Difference between revisions of "2020 AIME I Problems/Problem 13"

(Solution 11 (Simple geometry))
 
(21 intermediate revisions by 5 users not shown)
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pen dotstyle = black; /* point style */  
 
pen dotstyle = black; /* point style */  
 
real xmin = -10.645016481888238, xmax = 5.4445786933235505, ymin = 0.7766255516825293, ymax = 9.897545413994122;  /* image dimensions */
 
real xmin = -10.645016481888238, xmax = 5.4445786933235505, ymin = 0.7766255516825293, ymax = 9.897545413994122;  /* image dimensions */
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882);  
+
pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.8235294117647059,0.0396078431372549,0.07529411764705882);  
  
 
draw((-6.837129089839387,8.163360372429347)--(-6.8268938290378,5.895596632024835)--(-4.33118398380513,6.851781504978754)--cycle, linewidth(2) + rvwvcq);  
 
draw((-6.837129089839387,8.163360372429347)--(-6.8268938290378,5.895596632024835)--(-4.33118398380513,6.851781504978754)--cycle, linewidth(2) + rvwvcq);  
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Points are defined as shown. It is pretty easy to show that <math>\triangle AFE \sim \triangle AGH</math> by spiral similarity at <math>A</math> by some short angle chasing. Now, note that <math>AD</math> is the altitude of <math>\triangle AFE</math>, as the altitude of <math>AGH</math>. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that <math>AD/2 = \frac{\sqrt{18}}{2}</math>, the altitude of <math>\triangle AFE</math>. Similarly, the altitude of <math>\triangle AGH</math> is the altitude of <math>\triangle ABC</math>, or <math>\frac{3\sqrt{7}}{2}</math>. However, it's not too hard to see that <math>GB = HC = 1</math>, and therefore <math>[AGH] = [ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~awang11
 
Points are defined as shown. It is pretty easy to show that <math>\triangle AFE \sim \triangle AGH</math> by spiral similarity at <math>A</math> by some short angle chasing. Now, note that <math>AD</math> is the altitude of <math>\triangle AFE</math>, as the altitude of <math>AGH</math>. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that <math>AD/2 = \frac{\sqrt{18}}{2}</math>, the altitude of <math>\triangle AFE</math>. Similarly, the altitude of <math>\triangle AGH</math> is the altitude of <math>\triangle ABC</math>, or <math>\frac{3\sqrt{7}}{2}</math>. However, it's not too hard to see that <math>GB = HC = 1</math>, and therefore <math>[AGH] = [ABC]</math>. From here, we get that the area of <math>\triangle ABC</math> is <math>\frac{15\sqrt{7}}{14} \implies \boxed{036}</math>, by similarity. ~awang11
  
==Solution 2(coord bash + basic geometry)==
+
==Solution 2==
 +
Let <math>M_A</math>, <math>M_B</math>, <math>M_C</math> be the midpoints of arcs <math>BC</math>, <math>CA</math>, <math>AB</math>. By Fact 5, we know that <math>M_AB = M_AC = M_AI</math>, and so by Ptolmey's theorem, we deduce that <cmath>AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.</cmath>
 +
In particular, we have <math>AI = IM_A</math>.
 +
<asy>
 +
defaultpen(fontsize(10pt));
 +
size(200);
 +
pair A, B, C, D, E, F, I, P, MA, MB, MC;
 +
B = (0,0);
 +
C = (5,0);
 +
A = IP(Circle(B, 4), Circle(C, 6), 0);
 +
I = incenter(A, B, C);
 +
D = extension(A, I, B, C);
 +
P = midpoint(A--D);
 +
E = extension(P, rotate(90, P)*A, B, I);
 +
F = extension(P, rotate(90, P)*A, C, I);
 +
MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1);
 +
MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1);
 +
MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1);
 +
draw(A--B--C--cycle, orange);
 +
draw(circumcircle(A, B, C), red);
 +
draw(A--E--F--cycle, lightblue);
 +
draw(E--D--F, lightblue);
 +
draw(A--MA^^B--MB^^C--MC, heavygreen);
 +
draw(MA--MB--MC--cycle, magenta);
 +
dot("$A$", A, dir(120));
 +
dot("$B$", B, dir(220));
 +
dot("$C$", C, dir(320));
 +
dot("$D$", D, dir(230));
 +
dot("$E$", E, dir(330));
 +
dot("$F$", F, dir(250));
 +
dot("$I$", I, dir(80));
 +
dot("$M_A$", MA, dir(270));
 +
dot("$M_B$", MB, dir(60));
 +
dot("$M_C$", MC, dir(150));
 +
</asy>
 +
Now the key claim is that:
 +
 
 +
<b>Claim:</b> <math>\triangle DEF</math> and <math>\triangle M_AM_BM_C</math> are homothetic at <math>I</math> with ratio <math>2</math>.
 +
 
 +
<i>Proof.</i> First, we show that <math>D</math> is the midpoint of <math>M_AI</math>. Indeed, we have
 +
<cmath>\frac{ID}{DM_A} = \frac{BI}{BM_A}\cdot \frac{\sin\angle IBC}{\sin \angle CBM_A} = \frac{BI}{AI}\cdot\frac{\sin \angle B/2}{\sin \angle A/2} = 1</cmath>
 +
by Ratio lemma and Law of Sines.
 +
 
 +
Now observe that:
 +
* <math>\overline{M_BM_C}</math> is the perpendicular bisector of <math>\overline{AI}</math>,
 +
* <math>\overline{EF}</math> is the perpendicular bisector of <math>\overline{AD}</math>, and
 +
* <math>ID = AI/2</math>.
 +
 
 +
Combining these facts gives that <math>\overline{EF}</math> is a midline in <math>\triangle IM_BM_C</math>, which proves the claim. <math>\blacksquare</math>
 +
 
 +
To finish, we compute <math>[M_AM_BM_C]</math>, noting that <math>[AEF] = [DEF] = \tfrac{1}{4}[M_AM_BM_C]</math>.
 +
 
 +
By Heron's, we can calculate the circumradius <math>R = 8/\sqrt{7}</math>, and by Law of Cosines, we get
 +
<cmath>\begin{align*}\cos A &= \frac{9}{16}\implies \cos A/2 = \frac{5}{\sqrt{32}} \
 +
\cos B &= \frac{1}{8} \implies \cos B/2 = \frac{3}{4} \
 +
\cos C &= \frac{3}{4} \implies \cos C/2 = \sqrt{\frac{7}{8}}.\end{align*}</cmath>
 +
Then using <math>[XYZ] = 2R^2\sin X\sin Y\sin Z</math>, we can compute
 +
<cmath>[M_AM_BM_C] = 2\cdot \frac{64}{7}\cdot \frac{5}{\sqrt{32}}\cdot \frac{3}{4}\cdot \frac{\sqrt{7}}{\sqrt{8}} = \frac{30\sqrt{7}}{7}.</cmath>
 +
Thus <math>[AEF] = 15\sqrt{7}/14</math>, which gives a final answer of <math>\boxed{036}</math>.
 +
 
 +
~pinetree1
 +
 
 +
==Solution 3(coord bash + basic geometry)==
 
Let <math>\overline{BC}</math> lie on the x-axis and <math>B</math> be the origin. <math>C</math> is <math>(5,0)</math>. Use Heron's formula to compute the area of triangle <math>ABC</math>. We have <math>s=\frac{15}{2}</math>. and <math>[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}</math>. We now find the altitude, which is <math>\frac{\frac{15\sqrt{7}}{2}}{5}=\frac{3\sqrt{7}}{2}</math>, which is the y-coordinate of <math>A</math>. We now find the x-coordinate of <math>A</math>, which satisfies <math>x^2 + (\frac{3\sqrt{7}}{2})^{2}=16</math>, which gives <math>x=\frac{1}{2}</math> since the triangle is acute. Now using the Angle Bisector Theorem, we have <math>\frac{4}{6}=\frac{BD}{CD}</math> and <math>BD+CD=5</math> to get <math>BD=2</math>. The coordinates of D are <math>(2,0)</math>.
 
Let <math>\overline{BC}</math> lie on the x-axis and <math>B</math> be the origin. <math>C</math> is <math>(5,0)</math>. Use Heron's formula to compute the area of triangle <math>ABC</math>. We have <math>s=\frac{15}{2}</math>. and <math>[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}</math>. We now find the altitude, which is <math>\frac{\frac{15\sqrt{7}}{2}}{5}=\frac{3\sqrt{7}}{2}</math>, which is the y-coordinate of <math>A</math>. We now find the x-coordinate of <math>A</math>, which satisfies <math>x^2 + (\frac{3\sqrt{7}}{2})^{2}=16</math>, which gives <math>x=\frac{1}{2}</math> since the triangle is acute. Now using the Angle Bisector Theorem, we have <math>\frac{4}{6}=\frac{BD}{CD}</math> and <math>BD+CD=5</math> to get <math>BD=2</math>. The coordinates of D are <math>(2,0)</math>.
 
