the length of the altitude from <math>B</math> to <math>\overline{AC}</math>, and let <math>h_2</math> be the length of the altitude from <math>D</math> to <math>\overline{AC}</math>. Then, the triangle area equation becomes<br />
<br />
<cmath>\frac{h_1}{2}AP + \frac{h_2}{2}CP = \frac{h_1}{2}CP + \frac{h_2}{2}AP \rightarrow \left(h_1 - h_2\right)AP = \left(h_1 - h_2\right)CP \rightarrow AP = CP</cmath><br />
<br />
What an important finding! Note that the opposite sides <math>\overline{AB}</math> and <math>\overline{CD}</math> have equal length, and note that diagonal <math>\overline{DB}</math> bisects diagonal <math>\overline{AC}</math>. This is very similar to what happens if <math>ABCD</math> were a parallelogram with <math>AB = CD = 10</math>, so let's extend <math>\overline{DB}</math> to point <math>E</math>, such that <math>AECD</math> is a parallelogram. In other words, <cmath>AE = CD = 10</cmath> and <cmath>EC = DA = 2\sqrt{65}</cmath> Now, let's examine <math>\triangle ABE</math>. Since <math>AB = AE = 10</math>, the triangle is isosceles, and <math>\angle ABE \cong \angle AEB</math>. Note that in parallelogram <math>AECD</math>, <math>\angle AED</math> and <math>\angle CDE</math> are congruent, so <math>\angle ABE \cong \angle CDE</math> and thus <cmath>\text{m}\angle ABD = 180^\circ - \text{m}\angle CDB</cmath> Define <math>\alpha := \text{m}\angle CDB</math>, so <math>180^\circ - \alpha = \text{m}\angle ABD</math>. <br />
<br />
We use the Law of Cosines on <math>\triangle DAB</math> and <math>\triangle CDB</math>:<br />
<br />
<cmath>\left(2\sqrt{65}\right)^2 = 10^2 + BD^2 - 20BD\cos\left(180^\circ - \alpha\right) = 100 + BD^2 + 20BD\cos\alpha</cmath><br />
<br />
<cmath>14^2 = 10^2 + BD^2 - 20BD\cos\alpha</cmath><br />
<br />
Subtracting the second equation from the first yields<br />
<br />
<cmath>260 - 196 = 40BD\cos\alpha \rightarrow BD\cos\alpha = \frac{8}{5}</cmath><br />
<br />
This means that dropping an altitude from <math>B</math> to some foot <math>Q</math> on <math>\overline{CD}</math> gives <math>DQ = \frac{8}{5}</math> and therefore <math>CQ = \frac{42}{5}</math>. Seeing that <math>CQ = \frac{3}{5}\cdot BC</math>, we conclude that <math>\triangle QCB</math> is a 3-4-5 right triangle, so <math>BQ = \frac{56}{5}</math>. Then, the area of <math>\triangle BCD</math> is <math>\frac{1}{2}\cdot 10 \cdot \frac{56}{5} = 56</math>. Since <math>AP = CP</math>, points <math>A</math> and <math>C</math> are equidistant from <math>\overline{BD}</math>, so <cmath>\left[\triangle ABD\right] = \left[\triangle CBD\right] = 56</cmath> and hence <cmath>\left[ABCD\right] = 56 + 56 = \boxed{112}</cmath> -kgator<br />
<br />
<br />
Just to be complete -- <math>h_1</math> and <math>h_2</math> can actually be equal. In this case, <math>AP \neq CP</math>, but <math>BP</math> must be equal to <math>DP</math>. We get the same result. -Mathdummy.<br />
<br />
==Solution 3 (Another way to get the middle point)==<br />
<br />
So, let the area of <math>4</math> triangles <math>\triangle {ABP}=S_{1}</math>, <math>\triangle {BCP}=S_{2}</math>, <math>\triangle {CDP}=S_{3}</math>, <math>\triangle {DAP}=S_{4}</math>. Suppose <math>S_{1}>S_{3}</math> and <math>S_{2}>S_{4}</math>, then it is easy to show that <cmath>S_{1}\cdot S_{3}=S_{2}\cdot S_{4}.</cmath> Also, because <cmath>S_{1}+S_{3}=S_{2}+S_{4},</cmath> we will have <cmath>(S_{1}+S_{3})^2=(S_{2}+S_{4})^2.</cmath> So <cmath>(S_{1}+S_{3})^2=S_{1}^2+S_{3}^2+2\cdot S_{1}\cdot S_{3}=(S_{2}+S_{4})^2=S_{2}^2+S_{4}^2+2\cdot S_{2}\cdot S_{4}.</cmath> So <cmath>S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2.</cmath> So <cmath>S_{1}^2+S_{3}^2-2\cdot S_{1}\cdot S_{3}=S_{2}^2+S_{4}^2-2\cdot S_{2}\cdot S_{4}.</cmath> So <cmath>(S_{1}-S_{3})^2=(S_{2}-S_{4})^2.</cmath> As a result, <cmath>S_{1}-S_{3}=S_{2}-S_{4}.</cmath> Then, we have <cmath>S_{1}+S_{4}=S_{2}+S_{3}.</cmath> Combine the condition <cmath>S_{1}+S_{3}=S_{2}+S_{4},</cmath> we can find out that <cmath>S_{3}=S_{4},</cmath> so <math>P</math> is the midpoint of <math>\overline {AC}</math><br />
