self-host the Putnam diagram

diff --git a/content/2025/the-end-of-the-movie-sf-state-2024-putnam-competition-team-a-retrospective.md b/content/2025/the-end-of-the-movie-sf-state-2024-putnam-competition-team-a-retrospective.md
index 951f851770f9d5c1a11c015c805ea00090b1564f..ddff5b52ef89a6cf866dc719751d0dcffee08117 100644 (file)
--- a/content/2025/the-end-of-the-movie-sf-state-2024-putnam-competition-team-a-retrospective.md
+++ b/content/2025/the-end-of-the-movie-sf-state-2024-putnam-competition-team-a-retrospective.md
@@ -61,7 +61,7 @@ Luck seemed to deliver. On a skim, B1, B2, and B4 looked potentially tractable.
 
 > **B2.** Two convex quadrilaterals are called _partners_ if they have three vertices in common and they can be labeled ABCD and ABCE so that E is the reflection of D across the perpendicular bisector of the diagonal AC. Is there an infinite sequence of convex quadrilaterals such that each quadrilateral is a partner of its successor and no two elements of the sequence are congruent?
 
-![](https://cdn.artofproblemsolving.com/attachments/6/e/cc9da12a49043410c50733cb6843e5ec1005d3.jpeg)
+![]({static}/images/partner_quadrilaterals_putnam_2024_b2.jpeg)
 
 I imagined rotating the figure such that AC was the vertical axis and its bisector was the horizontal axis, and tried to imagine some way to perturb D and E to get a sequence of quadrilaterals that wouldn't be congruent (because the angles ∠CDA and ∠CEA were changing), but for which we could alternately take ABCD and ABCE so that successive shapes in the sequence would be partners. I couldn't see a way to make it work. Then I thought, what if perturb B instead?
 

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