Since we want the area of triangle <math>AEF</math>, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is <math>(\frac{5}{4}, \frac{3\sqrt{7}}{4})</math> and the slope of AD is <math>-\sqrt{7}</math>. The slope of the perpendicular bisector is <math>\frac{1}{\sqrt{7}}</math>. The equation is(in point slope form) <math>y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})</math>.
 
Since we want the area of triangle <math>AEF</math>, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is <math>(\frac{5}{4}, \frac{3\sqrt{7}}{4})</math> and the slope of AD is <math>-\sqrt{7}</math>. The slope of the perpendicular bisector is <math>\frac{1}{\sqrt{7}}</math>. The equation is(in point slope form) <math>y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})</math>.
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The area of AEF is equal to <math>\frac{EF \cdot \frac{AD}{2}}{2}</math> since AD is the altitude of that triangle with EF as the base, with <math>\frac{AD}{2}</math> being the height. <math>EF=\frac{5\sqrt{2}}{\sqrt{7}}</math> and <math>AD=3\sqrt{2}</math>, so <math>[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}</math> which gives <math>\boxed{036}</math>. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo
 
The area of AEF is equal to <math>\frac{EF \cdot \frac{AD}{2}}{2}</math> since AD is the altitude of that triangle with EF as the base, with <math>\frac{AD}{2}</math> being the height. <math>EF=\frac{5\sqrt{2}}{\sqrt{7}}</math> and <math>AD=3\sqrt{2}</math>, so <math>[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}</math> which gives <math>\boxed{036}</math>. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo
  
==Solution 3 (Coordinate Bash + Trig)==
+
==Solution 4 (Coordinate Bash + Trig)==
  
 
<asy>
 
<asy>
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The desired answer is <math>15+7+14=\boxed{036}</math> ~Imayormaynotknowcalculus
 
The desired answer is <math>15+7+14=\boxed{036}</math> ~Imayormaynotknowcalculus
  
==Solution 4 (Barycentric Coordinates)==
+
==Solution 5 (Barycentric Coordinates)==
  
 
<asy>
 
<asy>
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<b>Remark</b>: The area of <math>\triangle{AEF}</math> can also be computed using the <i>Barycentric Area Formula</i>, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.
 
<b>Remark</b>: The area of <math>\triangle{AEF}</math> can also be computed using the <i>Barycentric Area Formula</i>, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.
  
==Solution 5 (geometry+trig)==
+
==Solution 6 (geometry+trig)==
  
 
<asy>
 
<asy>
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(Added :) ~Math_Genius_164)
 
(Added :) ~Math_Genius_164)
  
== Solution 6 ==
+
== Solution 7 ==
 
[[Image:question13.png|frame|none|###px|]]
 
[[Image:question13.png|frame|none|###px|]]
 
First and foremost <math>\big[\triangle{AEF}\big]=\big[\triangle{DEF}\big]</math> as <math>EF</math> is the perpendicular bisector of <math>AD</math>. Now  note that quadrilateral <math>ABDF</math> is cyclic, because <math>\angle{ABF}=\angle{FBD}</math> and <math>FA=FD</math>. Similarly quadrilateral <math>AEDC</math> is cyclic, <cmath>\implies \angle{EDA}=\dfrac{C}{2}, \quad \angle{FDA}=\dfrac{B}{2}</cmath>
 
First and foremost <math>\big[\triangle{AEF}\big]=\big[\triangle{DEF}\big]</math> as <math>EF</math> is the perpendicular bisector of <math>AD</math>. Now  note that quadrilateral <math>ABDF</math> is cyclic, because <math>\angle{ABF}=\angle{FBD}</math> and <math>FA=FD</math>. Similarly quadrilateral <math>AEDC</math> is cyclic, <cmath>\implies \angle{EDA}=\dfrac{C}{2}, \quad \angle{FDA}=\dfrac{B}{2}</cmath>
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-tkhalid
 
-tkhalid
  
==Solution 7 -Trigonometry(only)==
+
==Solution 8 -Trigonometry(only)==
 
[[Image:Aime 13.png|frame|none|###px|]]
 
[[Image:Aime 13.png|frame|none|###px|]]
 
Trig values we use here:
 
Trig values we use here:
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Now let <math>AI \cap EF = G</math>. It is easy to angle chase <math>\angle GIE = 90- \frac{B}{2}</math> and <math>\angle GIF = 90- \frac{C}{2}</math>. Since <math>GI = GD - ID = \frac{3\sqrt2}{2}-\sqrt2=\frac{\sqrt2}{2}</math>, we compute that <cmath>EF = EG + FG = \frac{\sqrt2}{2}(\cot B/2 + \cot C/2) = \frac{\sqrt2}{2}(\frac{3}{\sqrt7}+\sqrt7) = \frac{5\sqrt{14}}{7}</cmath> which implies <cmath>[AEF] = AG*EF/2 = \frac{3\sqrt2}{2} * \frac{5\sqrt{14}}{7} / 2 = \frac{15\sqrt7}{14}</cmath> which gives an answer of <math>\boxed{36}</math>. ~Leonard_my_dude~
 
Now let <math>AI \cap EF = G</math>. It is easy to angle chase <math>\angle GIE = 90- \frac{B}{2}</math> and <math>\angle GIF = 90- \frac{C}{2}</math>. Since <math>GI = GD - ID = \frac{3\sqrt2}{2}-\sqrt2=\frac{\sqrt2}{2}</math>, we compute that <cmath>EF = EG + FG = \frac{\sqrt2}{2}(\cot B/2 + \cot C/2) = \frac{\sqrt2}{2}(\frac{3}{\sqrt7}+\sqrt7) = \frac{5\sqrt{14}}{7}</cmath> which implies <cmath>[AEF] = AG*EF/2 = \frac{3\sqrt2}{2} * \frac{5\sqrt{14}}{7} / 2 = \frac{15\sqrt7}{14}</cmath> which gives an answer of <math>\boxed{36}</math>. ~Leonard_my_dude~
  
==Solution 8 (Official MAA 1)==
+
==Solution 9 (Official MAA 1)==
 
Let <math>x = \angle BAD = \angle CAD</math>, <math>y = \angle CBE = \angle ABE</math>, and <math>z = \angle BCF = \angle ACF</math>. Notice that <math>x+y+z = 90^\circ</math>.
 
Let <math>x = \angle BAD = \angle CAD</math>, <math>y = \angle CBE = \angle ABE</math>, and <math>z = \angle BCF = \angle ACF</math>. Notice that <math>x+y+z = 90^\circ</math>.
  
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</cmath>The requested sum is <math>15+7+14 = 36</math>.
 
</cmath>The requested sum is <math>15+7+14 = 36</math>.
  
==Solution 9 (Official MAA 2)==
+
==Solution 10 (Official MAA 2)==
 
Let the point <math>M</math> be the midpoint of <math>\overline{AD}</math>, let <math>I</math> be the incenter of <math>\triangle ABC</math> which is the common point of lines <math>AD</math>, <math>BE</math>, and <math>CF</math>, and let <math>r</math> be the inradius of <math>\triangle ABC</math>. The semiperimeter of <math>\triangle ABC</math> is<cmath>s = \frac{AB + BC + CA}2 = \frac{15}2,</cmath>and Heron's Formula gives the area of <math>\triangle ABC</math> as<cmath>\sqrt{s(s-AB)(s-BC)(s-CA)} = \frac{15\sqrt7}4.</cmath>This area is also <math>rs</math> implying that <math>r = \frac{\sqrt7}2</math>. Stewart's Theorem gives <math>AD =3\sqrt2</math>. Because the ratio of the areas of <math>\triangle IBC</math> and <math>\triangle ABC</math> is <math>\frac{ID}{AD},</math> it follows that<cmath>ID = AD\cdot\frac{\frac{r\cdot BC}2}{\frac{15\sqrt7}4} = \sqrt2.</cmath>Thus <math>IM = MD - ID = \frac{\sqrt2}2</math>.
 
Let the point <math>M</math> be the midpoint of <math>\overline{AD}</math>, let <math>I</math> be the incenter of <math>\triangle ABC</math> which is the common point of lines <math>AD</math>, <math>BE</math>, and <math>CF</math>, and let <math>r</math> be the inradius of <math>\triangle ABC</math>. The semiperimeter of <math>\triangle ABC</math> is<cmath>s = \frac{AB + BC + CA}2 = \frac{15}2,</cmath>and Heron's Formula gives the area of <math>\triangle ABC</math> as<cmath>\sqrt{s(s-AB)(s-BC)(s-CA)} = \frac{15\sqrt7}4.</cmath>This area is also <math>rs</math> implying that <math>r = \frac{\sqrt7}2</math>. Stewart's Theorem gives <math>AD =3\sqrt2</math>. Because the ratio of the areas of <math>\triangle IBC</math> and <math>\triangle ABC</math> is <math>\frac{ID}{AD},</math> it follows that<cmath>ID = AD\cdot\frac{\frac{r\cdot BC}2}{\frac{15\sqrt7}4} = \sqrt2.</cmath>Thus <math>IM = MD - ID = \frac{\sqrt2}2</math>.
  
