[MAT 2022 #9] Add Me

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[MAT 2022 #9] Add Me

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#1 h Jul 23, 2022, 11:30 PM • 1 Y

Y by Mango247

Let $\triangle ABC$ be a scalene triangle with incenter $I$ , and let the midpoint of side $\overline{BC}$ be $M$ . Suppose that the feet of the altitudes from $I$ to $\overline{AC}$ and $\overline{AB}$ are $E$ and $F$ respectively, and suppose that there exists a point $D$ on $\overline{EF}$ such that $ADMI$ is a parallelogram. If $AB+AC = 24$ , $BC = 14$ , and the area of $\triangle ABC$ can be expressed as $m\sqrt{n}$ where $m$ and $n$ are positive integers such that $n$ is squarefree, find $m+n$ .

Aprameya Tripathy

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Taco12

#2 h Jul 23, 2022, 11:51 PM • 1 Y

Y by megarnie

bary ftw $\text{[math]}$

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bissue

#3 h Jul 23, 2022, 11:55 PM • 12 Y

Y by megarnie, I-_-I, fuzimiao2013, ApraTrip, centslordm, ihatemath123, insertionsort, Bradygho, mathleticguyyy, rayfish, Mango247, sargamsujit

:love:

:love:

:love:

:love:

:love:

PART I. Standard Observations

First make a few computational observations:

The semiperimeter is $s = \tfrac{AB + AC + BC}{2} = \tfrac{24 + 14}{2} = 19$ .
The lengths of $AE = AF$ are $s - BC = 19 - 14 = 5$ .

We can also make the following geometric observation:

Claim wrote:

Lines $AI$ and $MD$ are perpendicular to $EF$ . In particular, $D$ is necessarily the foot of the altitude from $M$ onto $EF$ .

Proof: Observe that since $\overline{AI}$ bisects $\angle A$ in $\triangle ABC$ , it also bisects $\angle A$ in $\triangle AEF$ . However, since $\triangle AEF$ is isosceles, the angle bisector $\overline{AI}$ must also be an altitude, so $AI \perp EF$ .

Consequently, since $ADMI$ is a parallelogram, $MD \perp EF$ as well. $\square$

Part II. wait what how did that just work

In order to learn more about $\triangle ABC$ , we'll focus primarily on the condition that $AI = MD$ (which follows from $AIMD$ being a parallelogram).

Suppose $X$ and $Y$ are the feet of the altitudes from $B$ and $C$ onto line $EF$ . Observe that:
$MD = \dfrac{BX + CY}{2}.$ This follows from the fact that $M$ is the midpoint of $\overline{BC}$ . We'll now compute $MD$ by computing $BX$ and $CY$ separately

Suppose that $AI$ has length $x$ . Observe that:
$\triangle BXF \sim \triangle AFI \hbox{ and } \triangle AEI \sim \triangle CYE.$ This follows from the fact that $BX \parallel AI \parallel CY$ , as all three are altitudes onto $EF$ . These similarities give these length ratios:
$\dfrac{BX}{BF} = \dfrac{AF}{AI} = \dfrac{5}{AI} \hbox{ and } \dfrac{CY}{CE} = \dfrac{AE}{AI} = \dfrac{5}{AI}.$ Therefore we may extract:
$BX = \dfrac{5}{AI} \times BF \hbox{ and } CY = \dfrac{5}{AI} \times CE.$ So then:
$MD = \dfrac{BX + CY}{2} = \dfrac{5}{2 \times AI} \times (BF + CE).$ Except notice that incircle tangencies form equal lengths! In particular, if we denote the third altitude from $I$ onto $\overline{BC}$ as $G$ , then $BF = BG$ and $CE = CG$ , meaning:
$MD = \dfrac{5}{2 \times AI} \times (BG + CG) = \dfrac{5}{2 \times AI} \times BC = \dfrac{35}{AI}.$ But we know that $MD$ should actually equal $AI$ , meaning that:
$AI = \dfrac{35}{AI} ~ \Longrightarrow ~ \boxed{AI = \sqrt{35}.}$
Part III. Conclusion

