AIME 2001 I · 第 7 题

Solution 1

Let $I$ be the incenter of $\triangle ABC$, so that $BI$ and $CI$ are angle bisectors of $\angle ABC$ and $\angle ACB$ respectively. Then, $\angle BID = \angle CBI = \angle DBI,$ so $\triangle BDI$ is isosceles, and similarly $\triangle CEI$ is isosceles. It follows that $DE = DB + EC$, so the perimeter of $\triangle ADE$ is $AD + AE + DE = AB + AC = 43$. Hence, the ratio of the perimeters of $\triangle ADE$ and $\triangle ABC$ is $\frac{43}{63}$, which is the scale factor between the two similar triangles, and thus $DE = \frac{43}{63} \times 20 = \frac{860}{63}$. Thus, $m + n = \boxed{923}$.

Solution 2

The semiperimeter of $ABC$ is $s = \frac{20 + 21 + 22}{2} = \frac{63}{2}$. By Heron's formula, the area of the whole triangle is $A = \sqrt{s(s-a)(s-b)(s-c)} = \frac{21\sqrt{1311}}{4}$. Using the formula $A = rs$, we find that the inradius is $r = \frac{A}{s} = \frac{\sqrt{1311}}6$. Since $\triangle ADE \sim \triangle ABC$, the ratio of the heights of triangles $ADE$ and $ABC$ is equal to the ratio between sides $DE$ and $BC$. From $A=\frac{1}{2}bh$, we find $h_{ABC} = \frac{21\sqrt{1311}}{40}$. Thus, we have

Solving for $DE$ gives $DE=\frac{860}{63},$ so the answer is $m+n=\boxed{923}$.

Or we have the area of the triangle as $S$. Using the ratio of heights to ratio of bases of $ADE$ and $ABC$ $\frac {\frac {2S}{20}-\frac {2S}{63}}{\frac {2S}{20}}= \frac {DE}{BC(20)}$ from that it is easy to deduce that $DE=\frac{860}{63}$.

Solution 3 (mass points)

Let $P$ be the incenter; then it is be the intersection of all three angle bisectors. Draw the bisector $AP$ to where it intersects $BC$, and name the intersection $F$.

Using the angle bisector theorem, we know the ratio $BF:CF$ is $21:22$, thus we shall assign a weight of $22$ to point $B$ and a weight of $21$ to point $C$, giving $F$ a weight of $43$. In the same manner, using another bisector, we find that $A$ has a weight of $20$. So, now we know $P$ has a weight of $63$, and the ratio of $FP:PA$ is $20:43$. Therefore, the smaller similar triangle $ADE$ is $43/63$ the height of the original triangle $ABC$. So, $DE$ is $43/63$ the size of $BC$. Multiplying this ratio by the length of $BC$, we find $DE$ is $860/63 = m/n$. Therefore, $m+n=\boxed{923}$.

Solution 4 (Faster)

More directly than Solution 2, we have

Solution 5

Diagram borrowed from Solution 3.

Let the angle bisector of $\angle{A}$ intersects $BC$ at $F$.

Applying the Angle Bisector Theorem on $\angle{A}$ we have

$$\frac{AB}{BF}=\frac{AC}{CF}$$ $$BF=BC\cdot(\frac{AB}{AB+AC})$$ $$BF=\frac{420}{43}$$ Since $BP$ is the angle bisector of $\angle{B}$, we can once again apply the Angle Bisector Theorem on $\angle{B}$ which gives

$$\frac{BA}{AP}=\frac{BF}{FP}$$ $$\frac{AP}{PF}=\frac{AB}{BF}=\frac{41}{20}$$ Since $\bigtriangleup ADE\sim\bigtriangleup ABC$ we have

$$\frac{DE}{BC}=\frac{AP}{AF}$$ $$DE=BC\cdot(\frac{AP}{(\frac{61}{41})\cdot AP})$$ Solving gets $DE=\frac{860}{63}$. Thus $m+n=860+63=\boxed{923}$.

