Try this beautiful Problem on Geometry based on GCF & Rectangle from AMC 10 A, 2010. You may use sequential hints to solve the problem.

**GCF & Rectangle** – AMC-10A, 2016- Problem 19

In rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $B E=E F=F C$. Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q$, respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$ where the greatest common factor of $r, s,$ and $t$ is $1 .$ What is $r+s+t ?$

,

- $7$
- $9$
- $12$
- $15$
- $20$

**Key Concepts**

Geometry

Rectangle

Diagonal

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2016 Problem-19

#### Check the answer here, but try the problem first

$20$

## Try with Hints

#### First Hint

Given that , rectangle $A B C D, A B=6$ and $B C=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that Segments $\overline{A E}$ and $\overline{A F}$ intersect $\overline{B D}$ at $P$ and $Q,$ respectively. The ratio $B P: P Q: Q D$ can be written as $r: s: t$. we have to find out $r+s+t ?$, where greatest common factor of \(r,s,t\) is \(1\)

Now $\triangle A P D \sim \triangle E P B$. From this relation we can find out a relation between \(DP\) and \(PB\)

Now can you finish the problem?

#### Second Hint

Now $\triangle A P D \sim \triangle E P B$\(\Rightarrow\) $\frac{D P}{P B}=\frac{A D}{B E}=3$ Therefore $P B=\frac{B D}{4}$.

SimIarly from the \(\triangle AQD \sim \triangle BQF\) \(\Rightarrow \)$\frac{D Q}{Q B}=\frac{3}{2}$

Therefore we can say that $D Q=\frac{3 \cdot B D}{5}$

Now can you finish the problem?

#### Third Hint

Therefore $r: s: t=\frac{1}{4}: \frac{2}{5}-\frac{1}{4}: \frac{3}{5}=5: 3: 12,$ so $r+s+t$=\(20\)

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=y-8Ru_qKDxk&t=6s