Now that we know the length of $AI$ , we may compute the inradius by looking at right triangle $AFI$ :
$AF^2 + FI^2 = AI^2 ~ \Longrightarrow ~ 5^2 + FI^2 = (\sqrt{35})^2 ~ \Longrightarrow FI = \hbox{ inradius } = \sqrt{10}.$ So the area is $rs = \boxed{19\sqrt{10}}$ , giving an answer of $\boxed{29.}$ $\blacksquare$

This post has been edited 2 times. Last edited by bissue, Jul 24, 2022, 12:47 AM

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#4 h Jul 24, 2022, 12:11 AM • 2 Y

Y by Arrowhead575, Taco12

WAIT WHAT THE ACTUAL HECK I GUESSED 10\sqrt{19} --> 29 LMHO

but i don't think i submitted by 4:00 rip

Fakesolve Solution

This post has been edited 1 time. Last edited by samrocksnature, Jul 24, 2022, 12:15 AM

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#5 h Jul 24, 2022, 12:21 AM

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bissue here writing five paragraph essays on math problems

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#6 h Jul 24, 2022, 12:55 AM

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NOOO I spent $20$ minutes on this problem and got the same observation but started some strange bash and got a weird radical

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#7 h Jul 24, 2022, 1:10 AM • 11 Y

Y by Arrowhead575, ApraTrip, centslordm, megarnie, insertionsort, I-_-I, samrocksnature, mathleticguyyy, rayfish, ike.chen, sargamsujit

Here's a slicker approach using Iran Lemma.

Let $BI,CI$ intersect $EF$ at $X,Y.$ So $X,Y$ lie on $(BC)$ by Iran. Easy angle chase gives $\angle XMY = \angle BAC$ and also notice $DM \perp XY$ so $D$ is the midpoint of $XY.$ Then by some similar triangle ratios

$\frac{DM}{7} = \frac{DM}{MY}= \frac{AF}{AI} = \frac{5}{AI}$
so $AI = DM = \sqrt{35},$ rest just computation. $\blacksquare$

Attachments:

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#8 h Jul 24, 2022, 1:15 AM

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$\text{[math]}$

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#9 h Jul 24, 2022, 1:55 AM

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But how many people would have known that lemma??
:noo:

:noo:

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#10 h Jul 24, 2022, 2:04 AM

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euler12345 wrote:

But how many people would have known that lemma??
:noo:

:noo:

I think its somewhat well known in oly geo.

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#11 h Jul 24, 2022, 2:22 AM • 3 Y

Y by samrocksnature, Taco12, Mango247

Congrats on this nice problem @ApraTrip!

First, note that $s=\frac{a+b+c}2=19, a=14,b=24-c,c=c$ . Now employ barycentrics on $\triangle ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$ . Clearly $M=(0,\tfrac12,\tfrac12), E=(s-c:0:s-a)=(19-c:0:5),F=(s-b:s-a:0)=(-5+c:5:0),I=(a:b:c)=(14:24-c:c)$ . Since $ADMI$ is a parallelogram, $A+M=D+I \Rightarrow D=(24:-5+c:19-c).$ However, we also know that $D,E,F$ are colinear so we must have $\begin{vmatrix}-5+c&5&0\\ 19-c&0&5\\ 24&-5+c&19-c\end{vmatrix}=0 \iff c^2-24c+133=0.$ By the quadratic formula, $c=12\pm \sqrt{11}$ and WLOG take $12+\sqrt{11}$ . Then $b=24-c=12-\sqrt{11}$ . Now, by Heron's $[ABC]=\sqrt{19(19-14)(19-12+\sqrt{11})(19-12-\sqrt{11})}=\sqrt{19(5)(7^2-11)}=19\sqrt{10}.$

This post has been edited 4 times. Last edited by franzliszt, Jul 24, 2022, 2:23 AM

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#12 h Jul 25, 2022, 3:10 PM

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We have $BC=BE+CF=14$ . Since $AB+AC=24$ and $BE+CF=14$ , we have $AE=AF=5$ and $s=19$ . Proceed as in post 3 by dropping altitudes from $B$ and $C$ on to $\overline{EF}$ . $BX+CY=2AI$ . We next note that because of parallel lines $AI$ , $MD$ , $BX$ , and $CY$ , we have equal angles. So $BX=BE\cos A/2$ and $CY=CF\cos A/2$ . And because $\cos A/2 = 5/AI$ , we have $2AI=70/AI$ , or $AI=\sqrt{35}$ . Thus, $r=\sqrt{10}$ , so $[ABC]=19r=19\sqrt{10}$ .

