https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Alleycat&feedformat=atom AoPS Wiki - User contributions [en] 2021-09-28T07:04:10Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=102728 2019 AMC 10A Problems/Problem 13 2019-02-15T01:41:44Z <p>Alleycat: </p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. <br /> Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3 (Outside Angles)==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==Solution 4==<br /> <br /> Notice that if &lt;math&gt;\angle BEC&lt;/math&gt; is &lt;math&gt;\text{90}&lt;/math&gt; degrees, then &lt;math&gt;\angle BEC&lt;/math&gt; and &lt;math&gt;\angle ACE&lt;/math&gt; must be &lt;math&gt;\text{20}&lt;/math&gt; degrees. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that &lt;math&gt;\angle EBD \cong \angle ECD = 20\text{degrees}&lt;/math&gt;. Thus &lt;math&gt;\angle CBF&lt;/math&gt; is&lt;math&gt; 70 - 20 = 50 \text{degrees}&lt;/math&gt;, and so &lt;math&gt;\angle BFC&lt;/math&gt; is &lt;math&gt;180 - 20 - 50 = 110\text{degrees}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> Cheap Solution: Create an accurate diagram and measure the angle using a protractor. If you were accurate, the answer is 110 degrees.<br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=102726 2019 AMC 10A Problems/Problem 13 2019-02-15T01:40:58Z <p>Alleycat: /* Solution 2*/</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. <br /> Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3 (Outside Angles)==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==Solution 4==<br /> <br /> Notice that if &lt;math&gt;\angle BEC&lt;/math&gt; is &lt;math&gt;\text{90}&lt;/math&gt; degrees, then &lt;math&gt;\angle BEC&lt;/math&gt; and &lt;math&gt;\angle ACE&lt;/math&gt; must be &lt;math&gt;\text{20}&lt;/math&gt; degrees. Using cyclic quadrilateral properties (or the properties of a subtended arc), we can find that &lt;math&gt;\angle EBD \cong \angle ECD = 20\text{degrees}&lt;/math&gt;. Thus &lt;math&gt;\angle CBF&lt;/math&gt; is&lt;math&gt; 70 - 20 = 50 \text{degrees}&lt;/math&gt;, and so &lt;math&gt;\angle BFC&lt;/math&gt; is &lt;math&gt;180 - 20 - 50 = 110\text{degrees}&lt;/math&gt;, which is &lt;math&gt;\boxed{\textbf{(D)}}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> angles FDA and FEA are both 90 degrees. angle CAB is &lt;math&gt;180=40+2CAB&lt;/math&gt; so CAB is 70 degrees. Then the answer is &lt;math&gt;360-90-90-70=110&lt;/math&gt;<br /> -liu4505<br /> <br /> ==See Also==<br /> Cheap Solution: Create an accurate diagram and measure the angle using a protractor. If you were accurate, the answer is 110 degrees.<br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101835 2019 AMC 10A Problems/Problem 13 2019-02-11T16:47:19Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt;<br /> unitsize(40);<br /> draw((-1,0)--(1,0)--(0,2.75)--cycle);<br /> draw(circumcircle((-1,0),(0,0),(0,2.75)));<br /> label(&quot;$A$&quot;,(1,0),SE);<br /> label(&quot;$C$&quot;,(0,2.75),N);<br /> label(&quot;$B$&quot;,(-1,0),SW);<br /> label(&quot;$E$&quot;,(0,0),S);<br /> label(&quot;$D$&quot;,(0.77,0.64),E);<br /> draw((0,0)--(0,2.75));<br /> draw((-1,0)--(0.77,0.64));<br /> &lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. <br /> Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat (Diagram by Brendanb4321)<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101834 2019 AMC 10A Problems/Problem 13 2019-02-11T16:46:05Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt;<br /> unitsize(40);<br /> draw((-1,0)--(1,0)--(0,2.75)--cycle);<br /> draw(circumcircle((-1,0),(0,0),(0,2.75)));<br /> label(&quot;$A$&quot;,(1,0),SE);<br /> label(&quot;$C$&quot;,(0,2.75),N);<br /> label(&quot;$B$&quot;,(-1,0),SW);<br /> label(&quot;$E$&quot;,(0,0),S);<br /> label(&quot;$D$&quot;,(0.77,0.64),E);<br /> label(&quot;$F$&quot;,intersection(B--D,C--E),NW);<br /> draw((0,0)--(0,2.75));<br /> draw((-1,0)--(0.77,0.64));<br /> &lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. <br /> Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat (Diagram by Brendanb4321, edited by alleycat)<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101833 2019 AMC 10A Problems/Problem 13 2019-02-11T16:45:49Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt;<br /> unitsize(40);<br /> draw((-1,0)--(1,0)--(0,2.75)--cycle);<br /> draw(circumcircle((-1,0),(0,0),(0,2.75)));<br /> label(&quot;$A$&quot;,(1,0),SE);<br /> label(&quot;$C$&quot;,(0,2.75),N);<br /> label(&quot;$B$&quot;,(-1,0),SW);<br /> label(&quot;$E$&quot;,(0,0),S);<br /> label(&quot;$D$&quot;,(0.77,0.64),E);<br /> label(&quot;$F$&quot;,intersectionpoint(B--D,C--E),NW);<br /> draw((0,0)--(0,2.75));<br /> draw((-1,0)--(0.77,0.64));<br /> &lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. <br /> Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat (Diagram by Brendanb4321, edited by alleycat)<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101832 2019 AMC 10A Problems/Problem 13 2019-02-11T16:43:16Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. <br /> Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt;<br /> <br /> <br /> Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101831 2019 AMC 10A Problems/Problem 13 2019-02-11T16:43:00Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt;<br /> <br /> So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. <br /> Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt;<br /> <br /> Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101830 2019 AMC 10A Problems/Problem 13 2019-02-11T16:42:44Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt;<br /> So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt;<br /> Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101829 2019 AMC 10A Problems/Problem 13 2019-02-11T16:42:13Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle BDA = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt; So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt; and &lt;math&gt;\angle BEC = 90^{\circ} = \angle BDA&lt;/math&gt;. Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt; Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101828 2019 AMC 10A Problems/Problem 13 2019-02-11T16:38:47Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle ADB = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt; So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt;. Thus, we know: &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt; Finally, we know: &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101827 2019 AMC 10A Problems/Problem 13 2019-02-11T16:38:18Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle ADB = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt; So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt;. Thus, we know &lt;cmath&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}.&lt;/cmath&gt; Finally, we know that &lt;cmath&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/cmath&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101826 2019 AMC 10A Problems/Problem 13 2019-02-11T16:37:48Z <p>Alleycat: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;cmath&gt;\angle ADB = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}.&lt;/cmath&gt; So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt;. Thus, we know &lt;math&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}&lt;/math&gt;. Finally, we know that &lt;math&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101825 2019 AMC 10A Problems/Problem 13 2019-02-11T16:35:34Z <p>Alleycat: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Alternatively, we could have used similar triangles. We start similarly to Solution 1.<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;math&gt;\angle ADB = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}&lt;/math&gt;. So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt;. Thus, we know &lt;math&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}&lt;/math&gt;. Finally, we know that &lt;math&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101824 2019 AMC 10A Problems/Problem 13 2019-02-11T16:35:01Z <p>Alleycat: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> <br /> <br /> Alternatively, we could have used similar triangles.<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;math&gt;\angle ADB = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}&lt;/math&gt;. So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt;. Thus, we know &lt;math&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}&lt;/math&gt;. Finally, we know that &lt;math&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 3==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101823 2019 AMC 10A Problems/Problem 13 2019-02-11T16:34:34Z <p>Alleycat: /* Solution 1 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> <br /> <br /> Alternatively, we could have used similar triangles.<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Therefore, &lt;math&gt;\angle ADB = 180^{\circ} - \angle BDC = 180^{\circ} - 90^{\circ} = 90^{\circ}&lt;/math&gt;. So, &lt;math&gt;\triangle BEF \sim BDA&lt;/math&gt; by AA Similarity, since &lt;math&gt;\angle EBF = \angle DBA&lt;/math&gt;. Thus, we know &lt;math&gt;\angle EFB = \angle DAB = \angle CAB = 70^{\circ}&lt;/math&gt;. Finally, we know that &lt;math&gt;\angle BFC = 180^{\circ} - \angle EFB = 180^{\circ} - 70^{\circ} = \boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~ alleycat<br /> <br /> ==Solution 2==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101822 2019 AMC 10A Problems/Problem 13 2019-02-11T16:17:33Z <p>Alleycat: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101821 2019 AMC 10A Problems/Problem 13 2019-02-11T16:17:12Z <p>Alleycat: /* Solution 3 */</p> <hr /> <div>==Solution 3==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_13&diff=101819 2019 AMC 10A Problems/Problem 13 2019-02-11T16:16:37Z <p>Alleycat: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> Let &lt;math&gt;\triangle ABC&lt;/math&gt; be an isosceles triangle with &lt;math&gt;BC = AC&lt;/math&gt; and &lt;math&gt;\angle ACB = 40^{\circ}&lt;/math&gt;. Construct the circle with diameter &lt;math&gt;\overline{BC}&lt;/math&gt;, and let &lt;math&gt;D&lt;/math&gt; and &lt;math&gt;E&lt;/math&gt; be the other intersection points of the circle with the sides &lt;math&gt;\overline{AC}&lt;/math&gt; and &lt;math&gt;\overline{AB}&lt;/math&gt;, respectively. Let &lt;math&gt;F&lt;/math&gt; be the intersection of the diagonals of the quadrilateral &lt;math&gt;BCDE&lt;/math&gt;. What is the degree measure of &lt;math&gt;\angle BFC ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> &lt;asy&gt; unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label(&quot;$A$&quot;,(1,0),SE);label(&quot;$C$&quot;,(0,2.75),N);label(&quot;$B$&quot;,(-1,0),SW);label(&quot;$E$&quot;,(0,0),S);label(&quot;$D$&quot;,(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64));&lt;/asy&gt;<br /> <br /> Drawing it out, we see &lt;math&gt;\angle BDC&lt;/math&gt; and &lt;math&gt;\angle BEC&lt;/math&gt; are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find &lt;math&gt;\angle ABC=70^{\circ}&lt;/math&gt;. We can find &lt;math&gt;\angle ECB=20^{\circ}&lt;/math&gt; and &lt;math&gt;\angle DBC=50^{\circ}&lt;/math&gt; by the triangle angle sum on &lt;math&gt;\triangle ECB&lt;/math&gt; and &lt;math&gt;\triangle DBC&lt;/math&gt;. <br /> <br /> &lt;cmath&gt;\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC\implies\angle DBC=50^{\circ}&lt;/cmath&gt;<br /> <br /> &lt;cmath&gt;\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB\implies\angle ECB=20^{\circ}&lt;/cmath&gt;<br /> <br /> Then, we take triangle &lt;math&gt;BFC&lt;/math&gt;, and find &lt;math&gt;\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110}.