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<math>\triangle IBC \sim \triangle IFE</math>. The height of <math>\triangle IBC</math> to <math>I</math> is <math>r=\frac{\sqrt7}2</math>, and the height of <math>\triangle IFE</math> to <math>I</math> is <math>IM=\frac{\sqrt2}2</math>, so <math>EF = BC\cdot \frac{IM}r = \frac{5\sqrt{14}}{7}</math>. The needed area of <math>\triangle AEF</math> is <math>\frac12\cdot EF\cdot \frac{AD}2 = \frac{15\sqrt7}{14}</math>, as above.
 
<math>\triangle IBC \sim \triangle IFE</math>. The height of <math>\triangle IBC</math> to <math>I</math> is <math>r=\frac{\sqrt7}2</math>, and the height of <math>\triangle IFE</math> to <math>I</math> is <math>IM=\frac{\sqrt2}2</math>, so <math>EF = BC\cdot \frac{IM}r = \frac{5\sqrt{14}}{7}</math>. The needed area of <math>\triangle AEF</math> is <math>\frac12\cdot EF\cdot \frac{AD}2 = \frac{15\sqrt7}{14}</math>, as above.
  
==Solution 10 Bash for life==  
+
==Solution 11 Bash for life==  
 
Firstly, it is easy to find <math>BD=2,CD=3</math> with angle bisector theorem.
 
Firstly, it is easy to find <math>BD=2,CD=3</math> with angle bisector theorem.
  
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~bluesoul
 
~bluesoul
  
==Solution 11 (Simple geometry)==
+
==Solution 12 (Simple geometry)==
 +
[[File:2020 AIME I 13a.png|350px|right]]
 +
Using the <i><b>Claim</i></b>  (below) we get <math>I</math> is orthocenter of <math>\triangle DEF,\angle EDG = \beta,</math>
 +
<math>\angle FDG = \gamma.</math> So area of <math>\triangle DEF</math> is
 +
<cmath>[\triangle DEF] =\frac {DG \cdot FE}{2} = \frac {AD^2}{8} (\tan \beta + \tan \gamma).</cmath>
 +
 
 +
Semiperimeter of <math>\triangle ABC \hspace{10mm} s = \frac{15}{2},</math> so the bisector
 +
<cmath>AD= \frac{2}{AB + AC}\sqrt{s(s-BC) \cdot AB\cdot AC} = 3\sqrt{2}.</cmath>
 +
 
 +
We get the inradius by applying Heron's formula
 +
<cmath>r = \sqrt{\frac {(s-AB)(s-BC)(s-AC)}{s}} = \sqrt {\frac {3\cdot 5 \cdot 7}{15 \cdot 4}}=\frac {\sqrt {7}}{2}.</cmath>
 +
 
 +
We use formulas for inradius and get
 +
<cmath> \tan \beta = \frac{r}{s – AC} = \frac {\sqrt {7}}{3} , \hspace{10mm} \tan \gamma  = \frac{r}{s – AB} = \frac {\sqrt {7}}{7}.</cmath>
 +
The area <math>\hspace{30mm} [\triangle DEF] = \frac{15 \sqrt 7}{14}.</math>
 
[[File:2020 AIME I 13.png|500px|right]]
 
[[File:2020 AIME I 13.png|500px|right]]
<i><b>Lemma</i></b>  
+
<i><b>Claim</i></b>  
  
Let <math>I</math> be incenter of <math>\triangle ABC.</math> Then bisector <math>BI,</math> perpendicular to bisector <math>CI</math> through point <math>D</math> and the perpendicular bisector of <math>AD</math> are concurrent.
+
Let <math>I</math> be incenter of <math>\triangle ABC.</math> Then bisector <math>\overline{BI},</math> perpendicular bisector of <math>\overline{AD},</math> and perpendicular dropped to bisector <math>\overline{CI}</math> from point <math>D</math> are concurrent.
  
 
<i><b>Proof</i></b>
 
<i><b>Proof</i></b>
  
Denote <math>\angle BAC = 2 \alpha,  \angle ABC = 2\beta, \angle ACB = 2 \gamma, \alpha + \beta + \gamma = 90^o.</math>
+
Denote <math>\angle BAC = 2 \alpha,  \angle ABC = 2\beta,\angle ACB = 2 \gamma.</math>
Denote <math>P</math>  the intersection point of <math>BC</math> and the tangent line to the circumcircle at point <math>A.</math> WLOC,<math>\gamma > \beta</math> (case  <math>\gamma = \beta</math> is trivial).
+
Then <math>\alpha + \beta + \gamma = 90^\circ.</math>
 +
Denote <math>P</math>  the intersection point of <math>BC</math> and the tangent line to the circumcircle at point <math>A.</math>
 +
 
 +
WLOC,<math>\hspace{20mm}\gamma > \beta</math> (case  <math>\gamma = \beta</math> is trivial).
  
 
<math>\angle PAC = \angle ABC = 2 \beta</math> (this angles are measured by half the arc <math>\overset{\Large\frown} {AC}</math>  of the circumcircle).
 
<math>\angle PAC = \angle ABC = 2 \beta</math> (this angles are measured by half the arc <math>\overset{\Large\frown} {AC}</math>  of the circumcircle).
  
 
<cmath>\angle APC = \angle ACD - \angle PAC = 2(\gamma - \beta),</cmath>  
 
<cmath>\angle APC = \angle ACD - \angle PAC = 2(\gamma - \beta),</cmath>  
<cmath>\angle ADP = 180^o – \angle DAC - \angle ACD = 180^o – \alpha – 2 \gamma = \alpha + 2 \beta = \angle DAP \implies AP = DP,</cmath> Therefore bisector of angle P coincite with the perpendicular bisector of <math>\overline{AD}</math>.
+
<cmath>\angle ADP = 180^o – \angle DAC - \angle ACD = 180^o – \alpha – 2 \gamma = \alpha + 2 \beta = \angle DAP \implies AP = DP.</cmath> Therefore bisector of angle P coincite with the perpendicular bisector of <math>\overline{AD}</math>.
 +
 
 +
By applying the Law of Sines to <math>\triangle ABP</math> we get <math>\hspace{20mm}\frac {BP}{AP} =  \frac {\sin 2 \gamma}{\sin 2 \beta.}</math>
 +
 
 +
Let <math>E</math> be crosspoint of <math>PG</math>  and bisector <math>BI.</math> By applying the Law of Sines to <math>\triangle BEP</math> we get <cmath>\frac {EP}{BP} = \frac {\sin \beta}{\sin(180^o – \beta – (\gamma – \beta))}= \frac {\sin \beta}{\sin \gamma}.</cmath>
 +
 
 +
Let <math>E'</math> be crosspoint of <math>PG</math> and the perpendicular dropped to bisector <math>\overline{CI}</math> from point <math>D.</math>
 +
<cmath>\frac {E'P}{DP} = \frac {\sin (90^\circ – \gamma)}{\sin(180^o - (90^o – \gamma) – (\gamma - \beta))}= \frac {\cos\gamma}{\cos \beta}.</cmath>
 +
<math>\hspace{50mm}\frac {E'P} {EP} = \frac {E'P}{DP} \cdot \frac {AP}{BP} \cdot \frac {BP}{EP} = \frac {\cos\gamma}{\cos \beta} \cdot \frac {\sin 2\beta}{\sin 2\gamma}\cdot \frac {\sin\gamma}{\sin \beta} = 1 \implies  E</math>  coincide with <math>E'.</math>
 +
 
 +
'''vladimir.shelomovskii@gmail.com, vvsss'''
 +
 
 +
==Solution 13 (Trig)==
 +
Denote <math>I</math> as the incenter of <math>\triangle ABC</math> and <math>M</math> as the intersecting point of <math>AD</math> and <math>EF</math>.
 +
 
 +
We solve this question based on this equation: <math>[\triangle AEF]=\dfrac{1}{2}\times AM\times (\dfrac{IM}{\tan{\angle EFI}}+\dfrac{IM}{\tan{\angle FEI}})</math>.
 +
 
 +
 
 +
Firstly, let's compute the trig value of <math>\angle EFI</math> and <math>\angle FEI</math>. Using the Cosine Law and half-angle formula, we obtain:
 +
<cmath>\cos{\angle B}=\dfrac{1}{8}, \sin{\dfrac{\angle B}{2}}=\dfrac{\sqrt{7}}{4}, \cos{\dfrac{\angle B}{2}}=\dfrac{3}{4}</cmath>
 +
<cmath>\cos{\angle C}=\dfrac{3}{4}, \sin{\dfrac{\angle C}{2}}=\dfrac{1}{\sqrt{8}}, \cos{\dfrac{\angle C}{2}}=\dfrac{\sqrt{7}}{\sqrt{8}}</cmath>
  
By applying the Law of Sines to <math>\triangle ABP</math> we get <math>\frac {BP}{AP} = \frac {\sin 2 \gamma}{\sin 2 \beta.}</math>
+
Noticing that by angle chasing, <math>\angle EFI= 90^{\circ}-\angle FIA=90^{\circ}-\angle CID=90^{\circ}-(180^{\circ}-\dfrac{\angle A}{2}-\angle B-\dfrac{\angle C}{2})= \dfrac{\angle B}{2}</math>.
 +
Similarly, <math>\angle FEI=\dfrac{\angle C}{2}</math>.
 +
Thus, <math>\tan{\angle EFI}=\dfrac{\sqrt{7}}{3}</math> and <math>\tan{\angle FEI}=\dfrac{1}{\sqrt{7}}</math>
  
Let <math>E</math> be crosspoint of <math>PG</math>  and bisector <math>BI.</math> By applying the Law of Sines to <math>\triangle APB</math> we get <cmath>\frac {EP}{BP} = \frac {\sin \beta}{\sin(180^o – \beta – (\gamma – \beta))}= \frac {\sin \beta}{\sin \gamma}.</cmath>
+
Then we find <math>AM</math>: by Stewart's theorem, <math>AM=\dfrac{AD}{2}=\dfrac{3\sqrt{2}}{2}</math>.
  
