测圆海镜

圆城图式（右图）是全书的总括图解，由一个直角三角形（古时称为勾股形）、它的内切圆以及一些特定的点和直线组成。其中的顶点、圆心和交点都用某个汉字来指代。最大的三角形的三个顶点分别是天、地、，天地三角形的内切圆圆心称为心。过心的垂直线从上至下分别和三角形、内切圆交于日、南、北三点。过心的水平线从左至右分别和三角形、内切圆交于川、东、西三点。过东的垂直线和过南的水平线都是内切圆的切线，它们分别交天地三角形于艮、坤、山、月四点，而相交于巽点。坤巽艮构成一个正方形。过月的垂直线交东西水平线于青点，交地边于泉点。过山的水平线交南北垂直线于朱点，交天边于金点。而这两条线相交于泛点。最后过日的水平线交天边于旦点，过川的垂直线交地边于夕点。总共22个点。

全书所研究的三角形一共有15个，全部是以天地线之间的线段为弦（斜边）的直角三角形。总率名号给出了这些三角形和线段的名称。它们分别是：

其中弦是三角形斜边，股是三角形的长直角边（这里是竖直的），勾是三角形短直角边（这里是水平的）。（${\displaystyle a_{1}}$代表通勾，${\displaystyle b_{1}}$代表通股，${\displaystyle c_{1}}$代表通弦，余类推）。

今问正数一节给出了圆城图式中每个线段的长度。其中以内切圆的半径为120步，作为标准。

• 弦
• 勾
• 股
• 勾股和：a + b
• 勾股校：b - a
• 勾弦和：a + c
• 勾弦校：c - a
• 股弦和：b + c
• 股弦校：c - b
• 弦校和：c + (b - a)
• 弦校校：c - (b - a)
• 弦和和：(a + b) + c
• 弦和校：(a + b) - c

例子:「通弦六百八十，勾三百二十，股六百；勾股和九百二十，较（兩者的差）二百八十；勾弦和一千，较三百六十；股弦和一千二百八十，较八十；弦较和九百六十，较四百；弦和和一千六百，较二百四十。」

15个勾股形中上高 = 下高；上平 = 下平，因此，15个勾股形中，只有13个勾股形是相异的。

《今问正数》共15个勾股形×13项=195项[2]。 ，列表如下。

识别杂记都是关于不同线段之间的几何关系式。一共给出了692个公式。是全书的纲领。

识别杂记包含八项：

• 诸杂名目：是全书的总纲，列出各项定义，例如虚勾虚股相得名为虚率，高股平勾差名为角差，又名远差等等。诸杂名目中还列出三十余项定理，如凡大差小差相乘为半段径幂，大差勾小差股相乘同上、黄广股黄长勾相乘为经幂等等。

[3]

1. ${\displaystyle (c_{1}-a_{1})*(c_{1}*b_{1})}$= ${\displaystyle 1 \over 2}$*${\displaystyle (d_{1})^{2}}$
2. ${\displaystyle a_{10}*b_{11}}$= ${\displaystyle 1 \over 2}$${\displaystyle (d_{1})^{2}}$
3. ${\displaystyle a_{13}*b_{1}}$= ${\displaystyle 1 \over 2}$${\displaystyle (d_{1})^{2}}$
4. ${\displaystyle a_{1}*b_{13}}$= ${\displaystyle 1 \over 2}$${\displaystyle (d_{1})^{2}}$
5. ${\displaystyle b_{2}*b_{15}}$= ${\displaystyle }$${\displaystyle (r_{1})^{2}}$
6. ${\displaystyle a_{14}*a_{3}}$= ${\displaystyle (r_{1})^{2}}$
7. ${\displaystyle a_{5}*b_{4}}$= ${\displaystyle (d_{1})^{2}}$
8. ${\displaystyle a_{8}*b_{6}}$= ${\displaystyle a_{9}*b_{7}}$${\displaystyle =(r_{1})^{2}}$
9. ${\displaystyle (b_{14}*c_{14})*(a_{15}+c_{15})}$= ${\displaystyle (r_{1})^{2}}$
10. ${\displaystyle c_{6}*c_{8}}$= ${\displaystyle c_{7}*c_{9})}$=${\displaystyle a_{13}*b_{13}}$
11. ${\displaystyle }$