&lt;/math&gt;<br /> <br /> ~Argonauts16 (Diagram by Brendanb4321)<br /> <br /> ==Solution 2==<br /> <br /> Through the property of angles formed by intersecting chords, we find that<br /> &lt;cmath&gt;m\angle BFC=\frac{m\overarc{BC}+m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Through the Outside Angles Theorem, we find that<br /> &lt;cmath&gt;m\angle CAB = \frac{m\overarc{BC}-m\overarc{DE}}{2}&lt;/cmath&gt;<br /> <br /> Adding the two equations gives us<br /> &lt;cmath&gt;m\angle BFC - m\angle CAB = m\overarc{BC}\implies m\angle BFC=m\overarc{BC} - m\angle CAB&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;\overarc{BC}&lt;/math&gt; is the diameter, &lt;math&gt;m\overarc{BC}=180&lt;/math&gt; and because &lt;math&gt;\triangle ABC&lt;/math&gt; is isosceles and &lt;math&gt;m\angle ACB=40&lt;/math&gt;, &lt;math&gt;m\angle CAB=70&lt;/math&gt;. Thus<br /> &lt;cmath&gt;m\angle BFC=180-70=\boxed{\textbf{(D) } 110}&lt;/cmath&gt;<br /> <br /> ~mn28407<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=97915 Angle Bisector Theorem 2018-09-24T03:06:19Z <p>Alleycat: /* Introduction */</p> <hr /> <div>{{WotWAnnounce|week=June 6-12}}<br /> <br /> == Introduction ==<br /> The '''Angle Bisector Theorem''' states that given [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt; and [[angle bisector]] AD, where D is on side BC, then &lt;math&gt; \frac cm = \frac bn &lt;/math&gt;. It follows that &lt;math&gt; \frac cb = \frac mn &lt;/math&gt;. Likewise, the [[converse]] of this theorem holds as well.<br /> <br /> &lt;asy&gt; size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label(&quot;$A$&quot;,A,(1,1));label(&quot;$B$&quot;,B,(-1,-1));label(&quot;$C$&quot;,C,(1,-1));label(&quot;$D$&quot;,D,(0,-1)); dot(A^^B^^C^^D,blue);label(&quot;$b$&quot;,(A+C)/2,(1,0));label(&quot;$c$&quot;,(A+B)/2,(0,1));label(&quot;$m$&quot;,(B+D)/2,(0,-1));label(&quot;$n$&quot;,(D+C)/2,(0,-1)); &lt;/asy&gt;<br /> <br /> == Examples ==<br /> <br /> # Let ABC be a triangle with angle bisector AD with D on line segment BC. If &lt;math&gt; BD = 2, CD = 5,&lt;/math&gt; and &lt;math&gt; AB + AC = 10 &lt;/math&gt;, find AB and AC.&lt;br&gt; '''''Solution:''''' By the angle bisector theorem, &lt;math&gt; \frac{AB}2 = \frac{AC}5&lt;/math&gt; or &lt;math&gt; AB = \frac 25 AC &lt;/math&gt;. Plugging this into &lt;math&gt; AB + AC = 10 &lt;/math&gt; and solving for AC gives &lt;math&gt; AC = \frac{50}7&lt;/math&gt;. We can plug this back in to find &lt;math&gt; AB = \frac{20}7 &lt;/math&gt;.<br /> # In triangle ABC, let P be a point on BC and let &lt;math&gt; AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 &lt;/math&gt;. Find the value of &lt;math&gt; m\angle BAP - m\angle CAP &lt;/math&gt;. &lt;br&gt; '''''Solution:''''' First, we notice that &lt;math&gt; \frac{AB}{BP}=\frac{AC}{CP} &lt;/math&gt;. Thus, AP is the angle bisector of angle A, making our answer 0.<br /> # Part '''(b)''', [[1959 IMO Problems/Problem 5]].<br /> <br /> == See also ==<br /> * [[Angle bisector]]<br /> * [[Geometry]]<br /> * [[Stewart's Theorem]]<br /> <br /> [[Category:Geometry]]<br /> <br /> [[Category:Theorems]]</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=Angle_Bisector_Theorem&diff=97914 Angle Bisector Theorem 2018-09-24T03:04:21Z <p>Alleycat: /* Introduction */</p> <hr /> <div>{{WotWAnnounce|week=June 6-12}}<br /> <br /> == Introduction ==<br /> The '''Angle Bisector Theorem''' states that given [[triangle]] &lt;math&gt;\triangle ABC&lt;/math&gt; and [[angle bisector]] AD, where D is on side BC, then &lt;math&gt; \frac cm = \frac bn &lt;/math&gt;. It follows that &lt;math&gt; \frac bc = \frac mn &lt;/math&gt;. Likewise, the [[converse]] of this theorem holds as well.<br /> <br /> &lt;asy&gt; size(200); defaultpen(fontsize(12)); real a,b,c,d; pair A=(1,9), B=(-11,0), C=(4,0), D; b = abs(C-A); c = abs(B-A); D = (b*B+c*C)/(b+c); draw(A--B--C--A--D,black); MA(B,A,D,2,green); MA(D,A,C,2,green); label(&quot;$A$&quot;,A,(1,1));label(&quot;$B$&quot;,B,(-1,-1));label(&quot;$C$&quot;,C,(1,-1));label(&quot;$D$&quot;,D,(0,-1)); dot(A^^B^^C^^D,blue);label(&quot;$b$&quot;,(A+C)/2,(1,0));label(&quot;$c$&quot;,(A+B)/2,(0,1));label(&quot;$m$&quot;,(B+D)/2,(0,-1));label(&quot;$n$&quot;,(D+C)/2,(0,-1)); &lt;/asy&gt;<br /> <br /> == Examples ==<br /> <br /> # Let ABC be a triangle with angle bisector AD with D on line segment BC. If &lt;math&gt; BD = 2, CD = 5,&lt;/math&gt; and &lt;math&gt; AB + AC = 10 &lt;/math&gt;, find AB and AC.&lt;br&gt; '''''Solution:''''' By the angle bisector theorem, &lt;math&gt; \frac{AB}2 = \frac{AC}5&lt;/math&gt; or &lt;math&gt; AB = \frac 25 AC &lt;/math&gt;. Plugging this into &lt;math&gt; AB + AC = 10 &lt;/math&gt; and solving for AC gives &lt;math&gt; AC = \frac{50}7&lt;/math&gt;. We can plug this back in to find &lt;math&gt; AB = \frac{20}7 &lt;/math&gt;.<br /> # In triangle ABC, let P be a point on BC and let &lt;math&gt; AB = 20, AC = 10, BP = \frac{20\sqrt{3}}3, CP = \frac{10\sqrt{3}}3 &lt;/math&gt;. Find the value of &lt;math&gt; m\angle BAP - m\angle CAP &lt;/math&gt;. &lt;br&gt; '''''Solution:''''' First, we notice that &lt;math&gt; \frac{AB}{BP}=\frac{AC}{CP} &lt;/math&gt;. Thus, AP is the angle bisector of angle A, making our answer 0.<br /> # Part '''(b)''', [[1959 IMO Problems/Problem 5]].