<br />
~Solution by <math>BladeRunnerAUG</math> (Frank FYC)<br />
<br />
==Solution 4 (With yet another way to get the middle point)==<br />
<br />
Denote <math>\angle APB</math> by <math>\alpha</math>. Then <math>\sin(\angle APB)=\sin \alpha = \sin(\angle APD)</math>.<br />
Using the formula for the area of a triangle, we get <cmath>\frac{1}{2} (AP\cdot BP+ CP\cdot DP)\sin\alpha=\frac{1}{2}(AP\cdot DP+ CP\cdot BP)\sin\alpha , </cmath> <br />
so <cmath>(AP-CP)(BP-DP)=0</cmath><br />
Hence <math>AP=CP</math> (note that <math>BP=DP</math> makes no difference here).<br />
Now, assume that <math>AP=CP=x</math>, <math>BP=y</math>, and <math>DP=z</math>. Using the cosine rule for <math>\triangle APB</math> and <math>\triangle BPC</math>, it is clear that <cmath>x^2+y^2-100=2 xy \cdot \cos{APB}=-(2 xy \cdot \cos{(\pi-CPB)})=-(x^2+y^2-196) </cmath> or <cmath>\begin{align}x^2+y^2=148\end{align}.</cmath><br />
Likewise, using the cosine rule for triangles <math>APD</math> and <math>CPD</math>, <cmath>\begin{align}\tag{2}x^2+z^2=180\end{align}.</cmath> It follows that <cmath>\begin{align}\tag{3}z^2-y^2=32\end{align}.</cmath> Since <math>\sin\alpha=\sqrt{1-\cos^2\alpha}</math>, <cmath>\sqrt{1-\frac{(x^2+y^2-100)^2}{4x^2y^2}}=\sqrt{1-\frac{(x^2+z^2-260)^2}{4x^2z^2}}</cmath> which simplifies to <cmath>\frac{48^2}{y^2}=\frac{80^2}{z^2} \qquad \Longrightarrow \qquad 5y=3z.</cmath> Plugging this back to equations <math>(1)</math>, <math>(2)</math>, and <math>(3)</math>, it can be solved that <math>x=\sqrt{130},y=3\sqrt{2},z=5\sqrt{2}</math>. Then, the area of the quadrilateral is <cmath>x(y+z)\sin\alpha=\sqrt{130}\cdot8\sqrt{2}\cdot\frac{14}{\sqrt{260}}=\boxed{112}</cmath><br />
--Solution by MicGu<br />
<br />
==Solution 5==<br />
As in all other solutions, we can first find that either <math>AP=CP</math> or <math>BP=DP</math>, but it's an AIME problem, we can take <math>AP=CP</math>, and assume the other choice will lead to the same result (which is true). <br />
<br />
From <math>AP=CP</math>, we have <math>[DAP]=[DCP]</math>, and <math>[BAP]=[BCP] \implies [ABD] = [CBD]</math>, therefore,<br />
<cmath>\begin{align}<br />
\nonumber \frac 12 \cdot AB\cdot AD\sin A &= \frac 12 \cdot BC\cdot CD\sin C \\<br />
\Longrightarrow \hspace{1in} 7\sin C &= \sqrt{65}\sin A <br />
\end{align}</cmath><br />
By Law of Cosines,<br />
<cmath>\begin{align}<br />
\nonumber 10^2+14^2-2\cdot 10\cdot 14\cos C &= 10^2+4\cdot 65-2\cdot 10\cdot 2\sqrt{65}\cos A \\<br />
\Longrightarrow \hspace{1in} -\frac 85 - 7\cos C &= \sqrt{65}\cos A \tag{2}<br />
\end{align}</cmath><br />
Square <math>(1)</math> and <math>(2)</math>, and add them, to get<br />
<cmath>\left(\frac 85\right)^2 + 2\cdot \frac 85 \cdot 7\cos C + 7^2 = 65 </cmath><br />
Solve, <math>\cos C = 3/5 \implies \sin C = 4/5</math>,<br />
<cmath>[ABCD] = 2[BCD] = BC\cdot CD\cdot \sin C = 14\cdot 10\cdot \frac 45 = \boxed{112}</cmath><br />
-Mathdummy<br />
<br />
==Solution 6==<br />
