asdf1434 Mock AIME #6

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P_Groudon

867 posts

P_Groudon

#1 h May 8, 2024, 7:09 PM • 1 Y

Y by Rounak_iitr

Let $ABCD$ be a trapezoid with $AB \parallel CD$ . Let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively, and let $E$ be the intersection of diagonals $AC$ and $BD$ . If $EM = 1$ , $EN = 3$ , $CD = 10$ , and $\frac{AD}{BC} = \frac{3}{4}$ , then $AD^2 + BC^2$ can be expressed as $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$ . Find $m + n$ .

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vanstraelen

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vanstraelen

#2 h May 9, 2024, 6:25 PM

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Given the trapezoid $ABCD\ :\ A(a,c),B(b,c),C(10,0),D(0,0)$ .
Then $M(\frac{a+b}{2},c)$ and $N(5,0)$ .
Point $E(\frac{10b}{10+b-a},\frac{10c}{10+b-a})$ .

Three conditions, three unknowns $\Rightarrow a=\frac{823}{375}, b=\frac{691}{125}$
and $AD^{2}=\frac{488}{25},BC^{2}=\frac{7808}{225}$ .

This post has been edited 1 time. Last edited by vanstraelen, May 9, 2024, 8:10 PM

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miyukina

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miyukina

#3 h May 9, 2024, 7:28 PM

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AD^2 = h^2 + x^2
= (3k)^2
BC^2 = h^2 + (20/3 – x)^2
= (4k)^2

7k^2= (20/3)^2 – 40x/3
40x/3 = 400/9 – 7k^2
===> x = 10/3 – 21k^2/40
4^2 = (21k^2/40)^2 + h^2
= (21k^2/40)^2 + (9k^2 – x^2)
= (21k^2/40)^2 + 9k^2 – (10/3 – 21k^2/40)^2
= 9k^2 – 100/9 + 7k^2/2
25k^2/2 = 16 + 100/9
= 244/9
===> k^2 = 488/225

Expression
= AD^2 + BC^2
= k^2 × (3^2 + 4^2)
= 488/225 × 25
= 488/9

Answer
= 488 + 9
= 497

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P_Groudon

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P_Groudon

#4 h May 9, 2024, 7:41 PM

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I got 497 too

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Zongyouyu

42 posts

Zongyouyu

#5 h Dec 15, 2024, 11:10 PM

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Why is $MN$ , $DB$ , and $AC$ concurrent?

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ethanzhang1001

1060 posts

ethanzhang1001

#6 h Dec 15, 2024, 11:19 PM

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Zongyouyu wrote:

Why is $MN$ , $DB$ , and $AC$ concurrent?

Let the intersection of $AC$ and $MN$ be at $E$ , and let $BD$ and $MN$ intersect at $E'$ . Notice that because $DN=CN$ and $BM=AM$ , then due to similar triangles having the same ratio, we have $\frac{AE}{CE}=\frac{AE'}{CE'}$ , implying $E$ is the same point as $E'$ .

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persamaankuadrat

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persamaankuadrat

#7 h Dec 16, 2024, 5:00 AM

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vanstraelen wrote:

Given the trapezoid $ABCD\ :\ A(a,c),B(b,c),C(10,0),D(0,0)$ .
Then $M(\frac{a+b}{2},c)$ and $N(5,0)$ .
Point $E(\frac{10b}{10+b-a},\frac{10c}{10+b-a})$ .

Three conditions, three unknowns $\Rightarrow a=\frac{823}{375}, b=\frac{691}{125}$
and $AD^{2}=\frac{488}{25},BC^{2}=\frac{7808}{225}$ .

How did you obtain $a,b$ from there? My equation is a complete mess

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Magnetoninja

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Magnetoninja

#8 h Dec 17, 2024, 12:14 AM

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By similar triangles $AB=\frac{10}{3}$ and then do pythagorean on right triangle with hypotenuse $AD$ and $BC$ and then you get a system and solve.

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clarkculus

221 posts

clarkculus

#9 h Jan 20, 2025, 9:59 PM • 1 Y

Y by centslordm

Outline: $AD, MN, BC$ concur at $P$ . Construct the line through $E$ parallel to $AB$ , intersecting $AD$ at $X$ and $BC$ and $Y$ . Then $XYN$ is the medial triangle of $PCD$ . Chase lengths then LoC on $XDN$ and $PXN$ .

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