<br /> <br /> == See also ==<br /> * [[Angle bisector]]<br /> * [[Geometry]]<br /> * [[Stewart's Theorem]]<br /> <br /> [[Category:Geometry]]<br /> <br /> [[Category:Theorems]]</div> Alleycat https://artofproblemsolving.com/wiki/index.php?title=Gmaas&diff=94363 Gmaas 2018-05-09T00:12:43Z <p>Alleycat: /* Gmaas Facts */</p> <hr /> <div>=== Gmaas Facts ===<br /> - None of these facts a true, except all of them<br /> <br /> - gm lion isn't even his final form.<br /> <br /> - piphi posts GMAAS pictures in his blog<br /> https://artofproblemsolving.com/community/c580181<br /> <br /> -Gmaas once farted. The result was the Big Bang.<br /> <br /> - Gmaas knows these digits of pi from memory: 3.1415926535897932384626433. Edit: He invented pi and pie.<br /> <br /> -since Gmaass invented pi, he knows that the digits are really 9.587979087879877897087r09780970987098790870987609870987908067897578786. digit-ese for &quot;Gmaass is the ruler of the earth.&quot;<br /> <br /> - Gmaas' wealth is unknown, but it is estimated to be way more than Scrooge's.<br /> <br /> - Gmaas has a summer house on Mars.<br /> <br /> -Gmaas attended Harvard, Yale, Stanford, MIT, UC Berkeley, Princeton, Columbia, and Caltech at the same time.<br /> <br /> -Gmaas also attended Hogwarts and was a prefect.<br /> <br /> -Mrs.Norris is Gmaas's archenemy.<br /> <br /> -Gmaas is a demigod and attends Camp Half-Blood over summer. He is the counselor for the Apollo cabin.<br /> <br /> -Gmaas painted the Mona Lisa, The Last Supper, and A Starry Night.<br /> <br /> -Gmaas actually attended all the Ivy Leagues.<br /> <br /> - I am Gmaas<br /> <br /> -I too am Gmaas<br /> <br /> -But it is I who is Gmaas<br /> <br /> -Gmass is us all<br /> <br /> - Gmaas was captured by the infamous j3370 in 2017 but was released due to sympathy. EDIT: j3370 only captured his concrete form, his abstract form cannot be processed by a feeble human brain<br /> <br /> -Gmaas's fur is purple and yellow and red and green and orange and blue and brown and pink all at the same time. <br /> <br /> -Gmaas crossed the event horizon of a black hole and ended up in the AoPS universe.<br /> <br /> -Gmaas crossed the Delaware River with Washington.<br /> <br /> -Gmaas also crossed the Atlantic with the pilgrims.<br /> <br /> -if you are able to capture a Gmaas hair, he will give you some of his gmaas power.<br /> <br /> -Chuck norris makes Gmaas jokes.<br /> <br /> -Gmaas is also the ruler of Oceania, Eastasia, and Eurasia (1984 reference)<br /> <br /> -Gmaas also owns Animal Farm. Napoleon was his servant.<br /> <br /> -Gmaas is the only one who knows where Amelia Earhart is.<br /> <br /> -Gmaas is the only cat that has been proven transcendental.<br /> <br /> - Gmaas happened to notice http://artofproblemsolving.com/community/c402403h1598015p9983782 and is not very happy about it.<br /> <br /> -Grumpy cat reads Gmaas memes. EDIT: Grumpy cat then steals them and claims they're his. Gmass in't very happy about that, either.<br /> <br /> - The real reason why AIME cutoffs aren't out yet is because Gmaas refused to grade them due to too much problem misplacement.<br /> <br /> -Gmaas dueled Grumpy Cat and lost. He wasn't trying.<br /> <br /> -Gmaas sits on the statue of pallas and says forevermore (the Raven refrence )<br /> <br /> - Gmaas does merely not use USD. He owns it.<br /> <br /> -Gmaas really knows that roblox is awful and does not play it seriously, thank god our lord is sane<br /> <br /> -In 2003, Gmaas used elliptical curves to force his reign over AoPS.<br /> <br /> -&quot;Actually, my name is spelled GMAAS&quot;<br /> <br /> - Gmaas is the smartest living being in the universe.<br /> <br /> -It was Gmaas who helped Monkey King on the Journey to the West.<br /> <br /> -Gmaas is the real creator of Wikipedia.<br /> <br /> -It is said Gmaas could hack any website he desires.<br /> <br /> -Gmaas is the basis of Greek mythology<br /> <br /> -Gmaas once sold Google to a man for around &lt;math&gt;12&lt;/math&gt; dollars!<br /> <br /> -Gmaas uses a HP printer.<br /> <br /> -Gmaas owns all AoPS staff including Richard Rusczyk<br /> <br /> -Gmaas was there when Yoda was born.<br /> <br /> - Gmaas's true number of lives left is unknown; however, Gmaas recently confirmed that he had at least one left. Why doesn't Gmaas have so many more lives than other cats? The power of Gmaas.<br /> <br /> - sseraj once spelled gmaas as gmaas on accident in Introduction to Geometry (1532).<br /> <br /> -Gmaas actively plays Roblox, and is a globally ranked professional gamer: https://www.roblox.com/users/29708533/profile...but he hates Roblox.<br /> <br /> -Gmaas has beaten Chuck Norris and The Rock and John Cena all together in a fight.<br /> <br /> -Gmaas is a South Korean, North Korean, Palestinian, Israeli, U.S., and Soviet citizen at the same time. EDIT: Gmaas has now been found out to be a citizen of every country in the world. Gmaas seems to enjoy the country of AOPS best, however.<br /> <br /> -&quot;i am sand&quot; destroyed Gmaas in FTW<br /> <br /> - sseraj posted a picture of gmaas with a game controller in Introduction to Geometry (1532).<br /> <br /> -Gmaas plays roblox mobile edition and likes Minecraft, Candy Crush, and Club Penguin Rewritten. He also &lt;math&gt;\boxed{\text{loves}}&lt;/math&gt; [b]Cat[/b]ch that fish.<br /> <br /> -Gmaas is Roy Moore's horse in the shape of a cat<br /> <br /> -Gmaas is a known roblox/club penguin rewritten player and is a legend at it. He has over &lt;math&gt;289547987693&lt;/math&gt; robux and &lt;math&gt;190348&lt;/math&gt; in CPR.