Either <math>PA=PC</math> or <math>PD=PB</math>. Let <math>PD=PB=s</math>. Applying Stewart's Theorem on <math>\triangle ABD</math> and <math>\triangle BCD</math>, dividing by <math>2s</math> and rearranging, <cmath>\tag{1}CP^2+s^2=148</cmath> <cmath>\tag{2}AP^2+s^2=180</cmath> Applying Stewart on <math>\triangle CAB</math> and <math>\triangle CAD</math>, <cmath>\tag{3} 5CP^2=3AP^2</cmath> Substituting equations 1 and 2 into 3 and rearranging, <math>s=BP=PD\sqrt{130}, CP=3\sqrt{2}, PA=5\sqrt{2}</math> . By Law of Cosines on <math>\triangle APB</math>, <math>\cos(\angle APB)=\frac{4\sqrt{65}}{65}</math> so <math>\sin(\angle APB)=\sin(\angle BPC)=\sin(\angle CPD)=\sin(\angle DPA)=\frac{7\sqrt{65}}{65}</math>. Using <math>[\triangle ABC]=\frac{ab\sin(\angle C)}{2}</math> to find unknown areas, <math>[ABCD]=[\triangle APB]+[\triangle BPC]+[\triangle CPD]+[\triangle DPA]=\boxed{112}</math>.<br />
<br />
-Solution by Gart<br />
<br />
==Solution 7==<br />
Now we prove P is the midpoint of <math>BD</math>. Denote the height from <math>B</math> to <math>AC</math> as <math>h_1</math>, height from <math>D</math> to <math>AC</math> as <math>h_2</math>.According to the problem, <math>AP* h_1 +CP* h_2 =CP* h_1 +AP* h_2 </math> implies <math> h_1 (AP-CP)= h_2 (AP-CP), h_1 = h_2 </math>. Then according to basic congruent triangles we get <math>BP=DP</math><br />
Firstly, denote that <math>CP=a,BP=b,CP=c,AP=d</math>. Applying Stewart theorem, getting that <math>100c+196b=(b+c)(bc+a^2); 100b+260c=(b+c)(bc+c^2); 100c+196b=100b+260c, 3b=5c</math>, denote <math>b=5x,c=3x</math><br />
Applying Stewart Theorem, getting <math>260a+100a=2a(a^2+25x^2); 196a+100a=2a(9x^2+a^2)</math> solve for a, getting <math>a=\sqrt{130},AP=5\sqrt{2}; CP=3\sqrt{2}</math><br />
Now everything is clear, we can find <math>cos\angle{BPA}=\frac{4}{\sqrt{65}}</math> using LOC, <math>sin\angle{BPA}=\frac{7\sqrt{65}}{65}</math>, the whole area is <math>\sqrt{130}*8\sqrt{2}*\frac{7\sqrt{65}}{65}=\boxed{112}</math><br />
<br />
~bluesoul<br />
<br />
==Solution 8 (Simple Geometry)==<br />
[[File:AIME-II-2018-12.png|400px|right]]<br />
<math>BP = PD</math> as in another solutions.<br />
<br />
Let <math>D'</math> be the reflection of <math>D</math> across <math>C</math>.<br />
Let points <math>E, E',</math> and <math>H</math> be the foot of perpendiculars on <math>AC</math> from <math>D,D'</math>, and <math>B</math> respectively.<br />
<cmath>\begin{align*}<br />
&\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\<br />
\Rightarrow &\qquad \triangle ABH = \triangle CDE = \triangle CD'E' \\<br />
\Rightarrow &\qquad \angle BAC = \angle ACD' \\<br />
\Rightarrow &\qquad \triangle ABC = \triangle AD'C \\<br />
\Rightarrow &\qquad BC = AD'.<br />
\end{align*}</cmath><br />
The area of quadrilateral <math>ABCD</math> is equal to the area of triangle <math>ADD'</math> with sides <math>AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20</math>.<br />
The semiperimeter is <math>s = 17 + \sqrt{65},</math> the area <cmath>[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.</cmath><br />
<br />
'''vladimir.shelomovskii@gmail.com, vvsss'''<br />
<br />
==Solution 9 (Mindless Law of Cosines Bash)==<br />
<br />
Use your favorite method to get that <math>P</math> is the midpoint of one of the two diagonals (suppose it's the midpoint of <math>\overline{AC}</math>). From here, let <math>x=AP=PC, y=BP, z=PD, a=\cos\theta</math> where <math>\theta</math> is the angle that the diagonals make. Then we have a system of four equations:<br />
<br />
\[\begin{align*}<br />
x^2+y^2+2xya=100 \\<br />
z^2+x^2+2xza=100 \\<br />
x^2+y^2-2xya=196 \\<br />
x^2+z^2-2xza=260 \\<br />
\end{align*}\]<br />
<br />
From these equations we get that <br />
<math>\begin{align*}<br />
xya=-24 \\<br />