Let <math>E'</math> be crosspoint of <math>PG</math> and the perpendicular for  bisector <math>CI</math> at point <math>D.</math>.
+
For <math>MI</math>, first noticing that <math>MI=MD-ID=\dfrac{3\sqrt{2}}{2}-ID</math>.
<cmath>\frac {E'P}{DP} = \frac {\sin (90^o – \gamma)}{\sin(180^o - (90^o – \gamma) – (\gamma - \beta))}= \frac {\cos\gamma}{\cos \beta}.</cmath>
+
Then, by Angel Bisector's theorem, we have <math>BD=2, CD=3</math>, and applying Sine Law:
<math>\frac {E'P} {EP} = \frac {E'P}{DP} \cdot \frac {AP}{BP} \cdot \frac {AP}{EP} = \frac {\cos\gamma}{\cos \beta} \cdot \frac {\sin 2\gamma}{\sin 2\beta}\cdot \frac {\sin\gamma}{\sin \beta} = 1 \implies  E</math> coincide with <math>E'.</math>
+
<cmath>\dfrac{ID}{\sin{\dfrac{\angle C}{2}}}=\dfrac{CD}{\sin{\angle CID}}</cmath>
 +
<cmath>\dfrac{ID}{\sin{\dfrac{\angle C}{2}}}=\dfrac{CD}{\cos{\dfrac{\angle B}{2}}}</cmath>
 +
Substituting the known values, we get <math>ID=\sqrt{2}</math>. Thus, <math>MI=\dfrac{\sqrt{2}}{2}</math>
  
<i><b>Solution</i></b>
+
Now, we already find all the unknown values in the very beginning equation. Substituting all the values, we obtain <math>[\triangle AEF]=\dfrac{15\sqrt{7}}{14}</math>, which gives a final answer of <math>\boxed{036}</math>.
  
We use Lemma and get <math>I</math> is orthocenter of <math>\triangle DEF, \angle EDG = \beta, \angle FDG = \gamma.</math> So area of <math>\triangle DEF</math> is <cmath>\frac {DG \cdot FE}{2} = \frac {AD^2}{8} (\tan \beta + \tan \gamma).</cmath>
+
~Ericcc
  
By applying Stewart's Theorem, we can get that <math>AD^2 = \frac {9}{2}.</math>
+
==Video Solution==
 +
https://youtu.be/NpB7zpy-yKM
 +
 
 +
~MathProblemSolvingSkills.com
  
By applying Heron's formula, we can get the inradius <math>r = \sqrt{\frac {(p-a)(p-b)(p-c)}{p}} = \frac {\sqrt {7}}{2}.</math>
 
  
 
==See Also==
 
==See Also==
  
 
{{AIME box|year=2020|n=I|num-b=12|num-a=14}}
 
{{AIME box|year=2020|n=I|num-b=12|num-a=14}}
 +
 +
[[Category:Intermediate Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 05:56, 3 August 2024

Problem

Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be written as $\tfrac{m\sqrt{n}}p,$ where $m$ and $p$ are relatively prime positive integers, and $n$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$

Diagram

[asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(18cm);  real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */  pen dotstyle = black; /* point style */  real xmin = -10.645016481888238, xmax = 5.4445786933235505, ymin = 0.7766255516825293, ymax = 9.897545413994122;  /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.8235294117647059,0.0396078431372549,0.07529411764705882);   draw((-6.837129089839387,8.163360372429347)--(-6.8268938290378,5.895596632024835)--(-4.33118398380513,6.851781504978754)--cycle, linewidth(2) + rvwvcq);  draw((-6.837129089839387,8.163360372429347)--(-8.31920210577661,4.188003838050227)--(-3.319253031309944,4.210570466954303)--cycle, linewidth(2) + rvwvcq);   /* draw figures */ draw((-6.837129089839387,8.163360372429347)--(-7.3192122908832715,4.192517163831042), linewidth(2) + wrwrwr);  draw((-7.3192122908832715,4.192517163831042)--(-2.319263216416622,4.2150837927351175), linewidth(2) + wrwrwr);  draw((-2.319263216416622,4.2150837927351175)--(-6.837129089839387,8.163360372429347), linewidth(2) + wrwrwr);  draw((xmin, -2.6100704119306224*xmin-9.68202796751058)--(xmax, -2.6100704119306224*xmax-9.68202796751058), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.3831314264278095*xmin + 8.511194202815297)--(xmax, 0.3831314264278095*xmax + 8.511194202815297), linewidth(2) + wrwrwr); /* line */ draw(circle((-6.8268938290378,5.895596632024835), 2.267786838055365), linewidth(2) + wrwrwr);  draw(circle((-4.33118398380513,6.851781504978754), 2.828427124746193), linewidth(2) + wrwrwr);  draw((xmin, 0.004513371749987873*xmin + 4.225551489816879)--(xmax, 0.004513371749987873*xmax + 4.225551489816879), linewidth(2) + wrwrwr); /* line */ draw((-7.3192122908832715,4.192517163831042)--(-4.33118398380513,6.851781504978754), linewidth(2) + wrwrwr);  draw((-6.8268938290378,5.895596632024835)--(-2.319263216416622,4.2150837927351175), linewidth(2) + wrwrwr);  draw((-6.837129089839387,8.163360372429347)--(-8.31920210577661,4.188003838050227), linewidth(2) + wrwrwr);  draw((xmin, 0.004513371749987873*xmin + 8.19421887771445)--(xmax, 0.004513371749987873*xmax + 8.19421887771445), linewidth(2) + wrwrwr); /* line */ draw((-3.837159645159393,8.176900349771794)--(-8.31920210577661,4.188003838050227), linewidth(2) + wrwrwr);  draw((-3.837159645159393,8.176900349771794)--(-5.3192326610966125,4.2015438153926725), linewidth(2) + wrwrwr);  draw((-6.837129089839387,8.163360372429347)--(-6.8268938290378,5.895596632024835), linewidth(2) + rvwvcq);  draw((-6.8268938290378,5.895596632024835)--(-4.33118398380513,6.851781504978754), linewidth(2) + rvwvcq);  draw((-4.33118398380513,6.851781504978754)--(-6.837129089839387,8.163360372429347), linewidth(2) + rvwvcq);  draw((-6.837129089839387,8.163360372429347)--(-8.31920210577661,4.188003838050227), linewidth(2) + rvwvcq);  draw((-8.31920210577661,4.188003838050227)--(-3.319253031309944,4.210570466954303), linewidth(2) + rvwvcq);  draw((-3.319253031309944,4.210570466954303)--(-6.837129089839387,8.163360372429347), linewidth(2) + rvwvcq);   /* dots and labels */ dot((-6.837129089839387,8.163360372429347),dotstyle);  label("$A$", (-6.8002301023571095,8.267690318323321), NE * labelscalefactor);  dot((-7.3192122908832715,4.192517163831042),dotstyle);  label("$B$", (-7.2808283997985,4.29753046989445), NE * labelscalefactor);  dot((-2.319263216416622,4.2150837927351175),linewidth(4pt) + dotstyle);  label("$C$", (-2.276337432963145,4.29753046989445), NE * labelscalefactor);  dot((-5.3192326610966125,4.2015438153926725),linewidth(4pt) + dotstyle);  label("$D$", (-5.274852897434433,4.287082680819637), NE * labelscalefactor);  dot((-6.8268938290378,5.895596632024835),linewidth(4pt) + dotstyle);  label("$F$", (-6.789782313282296,5.979624510939313), NE * labelscalefactor);  dot((-4.33118398380513,6.851781504978754),linewidth(4pt) + dotstyle);  label("$E$", (-4.292760724402025,6.93037331674728), NE * labelscalefactor);  dot((-8.31920210577661,4.188003838050227),linewidth(4pt) + dotstyle);  label("$G$", (-8.273368361905721,4.276634891744824), NE * labelscalefactor);  dot((-3.319253031309944,4.210570466954303),linewidth(4pt) + dotstyle);  label("$H$", (-3.2793251841451787,4.29753046989445), NE * labelscalefactor);  dot((-3.837159645159393,8.176900349771794),linewidth(4pt) + dotstyle);  label("$I$", (-3.7912668488110084,8.257242529248508), NE * labelscalefactor);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]