<br /> <br /> -This is all hypothetical EDIT: This is all factual <br /> <br /> -Gmaas's real name is Princess. He has a sibling named Rusty/Fireheart/Firestar<br /> (Warrior cats reference)<br /> <br /> - He is capable of salmon powers, according to PunSpark (ask him)<br /> <br /> The Gmaas told Richard Rusczyk to make AoPS<br /> <br /> -The Gmaas is everything. Yes, you are part of the Gmaas-Dw789<br /> <br /> -The Gmaas knows every dimension up to 9999999999999999999999999999999999999999999999999th dimension<br /> <br /> -Certain theories provide evidence that he IS darth plagueis the wise<br /> <br /> -Gmaas is &quot;TIRED OF PEOPLE ADDING TO HIS PAGE!!&quot; (Maas 45).<br /> <br /> -Gmaas has multiple accounts; some of them are pifinity, cyumi, squareman, Electro3.0, and lakecomo224<br /> <br /> -Gmaas has a penguin servant named sm24136 who runs GMAASINC. The penguin may or may not be dead. <br /> <br /> -Gmaas owns a TARDIS, and can sometimes be seen traveling to other times for reasons unknown.<br /> <br /> -Gmaas knows how to hack into top secret aops community pages<br /> <br /> -Gmaas was a river clant cat who crossed the event horizon of a black hole and came out the other end!<br /> <br /> - Gmaas is king of the first men, the anduls<br /> <br /> -Gmaas is a well known professor at MEOWston Academy<br /> <br /> -Gmaas is a Tuna addict, along with other, more potent fish such as Salmon and Trout<br /> <br /> - Gmaas won the reward of being cutest and fattest cat ever--he surpassed grumpy cat (He also out grumpied grumpy cat!!!)<br /> <br /> -Last sighting 1665 Algebra-A 3/9/18 at 9:08 PM<br /> <br /> - owner of sseraj, not pet<br /> <br /> - embodiment of life and universe and beyond <br /> <br /> - Watches memes<br /> <br /> -After Death became the GOD OF HYPERDEATH and obtained over 9000 souls <br /> <br /> -Gmaas's real name is Pablo Diego José Francisco de Paula Juan Nepomuceno María de los Remedios Cipriano de la Santísima Trinidad Ruiz y Picasso [STOP RICK ROLLING. (Source)]<br /> <br /> -Gmaas is a certified Slytherin.<br /> <br /> -Gmaas once slept on sseraj's private water bed, so sseraj locked him in the bathroom <br /> <br /> -Gmaas has superpowers that allow him to overcome the horrors of Mr. Toilet (while he was locked in the bathroom)<br /> <br /> - Gmaas once sat on an orange on a pile of AoPS books, causing an orange flavored equation explosion.<br /> <br /> -Gmaas once conquered the moon and imprinted his face on it until asteroids came.<br /> <br /> -Gmaas is a supreme overlord who must be given &lt;math&gt;10^{1000000000000000000000^{1000000000000000000000}}&lt;/math&gt; minecraft DIAMONDS<br /> <br /> - gmaas is the Doctor Who lord; he sports Dalek-painted cars and eats human finger cheese and custard, plus black holes.<br /> <br /> - Gmaas is everyone's favorite animal. <br /> <br /> - He lives with sseraj. <br /> <br /> -Gmaas is my favorite pokemon<br /> <br /> -Gmaas dislikes number theory but enjoys geometry.<br /> <br /> - Gmaas is cool<br /> <br /> - He is often overfed (with probability &lt;math&gt;\frac{3972}{7891}&lt;/math&gt;), or malnourished (with probability &lt;math&gt;\frac{3919}{7891}&lt;/math&gt;) by sseraj.<br /> <br /> - He has &lt;cmath&gt;\sum_{k=1}^{267795} [k(k+1)]+GMAAS+GMAAAAAAAS&lt;/cmath&gt; supercars, excluding the Purrari and the 138838383 Teslas. <br /> <br /> - He is an employee of AoPS.<br /> <br /> - He is a gmaas with yellow fur and white hypnotizing eyes.<br /> <br /> - He was born with a tail that is a completely different color from the rest of his fur.<br /> <br /> - His stare is very hypnotizing and effective at getting table scraps.<br /> <br /> - He sometimes appears several minutes before certain classes start as an admin. <br /> <br /> - He died from too many Rubik's cubes in an Introduction to Algebra A class, but got revived by the Dark Lord at 00:13:37 AM the next day.<br /> <br /> - It is uncertain whether or not he is a cat, or is merely some sort of beast that has chosen to take the form of a cat (specifically a Persian Smoke.) <br /> <br /> - Actually, he is a cat. He said so. And science also says so.<br /> <br /> - He is distant relative of Mathcat1234.<br /> <br /> - He is very famous now, and mods always talk about him before class starts.<br /> <br /> - His favorite food is AoPS textbooks because they help him digest problems.<br /> <br /> - Gmaas tends to reside in sseraj's fridge.<br /> <br /> - Gmaas once ate all sseraj's fridge food, so sseraj had to put him in the freezer.<br /> <br /> - The fur of Gmaas can protect him from the harsh conditions of a freezer.<br /> <br /> - Gmaas sightings are not very common. There have only been 8 confirmed sightings of Gmaas in the wild.<br /> <br /> - Gmaas is a sage omniscient cat.<br /> <br /> - He is looking for suitable places other than sseraj's fridge to live in.<br /> <br /> - Places where gmaas sightings have happened: <br /> ~The Royal Scoop ice cream store in Bonita Beach Florida<br /> <br /> ~MouseFeastForCats/CAT 8 Mouse Apartment 1083<br /> <br /> ~Alligator Swamp A 1072 <br /> <br /> ~Alligator Swamp B 1073<br /> <br /> ~Prealgebra A (1488)<br /> <br /> ~Introduction to Algebra A (1170)<br /> <br /> ~Introduction to Algebra B (1529)<br /> <br /> ~Welcome to Panda Town Gate 1076<br /> <br /> ~Welcome to Gmaas Town Gate 1221<br /> <br /> ~Welcome to Gmaas Town Gate 1125<br /> <br /> ~33°01'17.4&quot;N 117°05'40.1&quot;W (Rancho Bernardo Road, San Diego, CA)<br /> <br /> ~The other side of the ice in Antarctica<br /> <br /> ~Feisty Alligator Swamp 1115<br /> <br /> ~Introduction to Geometry 1221 (Taught by sseraj)<br /> <br /> ~Introduction to Counting and Probability 1142 <br /> <br /> ~Feisty-ish Alligator Swamp 1115 (AGAIN)<br /> <br /> ~Intermediate Counting and Probability 1137<br /> <br /> ~Intermediate Counting and Probability 1207<br /> <br /> ~Posting student surveys<br /> <br /> ~USF Castle Walls - Elven Tribe 1203<br /> <br /> ~Dark Lord's Hut 1210<br /> <br /> ~AMC 10 Problem Series 1200<br /> <br /> ~Intermediate Number Theory 1138<br /> <br /> ~Intermediate Number Theory 1476<br /> <br /> ~Introduction To Number Theory 1204. Date:7/27/16.