xza=-40 \\<br />
x^2+y^2-48=10 \\<br />
x^2+z^2-80=10<br />
\end{align*}</math><br />
From here we can see that <math>\frac{z}{y}={5}{3}, z^2-y^2=32,</math> so <math>z=5\sqrt{2}, y=3\sqrt{2}.</math> Furthermore, this implies <math>x=\sqrt{130}</math> and <math>xa=-4\sqrt{2},</math> which implies <math>a=\cos\theta=\frac{4}{\sqrt{65}}.</math> Then note that the area of the quadrilateral is <cmath>\frac{1}{2}\sin\theta (xy+xz+xz+xy)=\sin\theta (\sqrt{130}\cdot 3\sqrt{2}+\sqrt{130} \cdot 5\sqrt{2})=7\cdot (3\cdot 2+5\cdot 2)=7(6+10)=7\cdot 16=\boxed{112}.</cmath><br />
<br />
~Dhillonr25<br />
<br />
==See Also==<br />
{{AIME box|year=2018|n=II|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Awesomeming327.https://artofproblemsolving.com/wiki/index.php?title=User:Awesomeming327.&diff=139993User:Awesomeming327.2020-12-19T22:11:49Z<p>Awesomeming327.: AAAAAAAAAA</p>
<hr />
<div>Yes, awesomeming327 has a wiki page owo<br />
Here is a history of him<br />
<br />
Jan 30 2020: took the AMC 10A, got a 99/150; AIME Cutoff 103.5<br />
June 1 2020: started AMSP<br />
May 22 2020: Joined Mafia<br />
June 29 2020: Joined 1st Game forum<br />
Sep 14 2020: Joined The Arcade<br />
Oct 29 2020: Took the COMC, got a 62/80; CMO Cutoff Pending<br />
Nov 21 2020: Left The Arcade, Joined Road to Rewards</div>Awesomeming327.https://artofproblemsolving.com/wiki/index.php?title=Awesomeming327&diff=125849Awesomeming3272020-06-18T20:16:56Z<p>Awesomeming327.: Created page with "Hi. Bye."</p>
<hr />
<div>Hi. Bye.</div>Awesomeming327.https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_10B_Problems&diff=1034752002 AMC 10B Problems2019-02-19T23:36:10Z<p>Awesomeming327.: /* Problem 10 */</p>
<hr />
<div>==Problem 1==<br />
The ratio <math>\frac{2^{2001}\cdot3^{2003}}{6^{2002}}</math> is:<br />
<br />
<math> \mathrm{(A) \ } \frac{1}{6}\qquad \mathrm{(B) \ } \frac{1}{3}\qquad \mathrm{(C) \ } \frac{1}{2}\qquad \mathrm{(D) \ } \frac{2}{3}\qquad \mathrm{(E) \ } \frac{3}{2} </math><br />
<br />
[[2002 AMC 10B Problems/Problem 1|Solution]]<br />
<br />
== Problem 2 ==<br />
For the nonzero numbers <math>a, b,</math> and <math>c,</math> define<br />
<cmath>(a,b,c)=\frac{abc}{a+b+c}</cmath><br />
Find <math>(2,4,6)</math>.<br />
<br />
<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 24 </math><br />
<br />
[[2002 AMC 10B Problems/Problem 2|Solution]]<br />
<br />
== Problem 3 ==<br />
The arithmetic mean of the nine numbers in the set <math>\{9,99,999,9999,\ldots,999999999\}</math> is a <math>9</math>-digit number <math>M</math>, all of whose digits are distinct. The number <math>M</math> does not contain the digit<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 2\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 6\qquad \mathrm{(E) \ } 8 </math><br />
<br />
[[2002 AMC 10B Problems/Problem 3|Solution]]<br />
<br />
== Problem 4 ==<br />
<br />
What is the value of<br />
<br />
<cmath>(3x-2)(4x+1)-(3x-2)4x+1</cmath><br />
<br />
when <math>x=4</math>?<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } 12 </math><br />
<br />
[[2002 AMC 10B Problems/Problem 4|Solution]]<br />
<br />
== Problem 5 ==<br />
<br />
Circles of radius <math>2</math> and <math>3</math> are externally tangent and are circumscribed by a third circle, as shown in the figure. Find the area of the shaded region.<br />
<br />
<center><asy><br />
unitsize(5mm);<br />
defaultpen(linewidth(.8pt)+fontsize(10pt));<br />
dotfactor=4;<br />
<br />
real r1=3; real r2=2; real r3=5;<br />
pair A=(-2,0), B=(3,0), C=(0,0);<br />
pair X=(1,0), Y=(5,0);<br />
