Track / Stage 4 / 109 of 340 #369 of 1964

Problem 369

AMC 12 late, AIME early

Geometry Difficulty 4.7 Multiple choice

4. In a convex quadrilateral $ABCD$ , $E, F, G, H$ are the midpoints of $AB, BC, CD, DA$ respectively, and $EG$ intersects $FH$ at point $O$ . Let the areas of quadrilaterals $AEOH, BFOE, CGOF$ be $3, 4, 5$ respectively. Then the area of quadrilateral $DHOG$ is:

Pick one

Official solution

4.C.

As shown in Figure 3, connect $O A, O B, O C, O D$ . Then
$\begin{array}{l} S_{\triangle A E O}=S_{\triangle B E O}, \\ S_{\triangle B F O}=S_{\triangle C F O}, \\ S_{\triangle C O O}=S_{\triangle D C O}, \\ S_{\triangle D H O}=S_{\triangle A H O} . \end{array}$

Therefore, $S_{\text {quadrilateral }}$ EOH $+S_{\text {quadrilateral }}$ OCC
$=S_{\text {quadrilateral } B F O E}+S_{\text {quadrilateralDHOC }} \text {. }$

Thus, $S_{\text {quadrilateralDHOc }}=3+5-4=4$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.