From 461ae8186abf5d99dccf439016599ed826575aaf Mon Sep 17 00:00:00 2001 From: xieyuen Date: Fri, 3 Jul 2026 22:09:08 +0800 Subject: [PATCH 1/3] =?UTF-8?q?Correct=20OD=20value=20to=20=E2=88=9A2+1=20?= =?UTF-8?q?in=20problem=201?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Fix a typo in the geometry problem: the condition OD was incorrectly written as √2-1 and is corrected to √2+1 in content/post/exam/contests/2025…/paper/index.md (problem 9.1). This updates the stated necessary and sufficient condition for CE ⟂ AD. --- .../paper/index.md" | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git "a/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/paper/index.md" "b/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/paper/index.md" index f0533a9..067736c 100644 --- "a/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/paper/index.md" +++ "b/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/paper/index.md" @@ -40,7 +40,7 @@ tags: #### 9. (16分， 10+6) -1. 如图 1, 圆 $O$ 的一条半径 $OB = 1$, $A, D$ 在射线 $OB$ 上, 且 $OA = \sqrt{2}-1$, $E$ 是圆上一动点, 连接 $AE$ 并延长交圆 $O$ 于点 $F$, 连接 $DF$ 并延长交圆 $O$ 于点 $C$. 求证: $OD = \sqrt{2}-1$ 是 $CE \perp AD$ 的充要条件. +1. 如图 1, 圆 $O$ 的一条半径 $OB = 1$, $A, D$ 在射线 $OB$ 上, 且 $OA = \sqrt{2}-1$, $E$ 是圆上一动点, 连接 $AE$ 并延长交圆 $O$ 于点 $F$, 连接 $DF$ 并延长交圆 $O$ 于点 $C$. 求证: $OD = \sqrt{2}+1$ 是 $CE \perp AD$ 的充要条件. 2. 如图 2, $BC$ 是圆 $O$ 的一条直径, $A$ 在线段 $OB$ 上, $OA = a, OB = r$, 过点 $A$ 作直线交圆 $O$ 于 $D, E$, 直线 $CD, BE$ 交于点 $F$, 过 $F$ 作 $FG \perp BC$ 于点 $G$. 问: $BG$ 是否为定值? 若是, 求出该定值(用 $a, r$ 表示); 若不是, 说明理由. ![图1](figure1.png) From a9d7cd58fa6e946db44337caa50f521f6789c5c5 Mon Sep 17 00:00:00 2001 From: xieyuen Date: Fri, 3 Jul 2026 22:31:49 +0800 Subject: [PATCH 2/3] Add sufficiency proof to answer.md MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit Replace placeholder comments in content/post/exam/contests/.../answer.md with the sufficiency part of the geometric solution. Adds reasoning using OA·OD = 1 = OF^2, triangle similarity (△OAF ~ △OFD), angle equalities and symmetry from OC = OE = OF to conclude OD ⟂ CE. Cleans up commented-out stubs. --- .../answer.md" | 15 ++++++++++++--- 1 file changed, 12 insertions(+), 3 deletions(-) diff --git "a/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/answer.md" "b/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/answer.md" index bd4136d..50f94f4 100644 --- "a/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/answer.md" +++ "b/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/answer.md" @@ -47,11 +47,20 @@ tags: 答案暂不提供 - +**【充分性】**: + +连接 $OC, OE, OF, DE$, 延长 $DO$ 交 $CE$ 于点 $G$. +由 $OD = \sqrt{2}+1$ 可知 $OA \cdot OD = 1 = OF^2$, +则 $\triangle OAF \sim \triangle OFD$, 从而 $\angle OFA = \angle ODF, \angle OAF = \angle OFD$. +同理有 $\angle OEA = \angle ODE$. +记 $\angle OFA = \angle ODF = \alpha$, $\angle FAB = \beta$, 则 $\angle OAF = \angle OFD \implies \angle OFC = \beta$. +由 $OC=OE=OF$ 可得 $\angle OEF = \angle OFE = \alpha, \angle OCF = \angle CFO = \beta$, 结合 $\angle OAE = \angle FAD = \beta$ 有 $\angle GOE = \alpha + \beta = \angle GOC$, 故 $OD \perp CE$. + + ### 10. (20分, 4+8+8) From 0b6280a91dd56ba14a28b0f13e38618c8ed2bfb7 Mon Sep 17 00:00:00 2001 From: xieyuen Date: Fri, 3 Jul 2026 22:43:48 +0800 Subject: [PATCH 3/3] =?UTF-8?q?=E8=B0=83=E6=95=B4=E5=8E=9F=E5=88=9B?= =?UTF-8?q?=E9=A2=98=E5=BE=97=E5=88=86=E7=8E=8B=E9=83=A8=E5=88=86=E7=9A=84?= =?UTF-8?q?=E6=A0=BC=E5=BC=8F?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../award-winners/index.md" | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git "a/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/award-winners/index.md" "b/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/award-winners/index.md" index d8218b4..343b3f5 100644 --- "a/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/award-winners/index.md" +++ "b/content/post/exam/contests/2025\351\253\230\344\270\200\347\253\236\350\265\233/award-winners/index.md" @@ -55,8 +55,8 @@ tags: 在数协社干的原创试题中获得最高分的同学，在此予以表扬： -第9题圆与直线（满分16分）：黄思浩（1班）11分、刘星宇（1班）11分 -第10题函数与不等式（满分20分）：刘铠闻（1班）19分 +- 第9题圆与直线（满分16分）：黄思浩（1班）11分、刘星宇（1班）11分 +- 第10题函数与不等式（满分20分）：刘铠闻（1班）19分 ## 更多信息