AIME 2020 I · 第 13 题

Solution 1

Points are defined as shown (Actually Point G is a point on line BC such that $\triangle AGD$ has circumcenter F, while Point H is a point on line BC such that $\triangle AHD$ has circumcenter E). Since $\angle{AFE} = \angle{AGD}$, and $\angle{AEF} = \angle{AHD}$, it is a breakthrough to get $\triangle AFE \sim \triangle AGH$. We'd like to compare altitudes to help us compare their areas. Now, note that $AD/2$ is the altitude of $\triangle AFE$, and Stewart's theorem implies that $AD/2 = \frac{\sqrt{18}}{2}$. Similarly, the altitude of $\triangle AGH$ is the altitude of $\triangle ABC$, or $\frac{3\sqrt{7}}{2}$. However, by the symmetry axis $BE$ on quadrilateral $ABHE$, we have $BH = AB = 4$, and similarly $CG = CA = 6$, so $GB = HC = 1$, and therefore $[AGH] = [ABC]$. From here, we get that the area of $\triangle AFE$ is $\frac{15\sqrt{7}}{4}*(\frac{\frac{3\sqrt{7}}{2}}{\frac{\sqrt{18}}{2}})^2=\frac{15\sqrt{7}}{14} \implies \boxed{036}$, by similarity. ~awang11 ~gougutheorem

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 $AI = IM_A$.

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

$$\begin{aligned}\cos A &= \frac{9}{16}\implies \cos A/2 = \frac{5}{\sqrt{32}} \\ \cos B &= \frac{1}{8} \implies \cos B/2 = \frac{3}{4} \\ \cos C &= \frac{3}{4} \implies \cos C/2 = \sqrt{\frac{7}{8}}.\end{aligned}$$ 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)

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)

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)

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

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)

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

$$\begin{aligned} [\triangle AEF] = [\triangle DEF] &= \frac{DE\cdot DF\cdot\sin(\angle EDF)}{2}\\ &= \frac{DE\cdot DF\cdot\sin(y+z)}{2} = \frac{DE\cdot DF\cdot\cos x}{2}. \end{aligned}$$

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 $IM = MD - ID = \frac{\sqrt2}2$.

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)

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 $\hspace{30mm} [\triangle DEF] = \frac{15 \sqrt 7}{14}.$

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

Solution 14 (No Trig, No Coordinate Bashing, Simple Geometry)

[AUTHOR'S NOTE: Someone please help make a better diagram...]

By Angle Bisector Theorem, $BD=2,DC=3$. By Stewart's, $AD=3\sqrt2$, so $AK=\dfrac{3\sqrt2}{2}$. Thus, it suffices to find $FE$.

Notice that $\triangle AIK \cong \triangle DIK$, because $IK$ is the perpendicular bisector of $AD$. Also notice that $\triangle AIK \cong \triangle AJK$ by $SAS$. Thus, $AJDI$ is a rhombus, and $AI \parallel JD, AJ\parallel ID$.

Now notice $\triangle CJD \sim \triangle CAB$ by $AAA$. Since $\dfrac{CD}{CB}=\dfrac{JD}{AB}=\dfrac{3}{5}$, we have that $JD=ID=IA=AJ=\dfrac{12}{5}$. Thus, by Pythagorean Theorem, $KJ=\dfrac{3\sqrt{14}}{10},IJ=\dfrac{3\sqrt{14}}{5}.$

We now do some length chasing. $BI=4-\dfrac{12}{5}=\dfrac{8}{5},CJ=6-\dfrac{12}{5}=\dfrac{18}{5}.$ By Angle Bisector Theorem on $\triangle BID,$ we have $IL=\dfrac{16}{15}$. Similarly on $\triangle CJD$, $JM=\dfrac{72}{55}.$ We also have $AG=\dfrac{24}{11}$ and $AH=\dfrac{8}{3}$ by Angle Bisector Theorem on $\triangle ABC$. Thus, $GI=AI-AG=\dfrac{12}{5}-\dfrac{24}{11}=\dfrac{12}{55}$, and $JH=AH-AJ=\dfrac{8}{3}-\dfrac{12}{5}=\dfrac{4}{15}.$