path circleA=Circle(A,r1); path circleB=Circle(B,r2); path circleC=Circle(C,r3);<br />
fill(circleC,gray);<br />
fill(circleA,white);<br />
fill(circleB,white);<br />
draw(circleA); draw(circleB); draw(circleC);<br />
draw(A--X); draw(B--Y);<br />
<br />
pair[] ps={A,B}; dot(ps);<br />
<br />
label("$3$",midpoint(A--X),N);<br />
label("$2$",midpoint(B--Y),N);<br />
</asy></center><br />
<br />
<math> \mathrm{(A) \ } 3\pi\qquad \mathrm{(B) \ } 4\pi\qquad \mathrm{(C) \ } 6\pi\qquad \mathrm{(D) \ } 9\pi\qquad \mathrm{(E) \ } 12\pi </math><br />
<br />
[[2002 AMC 10B Problems/Problem 5|Solution]]<br />
<br />
== Problem 6 ==<br />
<br />
For how many positive integers <math>n</math> is <math>n^2-3n+2</math> a prime number?<br />
<br />
<math> \mathrm{(A) \ } \text{none}\qquad \mathrm{(B) \ } \text{one}\qquad \mathrm{(C) \ } \text{two}\qquad \mathrm{(D) \ } \text{more than two, but finitely many}\qquad \mathrm{(E) \ } \text{infinitely many} </math><br />
<br />
[[2002 AMC 10B Problems/Problem 6|Solution]]<br />
<br />
== Problem 7 ==<br />
<br />
Let <math>n</math> be a positive integer such that <math>\frac{1}{2}+\frac{1}{3}+\frac{1}{7}+\frac{1}{n}</math> is an integer. Which of the following statements is '''not''' true?<br />
<br />
<math> \mathrm{(A) \ } 2\text{ divides }n\qquad \mathrm{(B) \ } 3\text{ divides }n\qquad \mathrm{(C) \ } 6\text{ divides }n\qquad \mathrm{(D) \ } 7\text{ divides }n\qquad \mathrm{(E) \ } n>84 </math><br />
<br />
[[2002 AMC 10B Problems/Problem 7|Solution]]<br />
<br />
== Problem 8 ==<br />
<br />
Suppose July of year <math>N</math> has five Mondays. Which of the following must occur five times in the August of year <math>N</math>? (Note: Both months have <math>31</math> days.)<br />
<br />
<math>\textbf{(A)}\ \text{Monday} \qquad \textbf{(B)}\ \text{Tuesday} \qquad \textbf{(C)}\ \text{Wednesday} \qquad \textbf{(D)}\ \text{Thursday} \qquad \textbf{(E)}\ \text{Friday}</math><br />
<br />
[[2002 AMC 10B Problems/Problem 8|Solution]]<br />
<br />
== Problem 9 ==<br />
<br />
Using the letters <math>A</math>, <math>M</math>, <math>O</math>, <math>S</math>, and <math>U</math>, we can form five-letter "words". If these "words" are arranged in alphabetical order, then the "word" <math>USAMO</math> occupies position<br />
<br />
<math> \mathrm{(A) \ } 112\qquad \mathrm{(B) \ } 113\qquad \mathrm{(C) \ } 114\qquad \mathrm{(D) \ } 115\qquad \mathrm{(E) \ } 116 </math><br />
<br />
[[2002 AMC 10B Problems/Problem 9|Solution]]<br />
<br />
== Problem 10 ==<br />
<br />
Suppose that <math>a</math> and <math>b</math> are nonzero real numbers, and that the equation <math>x^2+ax+b=0</math> has solutions <math>a</math> and <math>b</math>. what is the pair <math>(a,b)</math>?<br />
<br />
<math> \mathrm{(A) \ } (-2,1)\qquad \mathrm{(B) \ } (-1,2)\qquad \mathrm{(C) \ } (1,-2)\qquad \mathrm{(D) \ } (2,-1)\qquad \mathrm{(E) \ } (4,4) </math><br />
<br />
[[2002 AMC 10B Problems/Problem 10|Solution]]<br />
<br />
== Problem 11 ==<br />
<br />
The product of three consecutive positive integers is <math>8</math> times their sum. What is the sum of the squares?<br />
<br />
<math> \mathrm{(A) \ } 50\qquad \mathrm{(B) \ } 77\qquad \mathrm{(C) \ } 110\qquad \mathrm{(D) \ } 149\qquad \mathrm{(E) \ } 194 </math><br />
<br />
[[2002 AMC 10B Problems/Problem 11|Solution]]<br />
<br />
== Problem 12 ==<br />
<br />
For which of the following values of <math>k</math> does the equation <math>\frac{x-1}{x-2} = \frac{x-k}{x-6}</math> have no solution for <math>x</math>?<br />
<br />
<math>\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math><br />
<br />
