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Difference between revisions of "1977 AHSME Problems/Problem 9"

(Created page with "==Solution== Solution by e_power_pi_times_i If arcs <math>AB</math>, <math>BC</math>, and <math>CD</math> are congruent, then <math>\measuredangle ACB = \measuredangle BDC =...")
 
(→‎Solution)
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==Solution==
 
==Solution==
 
Solution by e_power_pi_times_i
 
Solution by e_power_pi_times_i
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 +
<asy>
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size(120);
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path c = Circle((0, 0), 1);
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pair A = dir(20), B = dir(130), C = dir(240), D = dir(330);
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draw(c);
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pair F = 3(A-B) + B;
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pair G = 3(D-C) + C;
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pair E = intersectionpoints(B--F, C--G)[0];
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draw(B--E--C--A);
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label("$A$", A, NE);
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label("$B$", B, NW);
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label("$C$", C, SW);
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label("$D$", D, SE);
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label("$E$", E, E);
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//Credit to MSTang for the diagram
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</asy>
  
 
If arcs <math>AB</math>, <math>BC</math>, and <math>CD</math> are congruent, then <math>\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta</math>. Because <math>ABCD</math> is cyclic, <math>\measuredangle CAD = \measuredangle CBD = \theta</math>, and <math>\measuredangle ADB = \measuredangle ACB = \theta</math>. Then, <math>\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ</math>. <math>\theta = 55^\circ</math>. <math>\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{\textbf{(B) }15^\circ}</math>.
 
If arcs <math>AB</math>, <math>BC</math>, and <math>CD</math> are congruent, then <math>\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta</math>. Because <math>ABCD</math> is cyclic, <math>\measuredangle CAD = \measuredangle CBD = \theta</math>, and <math>\measuredangle ADB = \measuredangle ACB = \theta</math>. Then, <math>\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ</math>. <math>\theta = 55^\circ</math>. <math>\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{\textbf{(B) }15^\circ}</math>.

Revision as of 13:28, 18 November 2016

Solution

Solution by e_power_pi_times_i

[asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G)[0]; draw(B--E--C--A); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E); //Credit to MSTang for the diagram [/asy]

If arcs $AB$, $BC$, and $CD$ are congruent, then $\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta$. Because $ABCD$ is cyclic, $\measuredangle CAD = \measuredangle CBD = \theta$, and $\measuredangle ADB = \measuredangle ACB = \theta$. Then, $\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ$. $\theta = 55^\circ$. $\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{\textbf{(B) }15^\circ}$.

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