<br /> <br /> ~Algebra B 1112<br /> <br /> ~Intermediate Algebra 1561 7:17 PM 12/11/16<br /> <br /> ~Nowhere Else, Tasmania<br /> <br /> ~Earth Dimension C-137<br /> ~Geometry 1694 at 1616 PST military time. There was a boy riding him, and he seemed extremely miffed.<br /> <br /> <br /> - These have all been designated as the most glorious sections of Aopsland now (especially the USF castle walls), but deforestation is so far from threatens the wild areas (i.e. Alligator Swamps A&amp;B).<br /> <br /> - Gmaas has also been sighted in Olympiad Geometry 1148.<br /> <br /> - Gmaas has randomly been known to have sent his minions into Prealgebra 2 1163. However, the danger is passed, that class is over.<br /> <br /> - Gmaas once snuck into sseraj's email so he could give pianoman24 an extension in Introduction to Number Theory 1204. This was 1204 minutes after his sighting on 7/27/16.<br /> <br /> - Gmaas also has randomly appeared on top of the USF's Tribal Bases(he seems to prefer the Void Tribe). However, the next day there is normally a puddle in the shape of a cat's underbelly wherever he was sighted. Nobody knows what this does. <br /> <br /> EDIT: Nobody has yet seen him atop a tribal base yet.<br /> <br /> - Gmaas are often under the disguise of a penguin or cat. Look out for them.<br /> <br /> EDIT: Gmaas rarely disguises himself as a penguin.<br /> <br /> - He lives in the shadows. Is he a dream? Truth? Fiction? Condemnation? Salvation? AoPS site admin? He is all these things and none of them. He is... Gmaas.<br /> <br /> EDIT: He IS an AoPS site admin.<br /> <br /> - If you make yourself more than just a cat... if you devote yourself to an ideal... and if they can't stop you... then you become something else entirely. A LEGEND. Gmaas now belongs to the ages.<br /> <br /> - Is this the real life? Is this just fantasy? No. This is Gmaas, the legend.<br /> <br /> -Aha!! An impostor!! <br /> http://www.salford.ac.uk/environment-life-sciences/research/applied-archaeology/greater-manchester-archaeological-advisory-service<br /> (look at the acronym).<br /> <br /> -EDIT. The above fact is slightly irrelevant.<br /> <br /> - Gmaas might have been viewing (with a &lt;math&gt;\frac{99999.\overline{9}}{100000}&lt;/math&gt; chance) the Ultimate Survival Forum. He (or is he a she?) is suspected to be transforming the characters into real life. Be prepared to meet your epic swordsman self someday. If you do a sci-fi version of USF, then prepare to meet your Overpowered soldier with amazing weapons one day.<br /> <br /> - EDIT: Gmaas is a he.<br /> <br /> -Gmaas is love, Gmaas is life<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mew.<br /> <br /> - Gmaas is on the list of &quot;Elusive Creatures.&quot; If you have questions or want the full list, contact moab33.<br /> <br /> - Gmaas can be summoned using the &lt;math&gt;\tan(90)&lt;/math&gt; ritual. Draw a pentagram and write the numerical value of &lt;math&gt;\tan(90)&lt;/math&gt; in the middle, and he will be summoned.<br /> <br /> - EDIT: The above fact is incorrect. math101010 has done this and commented with screenshot proof at the below link, and Gmaas was not summoned.<br /> https://artofproblemsolving.com/community/c287916h1291232<br /> <br /> - EDIT EDIT: The above 'proof' is non-conclusive. math101010 had only put an approximation.<br /> <br /> - Gmaas's left eye contains the singularity of a black hole. (Only when everyone in the world blinks at the same time within a nano-nano second.)<br /> <br /> - EDIT: That has never happened and thus it does not contain the singularity of a black hole.<br /> <br /> - Lord Grindelwald once tried to make Gmaas into a Horcrux, but Gmaas's fur is Elder Wand protected and secure, as Kendra sprinkled holly into his fur.<br /> <br /> -Despite common belief, Harry Potter did not defeat Lord Voldemort. Gmaas did.<br /> <br /> - The original owner of Gmaas is Gmaas.<br /> <br /> - Gmaas was not the fourth Peverell brother, but he ascended into a higher being and now he resides in the body of a cat, as he was before. Is it a cat? We will know. (And the answer is YES.) EDIT: he wasn't the fourth Peverell brother, but he was a cousin of theirs, and he was the one who advised Ignotus to give up his cloak.<br /> <br /> - It is suspected that Gmaas may be ordering his cyber hairballs to take the forums, along with microbots.<br /> <br /> - The name of Gmaas is so powerful, it radiates Deja Mu.<br /> <br /> - Gmaas rarely frequents the headquarters of the Illuminati. He was their symbol for one yoctosecond, but soon decided that the job was too low for his power to be wasted on.<br /> <br /> - It has been wondered if Gmaas is the spirit of Obi-Wan Kenobi or Anakin Skywalker in a higher form, due to his strange capabilities and powers.<br /> <br /> - Gmaas has a habit of sneaking into computers, joining The Network, and exiting out of some other computer.<br /> <br /> - It has been confirmed that gmaas uses gmewal as his email service<br /> <br /> - Gmaas enjoys wearing gmean shorts<br /> <br /> - Gmaas has a bright orange tail with hot pink spirals. Or he had for 15 minutes. <br /> <br /> - Gmaas is well known behind his stage name, Michael Stevens (also known as Vsauce XD), or his page name, Purrshanks. EDIT: Crookshanks was his brother.