[[2002 AMC 10B Problems/Problem 12|Solution]]<br />
<br />
== Problem 13==<br />
<br />
Find the value(s) of <math>x</math> such that <math>8xy - 12y + 2x - 3 = 0</math> is true for all values of <math>y</math>.<br />
<br />
<math>\textbf{(A) } \frac23 \qquad \textbf{(B) } \frac32 \text{ or } -\frac14 \qquad \textbf{(C) } -\frac23 \text{ or } -\frac14 \qquad \textbf{(D) } \frac32 \qquad \textbf{(E) } -\frac32 \text{ or } -\frac14</math><br />
<br />
<br />
[[2002 AMC 10B Problems/Problem 13|Solution]]<br />
<br />
== Problem 14 ==<br />
<br />
The number <math>25^{64}\cdot 64^{25}</math> is the square of a positive integer <math>N</math>. In decimal representation, the sum of the digits of <math>N</math> is<br />
<br />
<math> \mathrm{(A) \ } 7\qquad \mathrm{(B) \ } 14\qquad \mathrm{(C) \ } 21\qquad \mathrm{(D) \ } 28\qquad \mathrm{(E) \ } 35 </math><br />
<br />
[[2002 AMC 10B Problems/Problem 14|Solution]]<br />
<br />
== Problem 15 ==<br />
<br />
The positive integers <math>A</math>, <math>B</math>, <math>A-B</math>, and <math>A+B</math> are all prime numbers. The sum of these four primes is<br />
<br />
<math> \mathrm{(A) \ } \text{even}\qquad \mathrm{(B) \ } \text{divisible by }3\qquad \mathrm{(C) \ } \text{divisible by }5\qquad \mathrm{(D) \ } \text{divisible by }7\qquad \mathrm{(E) \ } \text{prime}</math><br />
<br />
[[2002 AMC 10B Problems/Problem 15|Solution]]<br />
<br />
== Problem 16 ==<br />
<br />
For how many integers <math>n</math> is <math>\frac{n}{20-n}</math> the square of an integer?<br />
<br />
<math>\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 10</math><br />
<br />
<br />
[[2002 AMC 10B Problems/Problem 16|Solution]]<br />
<br />
== Problem 17 ==<br />
<br />
A regular octagon <math>ABCDEFGH</math> has sides of length two. Find the area of <math>\triangle ADG</math>.<br />
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<math>\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2</math><br />
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[[2002 AMC 10B Problems/Problem 17|Solution]]<br />
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== Problem 18 ==<br />
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Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?<br />
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<math>\textbf{(A) } 8\qquad \textbf{(B) } 9\qquad \textbf{(C) } 10\qquad \textbf{(D) } 12\qquad \textbf{(E) } 16</math><br />
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[[2002 AMC 10B Problems/Problem 18|Solution]]<br />
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== Problem 19 ==<br />
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Suppose that <math>\{a_n\}</math> is an arithmetic sequence with<br />
<cmath> a_1+a_2+\cdots+a_{100}=100 \text{ and } a_{101}+a_{102}+\cdots+a_{200}=200.</cmath><br />
What is the value of <math>a_2 - a_1 ?</math><br />
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<math> \mathrm{(A) \ } 0.0001\qquad \mathrm{(B) \ } 0.001\qquad \mathrm{(C) \ } 0.01\qquad \mathrm{(D) \ } 0.1\qquad \mathrm{(E) \ } 1 </math><br />
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[[2002 AMC 10B Problems/Problem 19|Solution]]<br />
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== Problem 20 ==<br />
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Let <math>a, b,</math> and <math>c</math> be real numbers such that <math>a-7b+8c=4</math> and <math>8a+4b-c=7.</math> Then <math>a^2-b^2+c^2</math> is<br />
<br />
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 4\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