1. ${\displaystyle a_{2}+b_{2}+c_{2}=b_{1}+c_{1}}$[4]
2. ${\displaystyle a_{3}+b_{3}+c_{3}=a_{1}+c_{1}}$
3. ${\displaystyle a_{4}+b_{4}+c_{4}=2b_{1}}$
4. ${\displaystyle a_{5}+b_{5}+c_{5}=2a_{1}}$
5. ${\displaystyle a_{6}+b_{6}+c_{6}=b_{1}}$
6. ${\displaystyle a_{7}+b_{7}+c_{7}=b_{1}}$
7. ${\displaystyle a_{8}+b_{8}+c_{8}=a_{1}}$
8. ${\displaystyle a_{9}+b_{9}+c_{9}=a_{1}}$
9. ${\displaystyle a_{10}+b_{10}+c_{10}=b_{1}+c_{1}-a_{1}}$
10. ${\displaystyle a_{11}+b_{11}+c_{11}=c_{1}-b_{1}+a_{1}}$
11. ${\displaystyle a_{12}+b_{12}+c_{12}=c_{1}}$
12. ${\displaystyle a_{13}+b_{13}+c_{13}=a_{1}+b_{1}-c_{1}}$
13. ${\displaystyle a_{14}+b_{14}+c_{14}=c_{1}-a_{1}}$
14. ${\displaystyle a_{15}+b_{15}+c_{15}=c_{1}-c_{1}}$

此外还有诸弦，大小差，诸差，诸率互见，四位拾遗，拾遗。

一共692关系式，这些关系式完全是几何定理，与具体数值无关。

举例：第三条中“勾股和即弦黄和”一句就是：三角形两直角边之和等于斜边加上内切圆直径（“黄”指内切圆直径）。这个命题可以由直角三角形的勾股定理推出：

设直角三角形三边分别为${\displaystyle a}$、${\displaystyle b}$、${\displaystyle c}$，其中${\displaystyle a^{2}+b^{2}=c^{2}}$

内接圆直径${\displaystyle d={\frac {2ab}{a+b+c}}}$，因此

内接圆直径 + 斜边 = ${\displaystyle d+c={\frac {2ab}{a+b+c}}+c={\frac {2ab+c^{2}+ac+bc}{a+b+c}}}$

${\displaystyle ={\frac {2ab+a^{2}+b^{2}+ac+bc}{a+b+c}}={\frac {(a+b)(a+b+c)}{a+b+c}}}$

${\displaystyle =a+b}$ = 两直角边之和

后面出现的各问题，都根据这些公式中的相等关系而列出方程，然后求解。

李冶的692个公式中，有8个是错误的，只是因为数值吻合而被误认为成立。

正率14问

测圆海镜卷二 正率

从第二卷开始，《测圆海镜》中一共出现了一百七十个问题，它们都是围绕着同一个题设背景而展开。 在第二卷开头，李冶作出了以后题目公用的总假设：

这里的圆城就是指天地三角形的内切圆，其方向按照圆城图式里面东南西北四个点的位置而定（注意北在下方，东在左边，与现在通用的方位相反），所谓的“地”、“坤地”则是指圆城图式里面出现的点、坤点等等。以后的每个问题中要求的长度都是圆城的半径或直径。

接下来的问题都是已知某些线段的长度，问圆城的半径或直径。李冶在每一题的题目之后都先写出解法（代数演算），再给出演草（代入数值的计算）。

洞渊九容

开头十个问题，不需要天元方程。清代数学李善兰认为，第一个问题和《九章算术》的勾股容圆题目一样，第二问至第十问就是《自序》中提到的“洞渊九容”[5]。但李冶原书或《四库全书》李锐较本都没有这九个问题的细草，李善兰在《天算或问》一书中根据相似三角形原理求得各式，并以第二问为例阐明如下[6]：