Solution 1

Points are defined as shown. It is pretty easy to show that $\triangle AFE \sim \triangle AGH$ by spiral similarity at $A$ by some short angle chasing. Now, note that $AD$ is the altitude of $\triangle AFE$, as the altitude of $AGH$. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that $AD/2 = \frac{\sqrt{18}}{2}$, the altitude of $\triangle AFE$. Similarly, the altitude of $\triangle AGH$ is the altitude of $\triangle ABC$, or $\frac{3\sqrt{7}}{2}$. However, it's not too hard to see that $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\triangle ABC$ is $\frac{15\sqrt{7}}{14} \implies \boxed{036}$, by similarity. ~awang11

Solution 2

Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that \[AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.\] In particular, we have $AI = IM_A$. [asy] defaultpen(fontsize(10pt)); size(200); pair A, B, C, D, E, F, I, P, MA, MB, MC; B = (0,0); C = (5,0); A = IP(Circle(B, 4), Circle(C, 6), 0); I = incenter(A, B, C); D = extension(A, I, B, C); P = midpoint(A--D); E = extension(P, rotate(90, P)*A, B, I); F = extension(P, rotate(90, P)*A, C, I); MA = IP(Line(A, I, 20), circumcircle(A, B, C), 1); MB = IP(Line(B, I, 20), circumcircle(A, B, C), 1); MC = IP(Line(C, I, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(A--E--F--cycle, lightblue); draw(E--D--F, lightblue); draw(A--MA^^B--MB^^C--MC, heavygreen); draw(MA--MB--MC--cycle, magenta); dot("$A$", A, dir(120)); dot("$B$", B, dir(220)); dot("$C$", C, dir(320)); dot("$D$", D, dir(230)); dot("$E$", E, dir(330)); dot("$F$", F, dir(250)); dot("$I$", I, dir(80)); dot("$M_A$", MA, dir(270)); dot("$M_B$", MB, dir(60)); dot("$M_C$", MC, dir(150)); [/asy] Now the key claim is that:

Claim: $\triangle DEF$ and $\triangle M_AM_BM_C$ are homothetic at $I$ with ratio $2$.

Proof. First, we show that $D$ is the midpoint of $M_AI$. Indeed, we have \[\frac{ID}{DM_A} = \frac{BI}{BM_A}\cdot \frac{\sin\angle IBC}{\sin \angle CBM_A} = \frac{BI}{AI}\cdot\frac{\sin \angle B/2}{\sin \angle A/2} = 1\] by Ratio lemma and Law of Sines.

Now observe that:

  • $\overline{M_BM_C}$ is the perpendicular bisector of $\overline{AI}$,
  • $\overline{EF}$ is the perpendicular bisector of $\overline{AD}$, and
  • $ID = AI/2$.

Combining these facts gives that $\overline{EF}$ is a midline in $\triangle IM_BM_C$, which proves the claim. $\blacksquare$

To finish, we compute $[M_AM_BM_C]$, noting that $[AEF] = [DEF] = \tfrac{1}{4}[M_AM_BM_C]$.

By Heron's, we can calculate the circumradius $R = 8/\sqrt{7}$, and by Law of Cosines, we get \begin{align*}\cos A &= \frac{9}{16}\implies \cos A/2 = \frac{5}{\sqrt{32}} \\ \cos B &= \frac{1}{8} \implies \cos B/2 = \frac{3}{4} \\ \cos C &= \frac{3}{4} \implies \cos C/2 = \sqrt{\frac{7}{8}}.\end{align*} Then using $[XYZ] = 2R^2\sin X\sin Y\sin Z$, we can compute \[[M_AM_BM_C] = 2\cdot \frac{64}{7}\cdot \frac{5}{\sqrt{32}}\cdot \frac{3}{4}\cdot \frac{\sqrt{7}}{\sqrt{8}} = \frac{30\sqrt{7}}{7}.\] Thus $[AEF] = 15\sqrt{7}/14$, which gives a final answer of $\boxed{036}$.

~pinetree1

Solution 3(coord bash + basic geometry)

Let $\overline{BC}$ lie on the x-axis and $B$ be the origin. $C$ is $(5,0)$. Use Heron's formula to compute the area of triangle $ABC$. We have $s=\frac{15}{2}$. and $[ABC]=\sqrt{\frac{15 \cdot 7 \cdot 5 \cdot 3}{2^4}}=\frac{15\sqrt{7}}{4}$. We now find the altitude, which is $\frac{\frac{15\sqrt{7}}{2}}{5}=\frac{3\sqrt{7}}{2}$, which is the y-coordinate of $A$. We now find the x-coordinate of $A$, which satisfies $x^2 + (\frac{3\sqrt{7}}{2})^{2}=16$, which gives $x=\frac{1}{2}$ since the triangle is acute. Now using the Angle Bisector Theorem, we have $\frac{4}{6}=\frac{BD}{CD}$ and $BD+CD=5$ to get $BD=2$. The coordinates of D are $(2,0)$. Since we want the area of triangle $AEF$, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is $(\frac{5}{4}, \frac{3\sqrt{7}}{4})$ and the slope of AD is $-\sqrt{7}$. The slope of the perpendicular bisector is $\frac{1}{\sqrt{7}}$. The equation is(in point slope form) $y-\frac{3\sqrt{7}}{4}=\frac{1}{\sqrt{7}}(x-\frac{5}{4})$. The slope of AB, or in trig words, the tangent of $\angle ABC$ is $3\sqrt{7}$. Finding $\sin{\angle ABC}=\frac{\frac{3\sqrt{7}}{2}}{4}=\frac{3\sqrt{7}}{8}$ and $\cos{\angle ABC}=\frac{\frac{1}{2}}{4}=\frac{1}{8}$. Plugging this in to half angle tangent, it gives $\frac{\frac{3\sqrt{7}}{8}}{1+\frac{1}{8}}=\frac{\sqrt{7}}{3}$ as the slope of the angle bisector, since it passes through $B$, the equation is $y=\frac{\sqrt{7}}{3}x$. Similarly, the equation for the angle bisector of $C$ will be $y=-\frac{1}{\sqrt{7}}(x-5)$. For $E$ use the B-angle bisector and the perpendicular bisector of AD equations to intersect at $(3,\sqrt{7})$. For $F$ use the C-angle bisector and the perpendicular bisector of AD equations to intersect at $(\frac{1}{2}, \frac{9}{2\sqrt{7}})$. The area of AEF is equal to $\frac{EF \cdot \frac{AD}{2}}{2}$ since AD is the altitude of that triangle with EF as the base, with $\frac{AD}{2}$ being the height. $EF=\frac{5\sqrt{2}}{\sqrt{7}}$ and $AD=3\sqrt{2}$, so $[AEF]=\frac{15}{2\sqrt{7}}=\frac{15\sqrt{7}}{14}$ which gives $\boxed{036}$. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo

Solution 4 (Coordinate Bash + Trig)

[asy]         size(8cm); defaultpen(fontsize(10pt));          pair A,B,C,I,D,M,T,Y,Z,EE,F;         A=(0,3sqrt(7));         B=(-1,0);         C=(9,0);         I=incenter(A,B,C);         D=extension(A,I,B,C);         M=(A+D)/2;          draw(B--EE,gray+dashed);         draw(C--F,gray+dashed);         draw(A--B--C--A);         draw(A--D);         draw(B--(5,sqrt(28)));         draw(M--(5,sqrt(28)));         draw(C--(0,9sqrt(7)/7));         draw(M--(0,9sqrt(7)/7));         dot("$A$",A,NW);         dot("$B$",B,SW);         dot("$C$",C,SE);         dot("$D$",D,S);         dot("$E$",(5,sqrt(28)),N);          dot("$M$",M,dir(70));         dot("$F$",(0,9sqrt(7)/7),N);          label("$2$",B--D,S);         label("$3$",D--C,S);         label("$6$",A--C,N);         label("$4$",A--B,W);     [/asy]

Let $B=(0,0)$ and $BC$ be the line $y=0$. We compute that $\cos{\angle{ABC}}=\frac{1}{8}$, so $\tan{\angle{ABC}}=3\sqrt{7}$. Thus, $A$ lies on the line $y=3x\sqrt{7}$. The length of $AB$ at a point $x$ is $8x$, so $x=\frac{1}{2}$.

We now have the coordinates $A=\left(\frac{1}{2},\frac{3\sqrt{7}}{2}\right)$, $B=(0,0)$ and $C=(5,0)$. We also have $D=(2,0)$ by the angle-bisector theorem and $M=\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$ by taking the midpoint. We have that because $\cos{\angle{ABC}}=\frac{1}{8}$, $\cos{\frac{\angle{ABC}}{2}}=\frac{3}{4}$ by half angle formula.

We also compute $\cos{\angle{ACB}}=\frac{3}{4}$, so $\cos{\frac{\angle{ACB}}{2}}=\frac{\sqrt{14}}{4}$.

Now, $AD$ has slope $-\frac{\frac{3\sqrt{7}}{2}}{2-\frac{1}{2}}=-\sqrt{7}$, so it's perpendicular bisector has slope $\frac{\sqrt{7}}{7}$ and goes through $\left(\frac{5}{4},\frac{3\sqrt{7}}{4}\right)$.

We find that this line has equation $y=\frac{\sqrt{7}}{7}x+\frac{4\sqrt{7}}{7}$.