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[[2002 AMC 10B Problems/Problem 20|Solution]]<br />
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== Problem 21 ==<br />
<br />
Andy's lawn has twice as much area as Beth's lawn and three times as much as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first?<br />
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<math> \mathrm{(A) \ } \text{Andy}\qquad \mathrm{(B) \ } \text{Beth}\qquad \mathrm{(C) \ } \text{Carlos}\qquad \mathrm{(D) \ } \text{Andy and Carlos tie for first.}\qquad \mathrm{(E) \ } \text{All three tie.} </math><br />
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[[2002 AMC 10B Problems/Problem 21|Solution]]<br />
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== Problem 22 ==<br />
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Let <math>\triangle{XOY}</math> be a right-angled triangle with <math>m\angle{XOY}=90^\circ</math>. Let <math>M</math> and <math>N</math> be the midpoints of the legs <math>OX</math> and <math>OY</math>, respectively. Given <math>XN=19</math> and <math>YM=22</math>, find <math>XY</math>.<br />
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<math> \mathrm{(A) \ } 24\qquad \mathrm{(B) \ } 26\qquad \mathrm{(C) \ } 28\qquad \mathrm{(D) \ } 30\qquad \mathrm{(E) \ } 32 </math><br />
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[[2002 AMC 10B Problems/Problem 22|Solution]]<br />
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== Problem 23 ==<br />
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Let <math>\{a_k\}</math> be a sequence of integers such that <math>a_1=1</math> and <math>a_{m+n}=a_m+a_n+mn,</math> for all positive integers <math>m</math> and <math>n.</math> Then <math>a_{12}</math> is<br />
<br />
<math> \mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89 </math><br />
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[[2002 AMC 10B Problems/Problem 23|Solution]]<br />
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== Problem 24 ==<br />
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Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius <math>20</math> feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point <math>10</math> vertical feet above the bottom?<br />
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<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15 </math><br />
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[[2002 AMC 10B Problems/Problem 24|Solution]]<br />
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== Problem 25 ==<br />
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When <math>15</math> is appended to a list of integers, the mean is increased by <math>2</math>. When <math>1</math> is appended to the enlarged list, the mean of the enlarged list is decreased by <math>1</math>. How many integers were in the original list?<br />
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<math> \mathrm{(A) \ } 4\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 6\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 8 </math><br />
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[[2002 AMC 10B Problems/Problem 25|Solution]]<br />
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== See also ==<br />
{{AMC10 box|year=2002|ab=B|before=[[2002 AMC 10A Problems]]|after=[[2003 AMC 10A Problems]]}}<br />
* [[AMC 10]]<br />
* [[AMC 10 Problems and Solutions]]<br />
* [[AMC Problems and Solutions]]<br />
* [[Mathematics competition resources]]<br />
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{{MAA Notice}}</div>Awesomeming327.