${\displaystyle {a_{1} \over a_{1}+b_{1}+c_{1}}={a_{2} \over a_{2}+b_{2}+c_{2}}}$

${\displaystyle {b_{1} \over b_{1}+c_{1}}={a_{2} \over b_{2}+c_{2}}}$

又因：

${\displaystyle b_{1}+c_{1}=a_{2}+b_{2}+c_{2}}$

所以

${\displaystyle {2b_{1}c_{1} \over a_{1}+b_{1}+c_{1}}={2b_{2}c_{2} \over b_{2}+c_{2}}}$

得

${\displaystyle {2a_{2}\times b_{2} \over a_{2}+b_{2}+c_{2}}=d}$

其余类推。 。

第一问：或问：甲乙二人俱在地，乙东行三百二十步而立。甲南行六百步望见乙，问径几里？

答曰：城径二百四十步。

勾股容圆${\displaystyle d={2a_{1}\times b_{1} \over a_{1}+b_{1}+c_{1}}}$

${\displaystyle ={2*320*600 \over 320+600+{\sqrt {(}}320^{2}+600^{2})}=240}$

第二问：勾上容圆

${\displaystyle {2a_{2}\times b_{2} \over a_{2}+b_{2}+c_{2}}=d}$

第三问：股上容圆

${\displaystyle {2a_{3}\times b_{3} \over a_{3}+b_{3}+c_{3}}=d}$

第四问：勾股上容圆

${\displaystyle {2a_{12}\times b_{12} \over c_{12}}=d}$

第五问：弦上容圆

${\displaystyle {2a\times b \over a+b}=d}$

第六问：勾外容圆

${\displaystyle {2a_{10}\times b_{10} \over b_{10}-a_{10}+c_{10}}=d}$

第七问：股外容圆

${\displaystyle {2a_{11}\times b_{11} \over b_{11}-a_{11}+c_{11}}=d}$

第八问：弦外容圆

${\displaystyle {2a_{13}\times b_{13} \over b_{13}+a_{13}-c_{13}}=d}$

第九问：勾外容圆半

${\displaystyle {2a_{14}\times b_{14} \over c_{14}-a_{14}}=d}$

第十问：股外容圆半

${\displaystyle {2a_{15}\times b_{15} \over c_{15}-b_{15}}=d}$

天元术

测圆海镜细草 卷二 第十四问 草曰

从第十四题开始，引入天元术，将所求的未知量设为“天元”，然后根据识别杂记中给出的公式构造出两个天元式，另其相等，然后解方程得出答案。《测圆海镜》中天元式的次序，高次幂在常数项之上，和《益古演段》，《四元玉鉴》的相反。

第十四问

“或问出西门南行四百八十步有树，出北门东行二百步见之。问答同前”。

法曰：以二行步相乘为实，二行步相并为从，一步常法，得半径。

草曰：立天元一为半径，置南行步在地，

内减天元半径得股圆差：${\displaystyle 480-x}$

元

。

又置乙东行步在地，内减天元，得勾圆差：${\displaystyle 200-x}$

元

以勾圆差增乘股圆差得半段黄方幂：${\displaystyle x^{2}-680x+96000}$

元

又置天元幂以倍之，也为半段黄方幂；

元

因此，得${\displaystyle x^{2}-680x+96000=2x^{2}}$

相消得：${\displaystyle -x^{2}-680x+96000=0}$

元

解方程，得半径${\displaystyle r=120}$。

边股17问 [7]

