2024 AMC 10A Problems/Problem 20
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Problem
Point 
lies outside regular pentagon 
so that 
is an equilateral triangle, as shown below. What is the degree measure of acute angle 
?
![[asy] import graph; size(7cm); real labelscalefactor = 0.75; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.089556028373145, xmax = 2.738320502950249, ymin = -1.3143096399343266, ymax = 1.2691431521185288; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); filldraw(arc((1.9562952014676112,0),0.25,168,192)--(1.9562952014676112,0)--cycle, mediumgrey); /* draw figures */ draw((-0.8090169943749473,0.5877852522924731)--(0.30901699437494745,0.9510565162951535)); draw((0.30901699437494745,0.9510565162951535)--(1,0)); draw((1,0)--(0.3090169943749473,-0.9510565162951536)); draw((0.3090169943749473,-0.9510565162951536)--(-0.8090169943749475,-0.587785252292473)); draw((-0.8090169943749475,-0.587785252292473)--(-0.8090169943749473,0.5877852522924731)); draw((0.30901699437494745,0.9510565162951535)--(0.3090169943749473,-0.9510565162951536)); draw((0.3090169943749473,-0.9510565162951536)--(1.9562952014676112,0)); draw((1.9562952014676112,0)--(0.30901699437494745,0.9510565162951535)); draw((-0.8090169943749473,0.5877852522924731)--(1.9562952014676112,0), dashed); draw((1.9562952014676112,0)--(-0.8090169943749475,-0.587785252292473), dashed); /* dots and labels */ dot((1,0),linewidth(4pt) + dotstyle); label("$A$", (1.013592864312142,0), E * labelscalefactor); dot((0.30901699437494745,0.9510565162951535),linewidth(4pt) + dotstyle); label("$B$", (0.26,1), NE * labelscalefactor); dot((-0.8090169943749473,0.5877852522924731),linewidth(4pt) + dotstyle); label("$C$", (-0.82,0.64), NW * 0.25); dot((-0.8090169943749475,-0.587785252292473),linewidth(4pt) + dotstyle); label("$D$", (-0.82,-0.64), SW * 0.25); dot((0.3090169943749473,-0.9510565162951536),linewidth(4pt) + dotstyle); label("$E$", (0.26,-1), SE * labelscalefactor); dot((1.9562952014676112,0),linewidth(4pt) + dotstyle); label("$X$", (1.9705619971429902, 0), E * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]](http://latex.artofproblemsolving.com/a/8/8/a885f409e123b4a6560ad0c31c863f12ac435565.png)
Solution
Note that 
, so it follows that 
is the circumcenter of 
. Therefore, 
, so by symmetry ![\[\angle CXD = \angle BXE - 2 \angle BXC = 60^{\circ} - 2 \cdot 18^{\circ} = \boxed{\textbf{(E)}~24^{\circ}}.\]](http://latex.artofproblemsolving.com/1/b/3/1b332aa25f58125e698e4aebde49c31958dc214d.png)
~ihatemath123
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