<br /> <br /> - Gmaas rekt sseraj at 12:54 June 4, 2016 UTC time zone. And then the Doctor chased him.<br /> <br /> - Gmaas watchers know that the codes above are NOT years. They are secret codes for the place. But if you've edited that section of the page, you know that.<br /> <br /> - Gmaas is a good friend of the TARDIS and the Millenium Falcon. <br /> <br /> - In the Dark Lord's hut, gmaas was seen watching Doctor Who. Anyone who has seen the Dark Lord's hut knows that both Gmaas and the DL (USF code name of the Dark Lord) love BBC. How Gmaas gave him a TV may be lost to history. And it has been lost.<br /> <br /> - The TV has been noticed to be invincible. Many USF weapons, even volcano rings, have tried (and failed) to destroy it. The last time it was seen was on a Kobold display case outside of a mine. The display case was crushed, and a report showed a spy running off with a non-crushed TV.<br /> <br /> -The reason why Dacammel left the USF is that gmaas entrusted his TV to him, and not wanting to be discovered by LF, Cobra, or Z9, dacammel chose to leave the USF, but is regretting it, as snakes keep spawning from the TV.<br /> <br /> - EDIT: The above fact is somewhat irrelevant.<br /> <br /> -EDIT EDIT. Dacammel gave the TV back to gmaas, and he left the dark side and their cookies alone. <br /> <br /> - Gmaas is a Super Duper Uper Cat Time Lord. He has &lt;math&gt;57843504&lt;/math&gt; regenerations and has used &lt;math&gt;3&lt;/math&gt;. &lt;cmath&gt;9\cdot12\cdot2\cdot267794=57843504&lt;/cmath&gt;. <br /> <br /> -Gmaas highly enjoys destroying squeaky toys until he finds the squeaky part, then destroys the squeaky part.<br /> <br /> - Gmaas loves to eat turnips. At &lt;math&gt;\frac{13}{32}&lt;/math&gt; of the sites he was spotted at, he was seen with a turnip.<br /> <br /> -Gmaas has a secret hidden garden full of turnips under sseraj's house.<br /> <br /> - Gmaas has three tails, one for everyday life, one for special occasions, and one that's invisible.<br /> <br /> -Gmaas is a dangerous creature. If you ever meet him, immediately join his army or you will be killed.<br /> <br /> -Gmaas is in alliance with the Cult of Skaro. How did he get an alliance with ruthless creatures that want to kill everything in sight? Nobody knows (except him), not even the leader of the Cult of Skaro.<br /> <br /> -Gmaas lives in Gallifrey and in Gotham City (he has sleepovers with Batman).<br /> <br /> -Gmaas is an excellent driver. EDIT: he was to one who designed the driver's license test, although he didn't bother with the permit test.<br /> <br /> -The native location of Gmaas is the twilight zone.<br /> <br /> -Donald Trump once sang &quot;All Hail the Chief&quot; to Gmaas, 3 days after being sworn in as US President.<br /> <br /> - Gmaas likes to talk with rrusczyk from time to time.<br /> <br /> - Gmaas can shoot fire from his smelly butt.<br /> <br /> - Gmaas is the reason why the USF has the longest thread on AoPS.<br /> <br /> - Gmass is an avid watcher of the popular T.V. show &quot;Bernie Sanders and the Gauntlet of DOOM&quot;<br /> <br /> - sseraj, in 1521 Introduction to Number Theory, posted an image of Gmaas after saying &quot;Who wants to see 5space?&quot; at around 5:16 PM Mountain Time, noting Gmaas was &quot;also 5space&quot;<br /> <br /> -EDIT: he also did it in Introduction to Algebra A once.<br /> <br /> - Gmaas is now my HD background on my Mac.<br /> <br /> - In 1521 Into to Number Theory, sseraj posted an image of a 5space Gmaas fusion. (First sighting) <br /> <br /> - Also confirmed that Gmaas doesn't like ketchup because it was the only food left the photo.<br /> <br /> - In 1447 Intro to Geometry, sseraj posted a picture of Gmaas with a rubik's cube suggesting that Gmaas's has an average solve time of &lt;math&gt;-GMAAS&lt;/math&gt; seconds.<br /> <br /> -Gmaas beat Superman in a fight with ease<br /> <br /> -Gmaas was an admin of Roblox<br /> <br /> -Gmaas traveled around the world, paying so much &lt;math&gt;MONEY&lt;/math&gt; just to eat :D<br /> <br /> -Gmaas is a confirmed Apex predator and should not be approached, unless in a domestic form.<br /> Summary:<br /> <br /> -When Gmaas subtracts &lt;math&gt;0.\overline{99}&lt;/math&gt; from &lt;math&gt;1&lt;/math&gt;, the difference is greater than &lt;math&gt;0&lt;/math&gt;.<br /> <br /> -Gmaas was shown to have fallen on Wed Aug 23 2017: https://ibb.co/bNrtmk https://ibb.co/jzUDmk<br /> <br /> -Gmaas died on August ,24, 2017, but fortunately IceParrot revived him after about 2 mins of being dead.<br /> <br /> -The results of the revival are top secret, and nobody knows what happened.<br /> <br /> -sseraj, in 1496 Prealgebra 2, said that Gmaas is Santacat.<br /> <br /> -sseraj likes to post a picture of gmaas in every class he passes by.<br /> <br /> -sseraj posted a picture of gmaas as an Ewok, suggesting he resides on the moon of Endor. Unfortunately, the moon of Endor is also uninhabitable ever since the wreckage of the Death Star changed the climate there. It is thought gmaas is now wandering space in search for a home.<br /> EDIT: What evidence is there Endor was affected? Other Ewoks still live there.<br /> EDIT EDIT: also, glass doesn't care. He can live there no matter what the climate is.<br /> <br /> -Gmaas is the lord of the pokemans<br /> <br /> -Gmaas can communicate with, and sometimes control any other cats, however this is very rare, as cats normally have a very strong will.<br /> <br /> -Picture of Gmaas http://i.imgur.com/PP9xi.png<br /> <br /> -Known by Mike Miller<br /> <br /> -Gmaas got mad at sseraj once, so he locked him in his own freezer<br /> <br /> -Then, sseraj decided to eat all of Gmaas's hidden turnips in the freezer as punishment<br /> <br /> -Gmaas is an obviously omnipotent cat.