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September 22, 2020

GCF & Rectangle | AMC 10A, 2016| Problem No 19

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 $r+s+t$=\(20\)

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