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#13 h Dec 23, 2022, 8:15 PM • 1 Y

Y by sargamsujit

The idea is to use the condition $AI = DM$ , as $D$ must be the foot of the altitude from $M$ to $\overline{EF}$ . Here, notice that $AI = \frac{AF}{\cos \frac A2} = \frac{AB+AC-BC}{2\cos \frac A2} = \frac 5{\cos \frac A2},$ and $DM$ is the average of the distances from $B, C$ to $\overline{EF}$ , which is $\frac{BF+CE}2 \cdot \cos \frac A2 = 7 \cos \frac A2.$ Thus equating the two, $\cos^2 \frac A2 = \frac 57$ , and $\cos A = \frac 37$ . To extract $AB \cdot AC$ , we can use the Law of Cosines to get $AB^2+AC^2-2 \cdot AB \cdot AC \cdot \frac 37 = 196 \implies (AB+AC)^2-\frac{20}7 \cdot AB \cdot AC = 196.$ Thus $AB \cdot AC = 133$ , and the area is $[ABC] = 133 \cdot \frac{2\sqrt{10}}7 \cdot \frac 12 = 19\sqrt{10}.$

This post has been edited 1 time. Last edited by HamstPan38825, Dec 23, 2022, 8:15 PM

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Taco12

#14 h Jan 28, 2023, 1:17 AM • 5 Y

Y by samrocksnature, megarnie, Mango247, Mango247, Mango247

Solution from OTIS:

Apply barycentric coordinates on $\triangle ABC$ . Note that $E=(19-c:0:5), F=(19-b:5:0)$ , so $EF$ has equation $5z(19-c)+5y(19-b)=25x$ . By Parallelogram Lemma, $D=\left(\frac{24}{38}, \frac{19-b}{38}, \frac{19-c}{38}\right)$ since $M=(0:1:1)$ . Plugging in $D$ into the equation for $EF$ gives $b,c = 12-\sqrt{11}, 12+\sqrt{11}$ in some order. Plugging into Heron's now gives $19\sqrt{10}$ .

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#15 h Jul 25, 2023, 6:27 AM

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Squareman's solution is absolutely unreal; here is a more reasonable approach.

Walk-Through:

Let $AI$ meet $EF$ and $(ABC)$ again at $N$ and $X$ respectively. We define $P$ and $Q$ as the projections of $X$ onto $AB$ and $AC$ respectively.
By Simson Lines, $M \in PQ$ . Also, symmetry implies $EF \perp \overline{ANIX} \perp \overline{MPQ}$ .
The parallelogram condition is clearly equivalent to $AI = MD = \text{dist}(M, EF)$ .
It's well-known
IMO 1987/2, Spiral Similarity Congruence
that $AP = AQ = \frac{AB + AC}{2} = 12$ .
Compute $AE = s - a = 5$ .
Now, $\cos^2 \left( \frac{A}{2} \right) = \frac{AN}{AI} = \frac{\text{dist}(A, EF)}{\text{dist}(M, EF)} = \frac{AE}{EP} = \frac{5}{7}$ where the penultimate equality follows from $EF \parallel \overline{MPQ}$ .
Use $\triangle AEI$ to obtain $IF = \sqrt{10}$ .
Extract $[ABC] = r \cdot s = 19 \sqrt{10}$ , whence $m + n = \boxed{029}$ .

This post has been edited 2 times. Last edited by ike.chen, Jul 25, 2023, 6:28 AM

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