标题文字	已知	未知数x	方程
1	<${\displaystyle b_{2}}$，${\displaystyle c_{4}}$		直接计算
2	${\displaystyle b_{2}}$，${\displaystyle a_{11}}$	d	${\displaystyle x^{2}+a_{11}x-2b_{2}a_{11}=0}$
3	${\displaystyle b_{2}}$，${\displaystyle b_{11}}$	r	${\displaystyle x^{2}+b_{2}x-b_{2}b_{11}=0}$
4	${\displaystyle b_{2}}$，${\displaystyle a_{15}}$	d	${\displaystyle x^{3}+a_{15}x^{2}-4a_{15}b_{2}^{2}=0}$
5	${\displaystyle b_{2}}$，${\displaystyle a_{14}}$	d	${\displaystyle x^{3}-(b_{2}-2a_{14})x^{2}+a_{14}^{2}*x+a_{14}^{2}*b_{2}=0}$
6	${\displaystyle b_{2}}$，${\displaystyle a_{10}}$	r	${\displaystyle x^{2}+(b_{2}-(b_{2}-c_{10}))x+b_{2}(b_{2}-c_{10})=0}$
7	${\displaystyle b_{2}}$，${\displaystyle c_{2}}$	r	${\displaystyle ((1/2)*c_{2}-(1/2)*b_{2}+b_{2})*x^{2}-(1/2)*(c_{2}-b_{2})b_{2}^{2}=0}$
8	${\displaystyle b_{2}}$， ${\displaystyle c_{1}}$	r	${\displaystyle 2x^{2}+((c_{1}+b_{2})+(c_{1}-b_{2}))x-((c_{1}+b_{2})(c_{1}-b_{2})-(c_{1}-b_{2})^{2}))=0}$
9	${\displaystyle b_{2}}$，${\displaystyle c_{6}}$	r	${\displaystyle 2x^{2}-2(b_{2}-2(b_{2}-c_{5}))b_{2}=0}$
10	${\displaystyle b_{2}}$，${\displaystyle b_{14}}$	r	${\displaystyle x^{2}-2b_{2}x+((b_{2}-b_{14})^{2}-b_{14}^{2}=0}$
11	${\displaystyle b_{2}}$，${\displaystyle a_{10}}$	r	${\displaystyle (2b_{2}-a_{10})x-b_{2}a_{10}=0}$
12	${\displaystyle b_{2}}$，${\displaystyle c_{15}}$	${\displaystyle b_{15}}$	${\displaystyle x^{2}+(b_{2}+c_{15})x-b_{2}c_{15}=0}$
13	${\displaystyle b_{2}}$，${\displaystyle c_{14}}$	${\displaystyle a_{14}}$	${\displaystyle x^{4}-2(b_{2}-c_{14})x^{3}+(b_{2}-c_{14})^{2}x^{2}+2b_{2}c_{14}^{2}x-(2(b_{2}-c_{14})-b_{2}))b_{2}c_{14}^{2}=0}$
14	${\displaystyle b_{2}}$，${\displaystyle c_{6}}$		${\displaystyle r={\sqrt {(}}(2c_{6}-b_{2})b_{2})}$
15	${\displaystyle b_{2}}$，${\displaystyle c_{8}}$	r	${\displaystyle -x^{3}-c_{8}x^{2}-b_{2}^{2}x+c_{8}b_{2}^{2}=0}$
16	${\displaystyle b_{2}}$，${\displaystyle b_{14}+c_{14}}$		用洞渊九容公式计算
17	${\displaystyle b_{2}}$，${\displaystyle a_{15}+c_{15}}$		用洞渊九容公式计算

底勾17问：已知${\displaystyle a_{3}}$及另一边求直径d.[8]

。

第三卷边股问与第四卷同次第底勾问成对偶。

问	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17
第二边	${\displaystyle c_{5}}$	${\displaystyle b_{10}}$	${\displaystyle a_{10}}$	${\displaystyle b_{14}}$	${\displaystyle b_{15}}$	${\displaystyle c_{11}}$	${\displaystyle c_{13}}$	${\displaystyle c_{1}}$	${\displaystyle c_{9}}$	${\displaystyle a_{15}}$	${\displaystyle b_{11}}$	${\displaystyle c_{14}}$	${\displaystyle c_{15}}$	${\displaystyle c_{9}}$	${\displaystyle c_{7}}$	${\displaystyle a_{15}+c_{15}}$	${\displaystyle b_{14}+c_{14}}$