As $\cos{\angle{CBI}}=\frac{3}{4}$, we have that line $BI$ has form $y=\frac{\sqrt{7}}{3}x$. Solving for the intersection point of these two lines, we get $x=3$ and thus $E=\left(3, \sqrt{7}\right)$

We also have that because $\cos{\angle{ICB}}=\frac{\sqrt{14}}{4}$, $CI$ has form $y=-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$.

Intersecting the line $CI$ and the perpendicular bisector of $AD$ yields $-\frac{x\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}=\frac{x\sqrt{7}}{7}+\frac{4\sqrt{7}}{7}$.

Solving this, we get $x=\frac{1}{2}$ and so $F=\left(\frac{1}{2},\frac{9\sqrt{7}}{14}\right)$.

We now compute $EF=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{5\sqrt{7}}{14}\right)^2}=\frac{5\sqrt{14}}{7}$. We also have $MA=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{3\sqrt{7}}{4}\right)^2}=\frac{3\sqrt{2}}{2}$.

As ${MA}\perp{EF}$, we have $[\triangle{AEF}]=\frac{1}{2}\left(\frac{3\sqrt{2}}{2}\times\frac{5\sqrt{14}}{7}\right)=\frac{15\sqrt{7}}{14}$.


The desired answer is $15+7+14=\boxed{036}$ ~Imayormaynotknowcalculus

Solution 5 (Barycentric Coordinates)

[asy]         size(8cm); defaultpen(fontsize(10pt));          pair A,B,C,I,D,M,T,Y,Z,EE,F;         A=(0,3sqrt(7));         B=(-1,0);         C=(9,0);         I=incenter(A,B,C);         D=extension(A,I,B,C);         M=(A+D)/2;          draw(B--EE,gray+dashed);         draw(C--F,gray+dashed);         draw(A--B--C--A);         draw(A--D);         draw(B--(5,sqrt(28)));         draw(M--(5,sqrt(28)));         draw(C--(0,9sqrt(7)/7));         draw(M--(0,9sqrt(7)/7));         dot("$A$",A,NW);         dot("$B$",B,SW);         dot("$C$",C,SE);         dot("$D$",D,S);         dot("$E$",(5,sqrt(28)),N);          dot("$M$",M,dir(70));         dot("$F$",(0,9sqrt(7)/7),N);          label("$2$",B--D,S);         label("$3$",D--C,S);         label("$6$",A--C,N);         label("$4$",A--B,W);     [/asy]

As usual, we will use homogenized barycentric coordinates.

We have that $AD$ will have form $3z=2y$. Similarly, $CF$ has form $5y=6x$ and $BE$ has form $5z=4x$. Since $A=(1,0,0)$ and $D=\left(0,\frac{3}{5},\frac{2}{5}\right)$, we also have $M=\left(\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$. It remains to determine the equation of the line formed by the perpendicular bisector of $AD$.

This can be found using EFFT. Let a point $T$ on $EF$ have coordinates $(x, y, z)$. We then have that the displacement vector $\overrightarrow{AD}=\left(-1, \frac{3}{5}, \frac{2}{5}\right)$ and that the displacement vector $\overrightarrow{TM}$ has form $\left(x-\frac{1}{2},y-\frac{3}{10},z-\frac{1}{5}\right)$. Now, by EFFT, we have $5^2\left(\frac{3}{5}\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(y-\frac{3}{10}\right)\right)+6^2\left(-1\times\left(z-\frac{1}{5}\right)+\frac{2}{5}\times\left(x-\frac{1}{2}\right)\right)+4^2\left(-1\times\left(y-\frac{3}{10}\right)+\frac{3}{5}\times\left(x-\frac{1}{2}\right)\right)=0$. This equates to $8x-2y-7z=2$.

Now, intersecting this with $BE$, we have $5z=4x$, $8x-2y-7z=2$, and $x+y+z=1$. This yields $x=\frac{2}{3}$, $y=-\frac{1}{5}$, and $z=\frac{8}{15}$, or $E=\left(\frac{2}{3},-\frac{1}{5},\frac{8}{15}\right)$.

Similarly, intersecting this with $CF$, we have $5y=6x$, $8x-2y-7z=2$, and $x+y+z=1$. Solving this, we obtain $x=\frac{3}{7}$, $y=\frac{18}{35}$, and $z=\frac{2}{35}$, or $F=\left(\frac{3}{7},\frac{18}{35},\frac{2}{35}\right)$.

We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being $\overrightarrow{FE}=\left(\frac{5}{21},-\frac{5}{7},\frac{10}{21}\right)$. We then have $FE^2=-25\left(-\frac{5}{7}\cdot\frac{10}{21}\right)-36\left(\frac{5}{21}\cdot\frac{10}{21}\right)-16\left(\frac{5}{21}\cdot-\frac{5}{7}\right)=\frac{50}{7}$, thus $FE=\frac{5\sqrt{14}}{7}$.

Our second displacement vector is $\overrightarrow{AM}=\left(-\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$. As a result, $AM^2=-25\left(\frac{3}{10}\cdot\frac{1}{5}\right)-36\left(-\frac{1}{2}\cdot\frac{1}{5}\right)-16\left(-\frac{1}{2}\cdot\frac{3}{10}\right)=\frac{9}{2}$, so $AM=\frac{3\sqrt{2}}{2}$.

As ${AM}\perp{EF}$, the desired area is $\frac{\frac{5\sqrt{14}}{7}\times\frac{3\sqrt{2}}{2}}{2}={\frac{15\sqrt{7}}{14}}\implies{m+n+p=\boxed{036}}$. ~Imayormaynotknowcalculus


Remark: The area of $\triangle{AEF}$ can also be computed using the Barycentric Area Formula, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.

Solution 6 (geometry+trig)

[asy]         size(8cm); defaultpen(fontsize(10pt));          pair A,B,C,I,D,M,T,Y,Z,EE,F;         A=(0,3sqrt(7));         B=(-1,0);         C=(9,0);         I=incenter(A,B,C);         D=extension(A,I,B,C);         M=(A+D)/2;          draw(B--EE,gray+dashed);         draw(C--F,gray+dashed);         draw(A--B--C--A);         draw(A--D);         draw(A--(5,sqrt(28)));         draw(A--(0,9sqrt(7)/7));         draw(D--(0,9sqrt(7)/7));         draw(D--(5,sqrt(28)));         draw(B--(5,sqrt(28)));         draw(M--(5,sqrt(28)));         draw(C--(0,9sqrt(7)/7));         draw(M--(0,9sqrt(7)/7));         dot("$A$",A,NW);         dot("$B$",B,SW);         dot("$C$",C,SE);         dot("$D$",D,S);         dot("$E$",(5,sqrt(28)),N);          dot("$M$",M,dir(70));         dot("$F$",(0,9sqrt(7)/7),N);          label("$2$",B--D,S);         label("$3$",D--C,S);         label("$6$",A--C,N);         label("$4$",A--B,W);     [/asy]

To get the area of $\triangle AEF$, we try to find $AM$ and $\angle EAF$.

Since $AD$ is the angle bisector, we can get that $BD=2$ and $CD=3$. By applying Stewart's Theorem, we can get that $AD=3\sqrt{2}$. Therefore $AM=\frac{3\sqrt{2}}{2}$.

Since $EF$ is the perpendicular bisector of $AD$, we know that $AE = DE$. Since $BE$ is the angle bisector of $\angle BAC$, we know that $\angle ABE = \angle DBE$. By applying the Law of Sines to $\triangle ABE$ and $\triangle DBE$, we know that $\sin \angle BAE = \sin \angle BDE$. Since $BD$ is not equal to $AB$ and therefore these two triangles are not congruent, we know that $\angle BAE$ and $\angle BDE$ are supplementary. Then we know that $\angle ABD$ and $\angle AED$ are also supplementary. Given that $AE=DE$, we can get that $\angle DAE$ is half of $\angle ABC$. Similarly, we have $\angle DAF$ is half of $\angle ACB$.

By applying the Law of Cosines, we get $\cos \angle ABC = \frac{1}{8}$, and then $\sin \angle ABC = \frac{3\sqrt{7}}{8}$. Similarly, we can get $\cos \angle ACB = \frac{3}{4}$ and $\sin \angle ACB = \frac{\sqrt{7}}{4}$. Based on some trig identities, we can compute that $\tan \angle DAE = \frac{\sin \angle ABC}{1 + \cos \angle ABC} = \frac{\sqrt{7}}{3}$, and $\tan \angle DAF = \frac{\sqrt{7}}{7}$.

Finally, the area of $\triangle AEF$ equals $\frac{1}{2}AM^2(\tan \angle DAE + \tan \angle DAF)=\frac{15\sqrt{7}}{14}$. Therefore, the final answer is $15+7+14=\boxed{036}$. ~xamydad

Remark: I didn't figure out how to add segments $AF$, $AE$, $DF$ and $DE$. Can someone please help add these segments?