<br /> <br /> -ehawk11 met him<br /> <br /> -sseraj is known to post pictures of Gmaas on various AoPS classrooms. It is not known if these photos have been altered with the editing program called 'Photoshop'.<br /> <br /> -sseraj has posted pictures of gmaas in '&quot;intro to algebra&quot;, before class started, with the title, &quot;caption contest&quot;. anyone who posted a caption mysteriously vanished in the middle of the night. <br /> EDIT: This has happened many times, including in Introduction to Geometry 1533, among other active classes. The person writing this (Poeshuman) did participate, and did not disappear. (You could argue Gmaas is typing this through his/her account...)<br /> <br /> - gmaas has once slept in your bed and made it wet<br /> <br /> -It is rumored that rrusczyk is actually Gmaas in disguise<br /> <br /> -Gmaas is suspected to be a Mewtwo in disguise<br /> <br /> -Gmaas is a cat but has characteristics of every other animal on Earth.<br /> <br /> -Gmaas is the ruler of the universe and has been known to be the creator of the species &quot;Gmaasians&quot;.<br /> <br /> -There is a rumor that Gmaas is starting a poll<br /> <br /> -Gmaas is a rumored past ThunderClan cat who ran away, founded GmaasClan, then became a kittypet.<br /> <br /> -There is a rumored sport called &quot;Gmaas Hunting&quot; where people try to successfully capture gmaas in the wild with video/camera/eyes. Strangely, no one has been able to do this, and those that have have mysteriously disappeared into the night. Nobody knows why. The person who is writing this(g1zq) has tried Gmaas Hunting, but has never been successful.<br /> <br /> - Gmaas burped and caused an earthquake.<br /> <br /> - Gmaas once drank from your pretty teacup.<br /> <br /> === Gmaas photos ===<br /> http://cdn.artofproblemsolving.com/images/f/f/8/ff8efef3a0d2eb51254634e54bec215b948a1bba.jpg<br /> <br /> === gmaas in Popular Culture ===<br /> <br /> - [s]Currently, is being written (by themoocow) about the adventures of gmaas. It is aptly titled, &quot;The Adventures of gmaas&quot;.[/s]Sorry, this was a rick roll troll.<br /> <br /> - BREAKING NEWS: tigershark22 has found a possible cousin to gmaas in Raymond Feist's book Silverthorn. They are mountain dwellers, gwali. Not much are known about them either, and when someone asked,&quot;What are gwali?&quot; the customary answer &quot;This is gwali&quot; is returned. Scientist 5space is now looking into it.<br /> <br /> - Sullymath and themoocow are also writing a book about Gmaas<br /> <br /> -Sighting of Gmaas: https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project<br /> <br /> -Oryx the mad god is actually gmass wearing a suit of armor. This explains why he is never truly killed<br /> <br /> - Potential sighting of gmaas [http://www.gmac.com/frequently-asked-questions/gmass-search-service.aspx]<br /> <br /> - Gmaas has been spotted in some Doctor Who and Phineas and Ferb episodes, such as Aliens of London, Phineas and Ferb Save Summer, Dalek, Rollercoaster, Rose, Boom Town, The Day of The Doctor, Candace Gets Busted, and many more.<br /> <br /> - Gmaas can be found in many places in Plants vs. Zombies Garden Warfare 2 and Bloons TD Battles<br /> <br /> - gmaas was an un-credited actor in the Doctor Who story Knock Knock, playing a Dryad. How he shrunk, we will never know.<br /> <br /> -oadaegan is also writing a story about him. He is continuing the book that was started by JpusheenS. when he is done he will post it here<br /> <br /> -Gmaas is a time traveler from 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 B.C.<br /> <br /> -No one knows if Gmass is a Mr. mime in a cat skin, the other way around, or just a downright combination of both.<br /> <br /> -In it, it mentions these four links as things Gmass is having trouble (specifically technical difficulties). What could it mean? Links:<br /> https://docs.google.com/document/d/1NZ0XcFYm80sA-fAoxnm7ulMCwdNU75Va_6ZjRHfSHV0<br /> https://docs.google.com/document/d/1ELN7ORauFFv1dwpU_u-ah_dFJHeuJ3szYxoeC1LlDQg/<br /> https://docs.google.com/document/d/1oy9Q3F7fygHw-OCWNEVE8d-Uob2dxVACFcGUcLmk3fA<br /> https://docs.google.com/document/d/1jzb9Q6FmDmrRyXwnik3e0sYw5bTPMo7aBwugmUbA13o<br /> <br /> <br /> - Another possible Gmaas sighting [https://2017.spaceappschallenge.org/challenges/warning-danger-ahead/and-you-can-help-fight-fires/teams/gmaas/project]<br /> <br /> -&lt;math&gt;Another&lt;/math&gt; sighting? [https://www.radarbox24.com/data/flights/GMAAS]<br /> <br /> -Gmaas has been sighted several times on the Global Announcements forum<br /> <br /> -Gmaas uses the following transportation: &lt;img&gt; http://cdn.artofproblemsolving.com/images/3/6/8/368da4e615ea3476355ee3388b39f30a48b8dd48.jpg &lt;/img&gt;<br /> <br /> - When Gmaas was mad, he started world wars 1 &amp; 2. It is only because of Gmaas that we have not had World War 3.<br /> <br /> - Gmaas is the only cat to have been proved irrational and transcendental, though we suspect all cats fall in the first category.<br /> <br /> - Gmaas plays Geometry Dash and shares an account with Springhill, his username is D3m0nG4m1n9.<br /> <br /> -Gmaas likes to whiz on the wilzo<br /> <br /> -Gmaas has been spotted in AMC 8 Basics<br /> <br /> -Gmaas is cool<br /> <br /> -Gmaas hemoon card that does over 9000000 dmg<br /> <br /> -Gmaas is a skilled swordsman who should not to be mistaken for Puss in Boots. Some say he even trained the mysterious and valiant Meta Knight.<br /> <br /> -Kirby once swallowed Gmaas. Gmaas had to spit him out.</div> Alleycat