Difference between revisions of "2021 CIME I Problems/Problem 1"

(created solution page)
 
(diagram)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
 +
 +
<asy>
 +
 +
import geometry;
 +
 +
point B = (0,0);
 +
point C = (16,0);
 +
point A = (0,16);
 +
point D = (16,16);
 +
point P = A * 1/3;
 +
point Q = C * 38/85 + D * 47/85;
 +
 +
// Square ABCD
 +
draw(A--B--C--D--A);
 +
dot(A);
 +
label("A",A,NW);
 +
dot(B);
 +
label("B",B,SW);
 +
dot(C);
 +
label("C",C,SE);
 +
dot(D);
 +
label("D",D,NE);
 +
 +
// Segment PQ
 +
draw(P--Q);
 +
dot(P);
 +
label("P",P,W);
 +
dot(Q);
 +
label("Q",Q,E);
 +
 +
</asy>
 +
 
<math>\boxed{123}</math>
 
<math>\boxed{123}</math>
  
 
== See also ==
 
== See also ==
 
{{CIME box|year=2021|n=I|before=First Problem|num-a=2}}
 
{{CIME box|year=2021|n=I|before=First Problem|num-a=2}}

Revision as of 18:12, 24 July 2024

Problem

Let $ABCD$ be a square. Points $P$ and $Q$ are on sides $AB$ and $CD,$ respectively$,$ such that the areas of quadrilaterals $APQD$ and $BPQC$ are $20$ and $21,$ respectively. Given that $\tfrac{AP}{BP}=2,$ then $\tfrac{DQ}{CQ}=\tfrac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. Find $a+b$.

Solution

[asy]  import geometry;  point B = (0,0); point C = (16,0); point A = (0,16); point D = (16,16); point P = A * 1/3; point Q = C * 38/85 + D * 47/85;  // Square ABCD draw(A--B--C--D--A); dot(A); label("A",A,NW); dot(B); label("B",B,SW); dot(C); label("C",C,SE); dot(D); label("D",D,NE);  // Segment PQ draw(P--Q); dot(P); label("P",P,W); dot(Q); label("Q",Q,E);  [/asy]

$\boxed{123}$

See also

2021 CIME I (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All CIME Problems and Solutions