(Added :) ~Math_Genius_164)

Solution 7

Question13.png

First and foremost $\big[\triangle{AEF}\big]=\big[\triangle{DEF}\big]$ as $EF$ is the perpendicular bisector of $AD$. Now note that quadrilateral $ABDF$ is cyclic, because $\angle{ABF}=\angle{FBD}$ and $FA=FD$. Similarly quadrilateral $AEDC$ is cyclic, \[\implies \angle{EDA}=\dfrac{C}{2}, \quad \angle{FDA}=\dfrac{B}{2}\] Let $A'$,$B'$, $C'$ be the $A$,$B$, and $C$ excenters of $\triangle{ABC}$ respectively. Then it follows that $\triangle{DEF} \sim \triangle{A'C'B'}$. By angle bisector theorem we have $BD=2 \implies \dfrac{ID}{IA}=\dfrac{BD}{BA}=\dfrac{1}{2}$. Now let the feet of the perpendiculars from $I$ and $A'$ to $BC$ be $X$ and $Y$ resptively. Then by tangents we have \[BX=s-AC=\dfrac{3}{2} \implies XD=2-\dfrac{3}{2}=\dfrac{1}{2}\] \[CY=s-AC \implies YD=3-\dfrac{3}{2}=\dfrac{3}{2} \implies \dfrac{ID}{DA'}=\dfrac{XD}{YD}=\dfrac{1}{3} \implies \big[\triangle{DEF}\big]=\dfrac{1}{16}\big[\triangle{A'C'B'}\big]\] From the previous ratios, $AI:ID:DA'=2:1:3 \implies AD=DA' \implies \big[\triangle{ABC}\big]=\big[\triangle{A'BC}\big]$ Similarly we can find that $\big[\triangle{B'AC}\big]=2\big[\triangle{ABC}\big]$ and $\big[\triangle{C'AB}\big]=\dfrac{4}{7}\big[\triangle{ABC}\big]$ and thus \[\big[\triangle{A'B'C'}\big]=\bigg(1+1+2+\dfrac{4}{7}\bigg)\big[\triangle{ABC}\big]=\dfrac{32}{7}\big[\triangle{ABC}\big] \implies \big[\triangle{DEF}\big]=\dfrac{2}{7}\big[\triangle{ABC}\big]=\dfrac{15\sqrt{7}}{14} \implies m+n+p = \boxed{036}\] -tkhalid

Solution 8 -Trigonometry(only)

Aime 13.png

Trig values we use here:

$\cos A = \frac{9}{16}$

$\cos \frac{A}{2} = \frac{5}{4\sqrt2}$

$\sin \frac{A}{2} = \frac{\sqrt7}{4\sqrt2}$

$\cos \frac{B}{2} = \frac{3}{4}$

$\cos \frac{C}{2} = \frac{\sqrt7}{2\sqrt2}$


First let the incenter be $I$. Let $M$ be the midpoint of minor arc $BC$ on $(ABC)$ and let $K$ be the foot of $M$ to $BC$.

We can find $AD$ using Stewart's Theorem: from Angle Bisector Theorem $BD = 2$ and $CD = 3$. Then it is easy to find that $AD = 3\sqrt3$.

Now we trig bash for $DI = MI - MD$. Notice that $MI = MB$ from the Incenter Excenter Lemma. We obtain that $MB = \frac{BK}{\cos \frac{A}{2}} = \frac{\frac{5}{2}}{\frac{5}{4\sqrt2}}=2\sqrt2$. To get $MD$ we angle chase to get $\angle KDM = \frac{A}{2}+C$. Then \[\cos(\frac{A}{2}+C) = \cos\frac{A}{2}\cos C - \sin\frac{A}{2}\sin C = \frac{1}{2\sqrt2} = \frac{\frac{1}{2}}{MD}\] gives $MD = \sqrt{2}$. This means $DI = \sqrt2$.

Now let $AI \cap EF = G$. It is easy to angle chase $\angle GIE = 90- \frac{B}{2}$ and $\angle GIF = 90- \frac{C}{2}$. Since $GI = GD - ID = \frac{3\sqrt2}{2}-\sqrt2=\frac{\sqrt2}{2}$, we compute that \[EF = EG + FG = \frac{\sqrt2}{2}(\cot B/2 + \cot C/2) = \frac{\sqrt2}{2}(\frac{3}{\sqrt7}+\sqrt7) = \frac{5\sqrt{14}}{7}\] which implies \[[AEF] = AG*EF/2 = \frac{3\sqrt2}{2} * \frac{5\sqrt{14}}{7} / 2 = \frac{15\sqrt7}{14}\] which gives an answer of $\boxed{36}$. ~Leonard_my_dude~

Solution 9 (Official MAA 1)

Let $x = \angle BAD = \angle CAD$, $y = \angle CBE = \angle ABE$, and $z = \angle BCF = \angle ACF$. Notice that $x+y+z = 90^\circ$.

In $\triangle ABD$, segment $\overline{BE}$ is the bisector of $\angle ABD$, and $E$ lies on the perpendicular bisector of side $\overline{AD}$. Therefore $E$ is the midpoint of arc $\stackrel{\textstyle\frown}{AD}$ on the circumcircle of $\triangle ABD$. It follows that $\angle BED = \angle BAD = x$ and $\angle EDA = \angle EBA = y$. Likewise, $ACDF$ is cyclic, $\angle CFD = \angle CAD = x$, and $\angle FDA = \angle FCA = z$. Because $\overline{EF}$ is the perpendicular bisector of $\overline{AD}$, triangles $AEF$ and $DEF$ are congruent, implying that \begin{align*} [\triangle AEF] = [\triangle DEF] &= \frac{DE\cdot DF\cdot\sin(\angle EDF)}{2}\\ &= \frac{DE\cdot DF\cdot\sin(y+z)}{2} = \frac{DE\cdot DF\cdot\cos x}{2}. \end{align*} [asy] unitsize(0.8 cm);  pair A, B, C, D, E, F, I; real angleC = aCos(3/4);  C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I);  draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(circumcircle(A,B,D)); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5));  dot("$A$", A, SW); dot("$B$", B, dir(0)); dot("$C$", C, N); dot("$D$", D, dir(0)); dot("$E$", E, N); dot("$F$", F, dir(0)); [/asy]

Applying the Law of Sines to $\triangle BED$ and $\triangle CFD$ gives\[DE = BD\cdot\frac{\sin y}{\sin x}\text{~ and ~} DF = CD\cdot\frac{\sin z}{\sin x}.\]By the Angle Bisector Theorem, $BD = 2$ and $CD = 3$. Combining the above information yields \[[\triangle AEF] = \frac{3\sin y\cdot\sin z\cdot\cos x}{\sin^2 x}.\]Applying the Law of Cosines to $\triangle ABC$ gives $\cos 2x = \frac9{16}$, $\cos 2y = \frac1{8}$, and $\cos 2z = \frac34$. By the Half Angle Formulas,\[\sin^2x = \frac7{32},~~ \cos x = \sqrt{\frac{25}{32}},~~ \sin y = \sqrt{\frac7{16}}, \text{~ and ~} \sin z = \sqrt{\frac18}.\]Therefore \[[\triangle AEF] = \frac{3\cdot\sqrt{\frac{7}{16}}\cdot\sqrt{\frac{1}{8}}\cdot\sqrt{\frac{25}{32}}} {\frac{7}{32}} = \frac{15\sqrt{7}}{14}.\]The requested sum is $15+7+14 = 36$.

Solution 10 (Official MAA 2)

Let the point $M$ be the midpoint of $\overline{AD}$, let $I$ be the incenter of $\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\triangle ABC$. The semiperimeter of $\triangle ABC$ is\[s = \frac{AB + BC + CA}2 = \frac{15}2,\]and Heron's Formula gives the area of $\triangle ABC$ as\[\sqrt{s(s-AB)(s-BC)(s-CA)} = \frac{15\sqrt7}4.\]This area is also $rs$ implying that $r = \frac{\sqrt7}2$. Stewart's Theorem gives $AD =3\sqrt2$. Because the ratio of the areas of $\triangle IBC$ and $\triangle ABC$ is $\frac{ID}{AD},$ it follows that\[ID = AD\cdot\frac{\frac{r\cdot BC}2}{\frac{15\sqrt7}4} = \sqrt2.\]Thus $IM = MD - ID = \frac{\sqrt2}2$.

[asy] unitsize(0.8 cm);  pair A, B, C, D, E, F, I, M; real angleC = aCos(3/4);  C = (0,0); A = 6*dir(270 - angleC/2); B = 5*dir(270 + angleC/2); I = incenter(A,B,C); D = extension(A, I, B, C); E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I); F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I); M = (A + D)/2;  draw(A--B--C--cycle); draw(A--interp(A,D,1.4)); draw(A--F--D--E--cycle); draw(E--F); draw(B--interp(B,E,1.5)); draw(C--interp(C,F,1.5));  dot("$A$", A, SW); dot("$B$", B, dir(0)); dot("$C$", C, N); dot("$D$", D, dir(0)); dot("$E$", E, N); dot("$F$", F, dir(0)); dot("$I$", I, dir(120)); dot("$M$", M, W); [/asy]

Note that $\angle EFI = 90^{\circ} - \angle FIA = 90^{\circ} - \angle CID = 90^{\circ} - \frac{\angle A}{2} - \frac{\angle C}{2}  = \frac{\angle B}{2} = \angle IBC$. Thus $\triangle IBC \sim \triangle IFE$. The height of $\triangle IBC$ to $I$ is $r=\frac{\sqrt7}2$, and the height of $\triangle IFE$ to $I$ is $IM=\frac{\sqrt2}2$, so $EF = BC\cdot \frac{IM}r = \frac{5\sqrt{14}}{7}$. The needed area of $\triangle AEF$ is $\frac12\cdot EF\cdot \frac{AD}2 = \frac{15\sqrt7}{14}$, as above.

