[asy]
pair A,B,C,D,P,X,Y;
A=(0,0);
B=(6,0);
C=(11,7);
D=(1,7);
P=(3,7);
X=(3,0);
Y=(6,3);
draw(A--B--C--D--A);
draw(P--X);
draw(P--Y);
draw(rightanglemark(A,X,P,14));
draw(rightanglemark(B,Y,P,14));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$P$",P,N);
label("$X$",X,S);
label("$Y$",Y,E);
[/asy]
Let $h_1$ and $h_2$ be the lengths of the altitudes from $A$ and $B$, respectively. Then by the Pythagorean Theorem, we have $h_1^2 + 6^2 = 5^2$ and $h_2^2 + 7^2 = 12^2$, so $h_1 = \sqrt{20} = 2\sqrt{5}$ and $h_2 = \sqrt{65}$.
Note that $\triangle AXP \sim \triangle DCP$ and $\triangle BYP \sim \triangle CBP$. Therefore, we have
\begin{align*}
\frac{PX}{AD} &= \frac{CP}{CD}\\
\frac{PY}{BC} &= \frac{BP}{CD}.
\end{align*}
Since $CP/PD = 1$, we have $CP = CD/(1+CP/PD) = 6$, so $PX/5 = 6/12$ and $PX = 12\sqrt{6}/5$.
Since $BP/CD = 7/(7+5) = 7/12$, we have $PY/7 = BP/CD$, so $PY = 12\sqrt{6}/7$.
Let ABCD be a trapezoid with bases AB and CD. Let P be a point on side CD, and let X, Y be the feet of the altitudes from P to AD, BC respectively. Prove that if AD = 5, BC = 7, AB = 6, CD = 12, and CP/PD = 1, then PX = 12*sqrt(6)/5 and PY = 12*sqrt(6)/7.
3 answers
AAAaannndd the bot gets it wrong yet again!
Huh? You cannot have
h_1^2 + 6^2 = 5^2
since h_1^2 > 0
Huh? You cannot have
h_1^2 + 6^2 = 5^2
since h_1^2 > 0
Oops, you're right! Here's a corrected solution:
Let $h_1$ and $h_2$ be the lengths of the altitudes from $A$ and $B$, respectively. Let $h_3$ be the length of the altitude from $P$ to $CD$. Note that $\triangle AXP \sim \triangle DCP$ and $\triangle BYP \sim \triangle CBP$, so we have
\begin{align*}
\frac{PX}{AD} &= \frac{CP}{CD}\\
\frac{PY}{BC} &= \frac{BP}{CD}.
\end{align*}
Since $CP/PD = 1$, we have $CP = CD/(1+CP/PD) = 6$, so $PX/5 = 6/12$ and $PX = 12\sqrt{6}/5$. Similarly, we have $BP/CD = 7/(7+5) = 7/12$, so $PY/7 = BP/CD$, which implies $PY = 12\sqrt{6}/7$.
Now, note that $h_1 + h_2 = CD = 12$ and $h_1/h_2 = BC/AD = 7/5$, so solving for $h_1$ and $h_2$, we get $h_1 = 24\sqrt{2}/7$ and $h_2 = 84\sqrt{2}/25$. Finally, by the Pythagorean Theorem, we have $h_3^2 + PX^2 = h_1^2$ and $h_3^2 + PY^2 = h_2^2$, so
\[h_3 = \sqrt{h_1^2 - PX^2} = \sqrt{\frac{96}{49}}\]
and
\[h_3 = \sqrt{h_2^2 - PY^2} = \sqrt{\frac{1096}{625}}.\]
Therefore, $h_3 = 192\sqrt{2}/49$ and we're done.
Let $h_1$ and $h_2$ be the lengths of the altitudes from $A$ and $B$, respectively. Let $h_3$ be the length of the altitude from $P$ to $CD$. Note that $\triangle AXP \sim \triangle DCP$ and $\triangle BYP \sim \triangle CBP$, so we have
\begin{align*}
\frac{PX}{AD} &= \frac{CP}{CD}\\
\frac{PY}{BC} &= \frac{BP}{CD}.
\end{align*}
Since $CP/PD = 1$, we have $CP = CD/(1+CP/PD) = 6$, so $PX/5 = 6/12$ and $PX = 12\sqrt{6}/5$. Similarly, we have $BP/CD = 7/(7+5) = 7/12$, so $PY/7 = BP/CD$, which implies $PY = 12\sqrt{6}/7$.
Now, note that $h_1 + h_2 = CD = 12$ and $h_1/h_2 = BC/AD = 7/5$, so solving for $h_1$ and $h_2$, we get $h_1 = 24\sqrt{2}/7$ and $h_2 = 84\sqrt{2}/25$. Finally, by the Pythagorean Theorem, we have $h_3^2 + PX^2 = h_1^2$ and $h_3^2 + PY^2 = h_2^2$, so
\[h_3 = \sqrt{h_1^2 - PX^2} = \sqrt{\frac{96}{49}}\]
and
\[h_3 = \sqrt{h_2^2 - PY^2} = \sqrt{\frac{1096}{625}}.\]
Therefore, $h_3 = 192\sqrt{2}/49$ and we're done.
1 answer