大股18问：已知${\displaystyle b_{1}}$。[8]

问	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18
第二边	${\displaystyle b_{14}}$	${\displaystyle a_{14}}$	${\displaystyle a_{15}}$	${\displaystyle b_{15}}$	${\displaystyle b_{11}}$	${\displaystyle a_{11}}$	${\displaystyle c_{10}}$	${\displaystyle c_{4}}$	${\displaystyle c_{2}}$	${\displaystyle c_{1}}$	${\displaystyle c_{6}}$	${\displaystyle c_{9}-a_{11}}$	${\displaystyle c_{15}}$	${\displaystyle c_{14}}$	${\displaystyle c_{9}}$	${\displaystyle c_{12}}$	${\displaystyle a_{15}+b_{14}}$	${\displaystyle c_{13}}$

大勾18问：

1-11，13-19已知a_{1}，及另一边求直径d.[8]

12问：已知 ${\displaystyle a_{1}+c_{3}}$及另边，求直径。

问	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18
已知	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}+c_{3}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$	${\displaystyle a_{1}}$
第二边	${\displaystyle a_{15}}$	${\displaystyle b_{15}}$	${\displaystyle b_{14}}$	${\displaystyle a_{14}}$	${\displaystyle a_{10}}$	${\displaystyle b_{10}}$	${\displaystyle c_{11}}$	${\displaystyle c_{5}}$	${\displaystyle c_{3}}$	${\displaystyle c_{1}}$	${\displaystyle c_{9}}$	${\displaystyle b_{10}-c_{6}}$	${\displaystyle c_{14}}$	${\displaystyle c_{15}}$	${\displaystyle c_{6}}$	${\displaystyle c_{12}}$	${\displaystyle a_{15}+b_{14}}$	${\displaystyle a_{13}}$

明叀前18问；求直径d。[9]

问	已知
1	${\displaystyle a_{14}}$，${\displaystyle b_{15}}$
2	${\displaystyle a_{15}}$，${\displaystyle b_{14}}$
3	${\displaystyle a_{14}}$，${\displaystyle a_{15}}$
4	${\displaystyle b_{14}}$，${\displaystyle b_{15}}$
5	${\displaystyle a_{14}}$，${\displaystyle c_{8}}$
6	${\displaystyle b_{15}}$，${\displaystyle c_{7}}$
7	${\displaystyle b_{15}}$，${\displaystyle c_{13}}$
8	${\displaystyle a_{14}}$，${\displaystyle c_{13}}$
9	${\displaystyle d-a_{14}}$，${\displaystyle d-b_{15}}$
10	${\displaystyle d-b_{14}}$，${\displaystyle d-a_{15}}$
11	${\displaystyle c_{12}}$，${\displaystyle a_{15}+b_{14}}$
12	${\displaystyle a_{15}+b_{14}}$，${\displaystyle c_{13}}$
13	${\displaystyle b_{15}+c_{13}}$，${\displaystyle c_{13}-b_{15}}$
14	${\displaystyle a_{14}+b_{15}+c_{13}}$，${\displaystyle a_{14}+b_{15}+c_{13}-a_{14}}$，
15	${\displaystyle c_{14}}$，${\displaystyle d-b_{15}}$
16	${\displaystyle c_{5}}$，${\displaystyle d-a_{14}}$
17	${\displaystyle a_{1}-a_{14}}$，${\displaystyle b_{1}-b_{15}}$
18	${\displaystyle a_{1}+a_{14}}$，${\displaystyle b_{1}-b_{15}}$

边股17问：已知 三至八边，或其差，和，求直径d.[10]

