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Dr. Barnhouse and the Bible

Dr. Barnhouse and the Bible

Dr. Donald Grey Barnhouse
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Demonstrating Dr. Barnhouse’s acute understanding of Romans and his heart for effective preaching, these messages skillful and reverently expound even the most difficult passages in a clear way. Dr. Barnhouse's concern for a universal appreciation of the epistle fuels this series and invites all listeners into a deeper understanding of the life-changing message of Romans.

运用多角度思维： （1）由于$M,N$为坐标轴上的点，因此通过直角三角形边长表示线段长是最直接、最基础的角度； （2）考虑到点$P$到直线$MN$的距离，通过构造平行线将距离问题转化为求平行线间距离的最大值是常用、有效的角度； （3）从解析几何的通用视角看，利用向量法或直接表示距离公式并结合三角换元、辅助角公式求最值是规范、易行的角度。 解法1：(利用直角三角形表示边长结合基本不等式) 设$M(a,0),N(0,b)$,由$PM\perp PN$, $P(2,2)$, $\overrightarrow{PM}=(a-2,-2)$, $\overrightarrow{PN}=(-2,b-2)$. $\therefore \overrightarrow{PM}\cdot \overrightarrow{PN}=-2(a-2)-2(b-2)=0$, 即$a+b=4$. 在$Rt\triangle OMN$中，$MN=\sqrt{a^2+b^2}$. 由$a+b=4$,得$a^2+b^2=(a+b)^2-2ab=16-2ab$. 由$ab\leqslant (\frac{a+b}{2})^2=4$,得$a^2+b^2\geqslant 16-8=8$, 当且仅当$a=b=2$时等号成立. 此时$MN$的最小值为$\sqrt{8}=2\sqrt{2}$. 解法2：(利用直线截距式及距离公式结合三角换元) 设直线$MN$的方程为$\frac{x}{a}+\frac{y}{b}=1$,由解法1可知$a+b=4$. 点$P(2,2)$到直线$MN$的距离为$d=\frac{| \frac{2}{a}+\frac{2}{b}-1 |}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}=\frac{| \frac{2(a+b)}{ab}-1 |}{\frac{\sqrt{a^2+b^2}}{ab}}=\frac{8-ab}{\sqrt{a^2+b^2}}$. 由$a+b=4$,设$a=2+2\cos\theta, b=2-2\cos\theta$ ($\theta \in (0,\frac{\pi}{2})$). 则$ab=4-4\cos^2\theta=4\sin^2\theta, a^2+b^2=8+8\cos^2\theta$. $d=\frac{8-4\sin^2\theta}{\sqrt{8+8\cos^2\theta}}=\frac{4+4\cos^2\theta}{\sqrt{8+8\cos^2\theta}}=\frac{1}{2}\sqrt{8+8\cos^2\theta}$. 当$\cos^2\theta \to 1$时，$d$的最大值为$\frac{1}{2}\sqrt{16}=2$, 但由$a,b>0$知$| \cos\theta |<1$,故 $d<2$. 同时，当$MN$最小时，$a=b=2, ab=4, a^2+b^2=8$, 此时$d=\frac{8-4}{\sqrt{8}}=\sqrt{2}$. 解法3：(利用平行线间距离结合斜率关系) 由$a+b=4$知，直线$MN$过定点$A(x_0,y_0)$,由$\frac{x_0}{a}+\frac{y_0}{b}=1$及$a+b=4$得 $x_0=y_0$, $A$为线段$MN$中点不可能(因$P,M,N$关系特殊). 实际上，$d=\frac{8-ab}{\sqrt{a^2+b^2}}$.令$t=ab$, $d=\frac{8-t}{\sqrt{16-2t}}$. 设$f(t)=\frac{(8-t)^2}{16-2t}=\frac{64-16t+t^2}{16-2t}$, 令$16-2t=u$,则$t=8-\frac{u}{2}$, $f(u)=\frac{64-16(8-\frac{u}{2})+(8-\frac{u}{2})^2}{u}=\frac{u^2/4}{u}=\frac{u}{4}$. 由于$0

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About Dr. Barnhouse and the Bible

Dr. Barnhouse & the Bible has been making God's Word plain for more than sixty years. His unique style springs from his careful speech, friendly manner, vivid analogies, and most of all from his faithful exposition of the Scriptures. He made the Bible relevant to the modern man. In fact his sermons have grown no less relevant to those who hear them today.

Dr. Barnhouse & the Bible is a ministry of the Alliance of Confessing Evangelicals. The Alliance exists to call the twenty-first century church to a modern reformation that recovers clarity and conviction about the great evangelical truths of the Gospel and that then seeks to proclaim these truths powerfully in our contemporary context.

About Dr. Donald Grey Barnhouse

Donald Grey Barnhouse, one of the twentieth century's outstanding American preachers, saw the need to spread God’s Word to a vast audience; he went on to start the radio broadcast which has become known as Dr. Barnhouse & the Bible. Dr. Barnhouse is best known for his many colorful illustrations of living the Christian life. His books include Teaching the Word of Truth, Life by the Son, God’s Methods for Holy Living, and more. Listen anytime at AllianceNet.org/Barnhouse.

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