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

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Solution by e_power_pi_times_i
 
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 = \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>.

Latest revision as of 12:30, 21 November 2016


Problem 9

[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]


In the adjoining figure $\measuredangle E=40^\circ$ and arc $AB$, arc $BC$, and arc $CD$ all have equal length. Find the measure of $\measuredangle ACD$.

$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }\left(\frac{45}{2}\right)^\circ\qquad \textbf{(E) }30^\circ$


Solution

Solution by e_power_pi_times_i

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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