~ Nafer

Solution 6

Let $A'$ be the foot of the altitude from $A$ to $\overline {BC}$ and $K$ be the foot of the altitude from $A$ to $\overline{DE}$. Evidently,

$$\frac{AK}{AA'} = 1- \frac{r}{AA'} = 1 - \frac{T/s}{T/BC}$$ where $r$ is the inradius, $T = [ABC]$, and $s$ is the semiperimeter. So,

$$\frac{AK}{AA'} = 1 - \frac{BC}{s} = 1 - \frac{20}{63}= \frac{43}{63}$$ Therefore, by similar triangles, we have $DE = BC * \frac{AK}{AA'} = 20 * \frac{AK}{AA'}= \boxed{\frac{860}{63}}$.

Solution 7

Label $P$ the point the angle bisector of $A$ intersects ${BC}$. First we find ${BP}$ and ${PC}$. By the Angle Bisector Theorem, $\frac{BP}{PC} = \frac{21}{22}$ and solving for each using the fact that ${BC} = 20$, we see that ${BP} = \frac{420}{43}$ and $PC = \frac{440}{43}$.

Because ${AP}$ is the angle bisector of $, we can simply calculate it using Stewarts,

$${AP} = 21*22 - \frac{440}{43}\cdot\frac{420}{43}$$ $${AP} = 21*22 - \frac{440\cdot420}{43^2}$$ Now we can calculate what ${AO}$ is. Using the formula to find the distance from a vertex to the incenter, ${AO} = \frac{43}{63} \cdot[21\cdot22 - \frac{420*440}{43^2}] = \frac{43^2\cdot22 - 20\cdot440}{43\cdot3}$.

Now because $\triangle{APE} ~ \triangle{ABC}$, we can find ${DE}$ by $\frac{AO}{AP} \cdot 20$. Dividing and simplifying, we see that $\frac{1}{21}\cdot\frac{43}{3}\cdot20 = \frac{860}{63}$. So the answer is $\boxed{923}$

~YBSuburbanTea

Solution 8 (vectors)

To solve this problem, we can use the fact that, in $\triangle ABC$, the vector representation of the incenter is $\overrightarrow I = \frac{a\overrightarrow A + b\overrightarrow B + c\overrightarrow C}{a+b+c}$ and that that the vector of the foot of the bisector of $\angle BAC$ on $\overline{BC}$ is $\overrightarrow P = \frac{b\overrightarrow B + c\overrightarrow C}{b+c}$, where $a=BC,$ $b=AC,$ and $c=AB$.

Let point $A$ be the origin of the coordinate plane. Then, $\overrightarrow A$ is the zero vector, so we can simplify our expression for $\overrightarrow I$ to $\frac{b\overrightarrow B + c\overrightarrow C}{a+b+c}$. Now, note that the vector components of $\overrightarrow I$ and $\overrightarrow P$ are the same, but they are multiplied by different scalars. Thus, the ratio of these scalars is the ratio of these vectors' magnitudes. Thus, we have $\frac{|\overrightarrow I|}{|\overrightarrow P|}=\frac{\tfrac1{a+b+c}}{\tfrac1{b+c}}=\frac{b+c}{a+b+c}=\frac{43}{63}$.

Let $D \in \overline{AB}$ and $E \in \overline{AC}$. Because $\triangle AIE \sim \triangle APC$, we have $\frac{AI}{AP}=\frac{AE}{AC}$. Further, because $\triangle ADE \sim \triangle ABC$, we have $\frac{AE}{AC}=\frac{DE}{BC}$. Thus, by transitivity, $\frac{AI}{AP}=\frac{DE}{BC}$. We know that $\frac{AI}{AP}=\frac{43}{63}$, so $DE=\frac{AI}{AD}\cdot BC = \frac{43}{63}\cdot 20 = \frac{860}{63}$.

Thus, our answer is $860+63=\boxed{923}$.