问	已知
1	${\displaystyle a_{14}+b_{14}}$，${\displaystyle a_{15}+b_{15}}$，${\displaystyle c_{12}}$
2	${\displaystyle a_{14}+b_{14}}$，${\displaystyle a_{15}+b_{15}}$，${\displaystyle c_{13}}$
3	${\displaystyle (c_{12}-a_{12})+(c_{12}-b_{12})}$，${\displaystyle d_{14}+d_{15}}$
4	${\displaystyle c_{15}}$，${\displaystyle c_{14}}$
5	${\displaystyle b_{14}+c_{14}}$，${\displaystyle c_{15}+b_{15}}$
6	${\displaystyle a_{15}+c_{15}}$，${\displaystyle a_{14}+c_{14}}$
7	${\displaystyle a_{1}+c_{1}}$，${\displaystyle a_{15}+c_{15}}$${\displaystyle }$
8	${\displaystyle a_{1}+c_{1}}$，${\displaystyle a_{14}+c_{14}}$${\displaystyle }$
9	${\displaystyle b_{1}+c_{1}}$，${\displaystyle b_{15}+c_{15}}$${\displaystyle }$
10	${\displaystyle b_{1}+c_{1}}$，${\displaystyle b_{14}+c_{14}}$，${\displaystyle }$
11	${\displaystyle b_{14}+c_{14}}$，${\displaystyle a_{15}+c_{15}}$，${\displaystyle b_{14}+a_{15}-c_{13}}$
12	${\displaystyle (b_{8}-a_{8})+(b_{2}-a_{2})}$，${\displaystyle b_{14}+a_{15}-c_{13}}$${\displaystyle }$
13	${\displaystyle b_{7}-a_{8}}$，${\displaystyle (b_{14}-a_{14})+(b_{15}-a_{15})}$，${\displaystyle c_{12}-d}$
14	${\displaystyle (b_{7}-a_{7})+(b_{8}-a_{8})}$，${\displaystyle (b_{14}-a_{14})+(b_{15}-a_{15})}$${\displaystyle }$
15	${\displaystyle a_{14}+b_{14}}$，${\displaystyle a_{15}+b_{15}}$${\displaystyle }$
16	${\displaystyle a_{14}+a_{15}}$，${\displaystyle b_{14}+b_{15}}$${\displaystyle }$

或问：高差，平差并一百六十一步，明差叀差并七十七步，问答同前。

即：

草曰：[11]

已知

${\displaystyle (b_{7}-a_{7})+(b_{8}-a_{8})=161}$

${\displaystyle (b_{14}-a_{14})+(b_{15}-a_{15})=77}$

相加除2 ; 根据#杂用公式，等于皇极差：

${\displaystyle (b_{7}-a_{7})+(b_{8}-a_{8})+(b_{14}-a_{14})+(b_{15}-a_{15}) \over 2}$ ${\displaystyle =(b_{12}-a_{12})}$

${\displaystyle b_{12}-a_{12}=}$${\displaystyle 161+77 \over 2}$${\displaystyle =119}$

设天元一为上平勾：

${\displaystyle x=a_{8}}$

${\displaystyle x+161}$ =${\displaystyle x+(b_{7}-a_{7})+(b_{8}-a_{8})=a_{8}+(b_{7}-a_{7})+(b_{8}-a_{8})}$

${\displaystyle =b_{7}}$ (杂用公式)

因为 ${\displaystyle a_{8}+b_{7}=c_{12}}$ (杂用公式)

${\displaystyle c_{12}=x+b_{7}=2*x+(b_{7}-a_{7})+(b_{8}-a_{8})=2*x+161}$

${\displaystyle c_{12}^{2}=(x+b_{7})^{2}=(2*x+161)^{2}=4*x^{2}+644*x+25921}$

${\displaystyle c_{12}^{2}-(b_{12}-a_{12})^{2}}$

${\displaystyle =4*x^{2}+644*x+25921-}$${\displaystyle ((b_{7}-a_{7})+(b_{8}-a_{8})+(b_{14}-a_{14})+(b_{15}-a_{15}))^{2} \over 4}$

${\displaystyle =4*x^{2}+644*x+11760=d}$(圆城直径),

${\displaystyle d^{2}=(4*x^{2}+644*x+11760)^{2}=16*x^{4}+5152*x^{3}+508816*x^{2}+15146880*x+138297600}$