It is easily shown that $\triangle GIF \sim \triangle JMF$, so by the side length ratios, we have

$$\dfrac{JM}{GI}=\dfrac{JF}{IF}=\dfrac{\frac{72}{55}}{\frac{12}{55}}=6.$$ Similarly, we can see that $\triangle EJH \sim \triangle EIL$, so by the side length ratios, we have

$$\dfrac{IL}{JK}=\dfrac{EI}{JE}=\dfrac{\frac{16}{15}}{\frac{4}{15}}=4.$$ So, we can see that the length of $FE$ is $\dfrac{6}{7}+\dfrac{1}{3}=\dfrac{25}{21}$ times the length of $IJ$. However, the length of $IJ$ is $\dfrac{3\sqrt{14}}{5}$, so $FE=\dfrac{5\sqrt{14}}{7}$, and the area of $\triangle AFE$ is $\dfrac{1}{2}\times \dfrac{5\sqrt{14}}{7}\times \dfrac{3\sqrt2}{2}=\dfrac{15\sqrt7}{14}.$ Thus, the desired answer is $15+7+14=\boxed{036}.$

~easond

Solution 15(Similar to Solutions with analytic geometry but this is pure coordinate bash)

Set $B$ to be at $(0, 0)$. Then $C$ will be at $(5, 0)$. Let $A$ be at $(x, y)$. Hence,

$x^{2} + y^{2} = 16$ $(x - 5)^{2} + y^{2} = 36$

Subtracting yields

$x^{2} - 10x + 25 - x^{2} = 20 \implies 10x = 5 \implies x = \frac{1}{2}$

Then $y = 16 - \frac{1}{4} = \frac{63}{4} \implies y = \frac{3\sqrt{7}}{2}$

Now by the Angle Bisector Theorem, $\frac{AB}{AC} = \frac{BD}{DC} \implies \frac{2}{3} = \frac{x}{5 - x}$ since $BD + DC = 5$. Now $10 - 2x = 3x \implies x = 2$. Hence $D$ is at $(2, 0)$. Let the point where $BE$ intersects $AC$ be $G$ and likewise $H$ for the point where $CF$ intersects $AB$. By the Angle Bisector Theorem,

$\frac{AB}{BC} = \frac{AG}{GB} \implies \frac{4}{5} = \frac{x}{6 - x}$. Solving yields $x = \frac{8}{3}$ and so $AG = \frac{8}{3}$ and $GC = \frac{10}{3}$. Doing similar things to finding the length of $AH$, we get $AH = \frac{24}{11}$ and $HB = \frac{20}{11}$.

Now we find the equation of line $AC$. The slope is just $\frac{\frac{3\sqrt{7}}{2}}{\frac{1}{2} - 5} = -\frac{\sqrt{7}}{3}$. Hence the equation of $AC$ is $y = -\frac{x\sqrt{7}}{3} + b$ and we can plug in point $(5, 0)$ because it lies on line $AC$ to get $0 = -\frac{5\sqrt{7}}{3} + b$ and solve to get the equation of line $AC$ is just $y = -\frac{x\sqrt{7}}{3} + \frac{5\sqrt{7}}{3}$.

Therefore, any point on line $AC$ must have the form $(x, -\frac{x\sqrt{7}}{3} + \frac{5\sqrt{7}}{3})$. Let $G$ be at point $(x, -\frac{x\sqrt{7}}{3} + \frac{5\sqrt{7}}{3})$. We know $GC$ is just $\frac{10}{3}$ so the distance between points $(x, -\frac{x\sqrt{7}}{3} + \frac{5\sqrt{7}}{3})$ and $(5, 0)$ is just $\frac{10}{3}$. Solving, we get $x = \frac{5}{2}, \frac{15}{2}$ but we know $x$ can at most be $5$ hence only $x = \frac{5}{2}$ works. Thus, $G$ will be at point $(\frac{5}{2}, \frac{5\sqrt{7}}{6})$.