Solution 11 Bash for life

Firstly, it is easy to find $BD=2,CD=3$ with angle bisector theorem.

Using LOC and some trig formulas we get all those values: $cos\angle{B}=\frac{1}{8},sin\angle{B}=\frac{3\sqrt{7}}{8},cos\angle{\frac{B}{2}}=\frac{3}{4},cos\angle{\frac{{ACB}}{2}}=\frac{\sqrt{14}}{4}$ Now we find the coordinates of points $A,E,F$ and we apply shoelace theorem later. Point A's coordinates is $(4*\frac{1}{8},4*\frac{3\sqrt{7}}{8})=(\frac{1}{2},\frac{3\sqrt{7}}{2})$, Let $AJ$ is perpendicular to $BC$, $tan\angle{JAD}=\frac{2-\frac{1}{2}}{\frac{3\sqrt{7}}{2}}=\frac{\sqrt{7}}{7}$, which means the slope of $FE$ is $\frac{\sqrt{7}}{7}$. Find the coordinate of $M$, it is easy, $(\frac{5}{4},\frac{3\sqrt{7}}{4})$, the function $EF$ is $y=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}$. Now find the intersection of $EF,EB$, $\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}=\frac{\sqrt{7}x}{3}$, getting that $x=3,E(3,\sqrt{7})$ Now we look at line segment $CF$, since $cos\angle{\frac{ACB}{2}}=\frac{\sqrt{14}}{4},tan\angle{\frac{ACB}{2}}=\frac{\sqrt{7}}{7}$. Since the line passes $(5,0)$, we can set the equation to get the $CF:y=-\frac{\sqrt{7}}{7}+\frac{5\sqrt{7}}{7}$, find the intersection of $CF,EF$,$-\frac{\sqrt{7}x}{7}+\frac{5\sqrt{7}}{7}=\frac{\sqrt{7}x}{7}+\frac{4\sqrt{7}}{7}$, getting that $F:(\frac{1}{2},\frac{9\sqrt{7}}{14})$ and in the end we use shoelace theorem with coordinates of $A,F,E$ getting the area $\frac{15\sqrt{7}}{14}$ leads to the final answer $\boxed{36}$

~bluesoul

Solution 12 (Simple geometry)

2020 AIME I 13a.png

Using the Claim (below) we get $I$ is orthocenter of $\triangle DEF,\angle EDG = \beta,$ $\angle FDG = \gamma.$ So area of $\triangle DEF$ is \[[\triangle DEF] =\frac {DG \cdot FE}{2} = \frac {AD^2}{8} (\tan \beta + \tan \gamma).\]

Semiperimeter of $\triangle ABC \hspace{10mm} s = \frac{15}{2},$ so the bisector \[AD= \frac{2}{AB + AC}\sqrt{s(s-BC) \cdot AB\cdot AC} = 3\sqrt{2}.\]

We get the inradius by applying Heron's formula \[r = \sqrt{\frac {(s-AB)(s-BC)(s-AC)}{s}} = \sqrt {\frac {3\cdot 5 \cdot 7}{15 \cdot 4}}=\frac {\sqrt {7}}{2}.\]

We use formulas for inradius and get \[\tan \beta = \frac{r}{s – AC} = \frac {\sqrt {7}}{3} , \hspace{10mm} \tan \gamma  = \frac{r}{s – AB} = \frac {\sqrt {7}}{7}.\] The area $\hspace{30mm} [\triangle DEF] = \frac{15 \sqrt 7}{14}.$

2020 AIME I 13.png

Claim

Let $I$ be incenter of $\triangle ABC.$ Then bisector $\overline{BI},$ perpendicular bisector of $\overline{AD},$ and perpendicular dropped to bisector $\overline{CI}$ from point $D$ are concurrent.

Proof

Denote $\angle BAC = 2 \alpha,  \angle ABC = 2\beta,\angle ACB = 2 \gamma.$ Then $\alpha + \beta + \gamma = 90^\circ.$ Denote $P$ the intersection point of $BC$ and the tangent line to the circumcircle at point $A.$

WLOC,$\hspace{20mm}\gamma > \beta$ (case $\gamma = \beta$ is trivial).

$\angle PAC = \angle ABC = 2 \beta$ (this angles are measured by half the arc $\overset{\Large\frown} {AC}$ of the circumcircle).

\[\angle APC = \angle ACD - \angle PAC = 2(\gamma - \beta),\] \[\angle ADP = 180^o – \angle DAC - \angle ACD = 180^o – \alpha – 2 \gamma = \alpha + 2 \beta = \angle DAP \implies AP = DP.\] Therefore bisector of angle P coincite with the perpendicular bisector of $\overline{AD}$.

By applying the Law of Sines to $\triangle ABP$ we get $\hspace{20mm}\frac {BP}{AP} =  \frac {\sin 2 \gamma}{\sin 2 \beta.}$

Let $E$ be crosspoint of $PG$ and bisector $BI.$ By applying the Law of Sines to $\triangle BEP$ we get \[\frac {EP}{BP} = \frac {\sin \beta}{\sin(180^o – \beta – (\gamma – \beta))}= \frac {\sin \beta}{\sin \gamma}.\]

Let $E'$ be crosspoint of $PG$ and the perpendicular dropped to bisector $\overline{CI}$ from point $D.$ \[\frac {E'P}{DP} = \frac {\sin (90^\circ – \gamma)}{\sin(180^o - (90^o – \gamma) – (\gamma - \beta))}= \frac {\cos\gamma}{\cos \beta}.\] $\hspace{50mm}\frac {E'P} {EP} = \frac {E'P}{DP} \cdot \frac {AP}{BP} \cdot \frac {BP}{EP} = \frac {\cos\gamma}{\cos \beta} \cdot \frac {\sin 2\beta}{\sin 2\gamma}\cdot \frac {\sin\gamma}{\sin \beta} = 1 \implies  E$ coincide with $E'.$

vladimir.shelomovskii@gmail.com, vvsss

Solution 13 (Trig)

Denote $I$ as the incenter of $\triangle ABC$ and $M$ as the intersecting point of $AD$ and $EF$.

We solve this question based on this equation: $[\triangle AEF]=\dfrac{1}{2}\times AM\times (\dfrac{IM}{\tan{\angle EFI}}+\dfrac{IM}{\tan{\angle FEI}})$.


Firstly, let's compute the trig value of $\angle EFI$ and $\angle FEI$. Using the Cosine Law and half-angle formula, we obtain: \[\cos{\angle B}=\dfrac{1}{8}, \sin{\dfrac{\angle B}{2}}=\dfrac{\sqrt{7}}{4}, \cos{\dfrac{\angle B}{2}}=\dfrac{3}{4}\] \[\cos{\angle C}=\dfrac{3}{4}, \sin{\dfrac{\angle C}{2}}=\dfrac{1}{\sqrt{8}}, \cos{\dfrac{\angle C}{2}}=\dfrac{\sqrt{7}}{\sqrt{8}}\]

Noticing that by angle chasing, $\angle EFI= 90^{\circ}-\angle FIA=90^{\circ}-\angle CID=90^{\circ}-(180^{\circ}-\dfrac{\angle A}{2}-\angle B-\dfrac{\angle C}{2})= \dfrac{\angle B}{2}$. Similarly, $\angle FEI=\dfrac{\angle C}{2}$. Thus, $\tan{\angle EFI}=\dfrac{\sqrt{7}}{3}$ and $\tan{\angle FEI}=\dfrac{1}{\sqrt{7}}$

Then we find $AM$: by Stewart's theorem, $AM=\dfrac{AD}{2}=\dfrac{3\sqrt{2}}{2}$.

For $MI$, first noticing that $MI=MD-ID=\dfrac{3\sqrt{2}}{2}-ID$. Then, by Angel Bisector's theorem, we have $BD=2, CD=3$, and applying Sine Law: \[\dfrac{ID}{\sin{\dfrac{\angle C}{2}}}=\dfrac{CD}{\sin{\angle CID}}\] \[\dfrac{ID}{\sin{\dfrac{\angle C}{2}}}=\dfrac{CD}{\cos{\dfrac{\angle B}{2}}}\] Substituting the known values, we get $ID=\sqrt{2}$. Thus, $MI=\dfrac{\sqrt{2}}{2}$

Now, we already find all the unknown values in the very beginning equation. Substituting all the values, we obtain $[\triangle AEF]=\dfrac{15\sqrt{7}}{14}$, which gives a final answer of $\boxed{036}$.

~Ericcc

Video Solution

https://youtu.be/NpB7zpy-yKM

~MathProblemSolvingSkills.com


See Also

2020 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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