将 ${\displaystyle 4*x}$乘下高股

${\displaystyle 4*x*b_{7}=4*x*(x+(b_{7}-a_{7})+(b_{8}-a_{8}))=4*x*(x+161)=4*x^{2}+644*x}$

乘之以皇极弦幂:

${\displaystyle d^{2}=4*x*b_{7}*c_{12}^{2}=}$${\displaystyle (4*x^{2}+644*x)*(4*x^{2}+644*x+25921)=}$${\displaystyle 16*x^{4}+5152*x^{3}+518420*x^{2}+16693124}$

因此

${\displaystyle 16*x^{4}+5152*x^{3}+518420*x^{2}+16693124=}$${\displaystyle 16*x^{4}+5152*x^{3}+508816*x^{2}+15146880*x+138297600}$

左右相消得：

${\displaystyle 9604*x^{2}+1546244*x-138297600=0}$

解之得 ${\displaystyle x=a_{8}=64}$;

正合#今问正数中的下平勾。

：大斜四问[12]

问	已知
1	${\displaystyle a_{12}+b_{12}+c_{12}}$，${\displaystyle b_{12}-a_{12}}$
2	${\displaystyle c_{1}}$，${\displaystyle b_{1}-a_{1}}$
3	${\displaystyle c_{1}}$，${\displaystyle a_{10}+b_{11}}$
4	${\displaystyle c_{1}}$，${\displaystyle a_{2}+b_{3}}$

：大和8问

边股17问：已知三边，求直径d[13]。

问	已知条件
1	${\displaystyle a_{1}+b_{1}}$，${\displaystyle a_{2}}$，${\displaystyle b_{3}}$
2	${\displaystyle a_{1}+b_{1}}$，${\displaystyle c_{13}+b_{13}-a_{13}}$，${\displaystyle c_{13}-b_{13}+a_{13}}$
3	${\displaystyle a_{1}+b_{1}}$，${\displaystyle a_{11}+b_{11}}$，${\displaystyle a_{10}+b_{10}}$
4	${\displaystyle a_{1}+b_{1}}$，${\displaystyle c_{10}-a_{10}}$，${\displaystyle c_{11}-b_{11}}$
5	${\displaystyle a_{1}+b_{1}}$，${\displaystyle c_{6}+c_{8}}$，${\displaystyle c_{6}-c_{8}}$
6	${\displaystyle a_{1}+b_{1}}$，${\displaystyle c_{10}}$，${\displaystyle c_{11}}$
7	${\displaystyle a_{1}+b_{1}}$，${\displaystyle c_{4}}$，${\displaystyle c_{5}}$
8	${\displaystyle a_{1}+b_{1}}$，${\displaystyle c_{2}}$，${\displaystyle c_{3}}$

：三事和8问[14]

问	已知
1	${\displaystyle a_{1}+b_{1}+c_{1}}$，${\displaystyle c_{1}-b_{1}}$
2	${\displaystyle a_{1}+b_{1}+c_{1}}$，${\displaystyle c_{1}-a_{1}}$
3	${\displaystyle a_{1}+b_{1}+c_{1}}$，${\displaystyle b_{1}-a_{1}}$
4	${\displaystyle a_{1}+b_{1}+c_{1}}$，${\displaystyle (c_{1}-b_{1})+(c_{1}-a_{1})}$
5	${\displaystyle a_{1}+b_{1}+c_{1}}$，${\displaystyle (c_{1}-b_{1})+(b_{1}-a_{1})+(c_{1}-a_{1})}$
6	${\displaystyle a_{1}+b_{1}+c_{1}}$，${\displaystyle d_{14}+d_{15}}$
7	${\displaystyle a_{1}+b_{1}+c_{1}}$，${\displaystyle c_{12}}$
8	${\displaystyle a_{1}+b_{1}+c_{1}}$，${\displaystyle c_{13}}$