The same process applies for finding the coordinates of point $H$. We end up with $H$ being at point $(\frac{5}{22}, \frac{15\sqrt{7}}{22})$.

We now find the equation of $BE$. Note that the slope is just $\frac{\frac{5\sqrt{7}}{6}}{\frac{5}{2}} = \frac{\sqrt{7}}{3}$. We know that $B$ is at the origin and so the equation of line $BE$ is just $y = \frac{x\sqrt{7}}{3}$.

Then find the equation of $CF$ in a similar way. We get $y = -\frac{x\sqrt{7}}{7} + \frac{5\sqrt{7}}{7}$.

Finally, the last step is finding the equation of the perpendicular bisector of $AD$. Note that the slope of $AD$ is just $\frac{\frac{3\sqrt{7}}{2}}{\frac{1}{2} - 2} = -\sqrt{7}$. Hence the slope of the perpendicular bisector is just $\frac{\sqrt{7}}{7}$ because that is the negative reciprocal of $-\sqrt{7}$. We also know that because this is the bisector, it passes through the midpoint of $AD$ which is just $(\frac{\frac{1}{2} + 2}{2}, \frac{3\sqrt{7}}{4}) = (\frac{5}{4}, \frac{3\sqrt{7}}{4})$. Therefore $y = -x\sqrt{7} + b$ and we know $(\frac{5}{4}, \frac{3\sqrt{7}}{4})$ is on it and plugging this in and solving and simplifying, we end up with the equation of the perpendicular bisector of $AD$:

$y = \frac{x\sqrt{7}}{7} + \frac{4\sqrt{7}}{7}$.

To find the coordinates of $E$, note that $BE$ intersects the perpendicular bisector of $AB$ at $E$ so we just need to solve $\frac{x\sqrt{7}}{3} = \frac{x\sqrt{7}}{7} + \frac{4\sqrt{7}}{7}$ and we get $x = 3$ and so point $E$ is on $(3, \sqrt{7})$.

Similarly, to find the coordinates of $F$, note that $CF$ intersects the perpendicular bisector of $AD$ at point $F$ so we just need to solve $-\frac{x\sqrt{7}}{7} + \frac{5\sqrt{7}}{7} = \frac{x\sqrt{7}}{7} + \frac{4\sqrt{7}}{7}$ and we get $x = \frac{1}{2}$. Thus point $F$ is on $(\frac{1}{2}, \frac{9\sqrt{7}}{14})$.

Now we have everything we need. We know $A$ is on $(\frac{1}{2}, \frac{3\sqrt{7}}{2})$, $E$ is on $(3, \sqrt{7})$, and $F$ is on $(\frac{1}{2}, \frac{9\sqrt{7}}{14})$.

We apply Shoelace Theorem to finish the problem off:

$[AEF] = \frac{1}{2} \mid (\frac{1}{2} \cdot \frac{9\sqrt{7}}{14} + \frac{1}{2} \cdot \sqrt{7} + 3 \cdot \frac{3\sqrt{7}}{2}) - (\frac{3\sqrt{7}}{2} \cdot \frac{1}{2} + \frac{9\sqrt{7}}{14} \cdot 3 + \sqrt{7} \cdot \frac{1}{2}) \mid = \frac{1}{2} \mid \frac{9\sqrt{7}}{28} + \frac{\sqrt{7}}{2} + \frac{9\sqrt{7}}{2} - \frac{3\sqrt{7}}{4} - \frac{27\sqrt{7}}{14} - \frac{\sqrt{7}}{2} \mid = \frac{1}{2} \mid \frac{60\sqrt{7}}{28} \mid = \frac{15\sqrt{7}}{14}$.

Hence, $[AEF] = \frac{15\sqrt{7}}{14}$. Then the answer is just $15 + 7 + 14 = \boxed{036}$.

~ilikemath247365

Video Solution

https://youtu.be/NpB7zpy-yKM

~MathProblemSolvingSkills.com